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Physics Grade 12
Oscillatory Motion
At the end of this chapter, the student should be able to:
1.1. Explain Periodic Motion.
1.2. Describe Simple Harmonic motion (SHM).
1.3. Formulate Simple Harmonic Motion Equations.
1.4. Define Simple Pendulum.
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Physics Grade 12
1.0 Introduction
This chapter describes the periodic motion, simple harmonic motion, period and
frequency, potential and kinetic energy of SHM, simple pendulum and solving related
problems.
1.1 Periodic Motion
Periodic motion is a motion of an object that regularly repeats to a given position after
a fixed time interval. With a little thought, we can identify several types of periodic
motion in everyday life. The Earth returns to the same position in its orbit around the
Sun each year, resulting in the variation among the four seasons. The Moon returns to
the same relationship with the Earth and the Sun, resulting in a full Moon
approximately once a month.
Definition:
Periodic motion is a motion of an object that regularly returns to a given position after
a fixed time interval.
1.1.1 Period and Frequency of Periodic Motion

Period (T): is the time taken to make one complete cycle. SI unit of period is second (s).
Period =

Frequency (f): is the number of cycles made in one second. SI unit of frequency
is called Hertz (Hz).
Frequency =
f=
1
T
T=
1
f
Example 1
The suspended mass makes 30 complete oscillations in 15 s. What is the period and
frequency of the motion?
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Physics Grade 12
Solution
, Time taken= 15s, T=?, f=?
Given:
b. Period =
=15 s/30=0.5s
a. Frequency =
=30/15=2Hz
1.2 Simple Harmonic motion
You are familiar with many examples of repeated motion in your daily life. If an
object returns to its original position a number of times, we call its motion repetitive.
Typical examples of repetitive motion of the human body are heartbeat and breathing.
Many objects move in a repetitive way, a swing, a rocking chair, and a clock
pendulum.
Special kind of periodic motion occurs in mechanical systems when the force acting
on an object is proportional to the position of the object relative to some equilibrium
position. If this force is always directed toward the equilibrium position, the motion is
called simple harmonic motion (SHM).
Definition:
 Simple Harmonic Motion is motion under the influence of a type of a force
described by Hooke’s law.

Simple Harmonic motion (SHM) is a motion in which a body moves back and forth
over a fixed path, returning to each position and velocity after a definite interval of
time.
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Physics Grade 12
Pendulum
Curved path
A
Mass attached to spring
1.2.1 Mass Attached to a Spring
As a model for simple harmonic motion, consider a block of mass m attached to the
end of a spring, with the block free to move on a horizontal surface. Hooke’s law
states that the restoring force applied by a spring is proportional to the displacement of
the spring and opposite in direction.
F = -kx
An object moves with simple harmonic motion whenever its acceleration is
proportional to its position and is oppositely directed to the displacement from
equilibrium.The acceleration of SHM is zero at equilibrium position.
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Physics Grade 12
a=-
a= acceleration
m= Mass
x= displacement
K= spring constant
Example 2
A 4-kg mass suspended from a spring produces a displacement of 20 cm. What is the
spring constant?
Solution
Example 3
A load of 50 N attached to a spring hanging vertically stretches the spring 5.0 cm. The
spring is now placed horizontally on a table and stretched 11.0 cm. What force is
required to stretch the spring?
Solution
Step1: Finding the spring constant
50/ (0.05) =1000 N/m
Step2: Finding the force required to stretch the spring
F = kx = (1000 N/m) (0.11m) =110 N
Example 4
A 2-kg mass hangs at the end of a spring whose constant is k = 400 N/m. The mass is
displaced a distance of 12 cm and released. A) What is the acceleration at the instant
the displacement is x = +7 cm? B) What is the
maximum acceleration at the
displacement of 12cm?
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Physics Grade 12
Solution
a
kx
m
a)
a= (-400N/m 0.07m)/2kg=
b)
am= (-400N/m 0.12m)/2kg=
s2
m/s2
Note: When the displacement is +7 cm (downward), the acceleration is -14.0 m/s2
(upward) independent of motion direction.
1.2.2 Period and frequency in SHM
The mass on the spring makes one full oscillation (2 radians) in one period T, so the
angular frequency can be found by:

2
 2 f
T
Another relationship between the angular frequency of a mass oscillating on a spring
and the spring constant k is:
 
k
m
We see from this equation that the higher the spring constant k, the stiffer the spring,
and the greater the angular frequency of oscillation. A smaller mass will also increase
the angular frequency for a particular spring.
If we set the two equations above for  equal to each other and solve for the period T
of oscillation, we get.
f
T  2
m
k
1 
1 k


T 2 2 m
N.B.
 Period of a SHM depends on the mass of the vibrating body and the force constant
of the spring and independent of the amplitude of the motion.
 Frequency is only dependent on the mass of the object and the force constant of the
spring.
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Physics Grade 12
Example 5
A mass of 0.5 kg oscillates on the end of a spring on a horizontal surface with
negligible friction with the spring constant of 50 N/m. Determine:
a. angular frequency of the oscillation
b. period of the oscillation
c. frequency of the oscillation
Solution
50 N / m
k
rad

 10
m
0.5 kg
s
a.  
=
b. T =
c.
f 
= 6.25s
 10 rad / s

 1.6 Hz
2
2
Example 6
A 7-kg mass is hung from the bottom end of a vertical spring fastened to an overhead
beam. The mass is set into vertical oscillations having a period of 2.60 s. Find the
force constant of the spring.
Solution
K=
=
(7)/(2.6)2=41N/m
Example 7
A block of mass m attached at the end of a spring of force constant 100N/m vibrates in
SHM with a frequency of 10/π Hz. Find the mass of the block.
Solution
m=
= 100(π/10)2/(
= 4 kg
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Physics Grade 12
1.2.3 Energy of the Simple Harmonic Motion
The potential energy is equal to the product of average applied force and the
displacement.
P.E= Kx
2
The potential energy is at maximum for the amplitude.
2
P.E= KA
As an object accelerates towards the mean position, the energy is transferred from
potential energy and to kinetic energy.
K.E= mv
2
v
k
m
A2  x 2
v
k
A
m
The total mechanical energy of a simple
harmonic motion is a constant and is
proportional to the square of the amplitude.
Thus, whatever potential energy is lost
must be gained by kinetic energy, and viceversa. Total Energy = Potential Energy +
Kinetic Energy = a constant
E= K.E+ P.E
E= mv2 +
Kx2
Example 8
The mass m is now stretched a distance of 8 cm and held. What is the potential
energy? (k = 196 N/m)
Solution
P.E= Kx2=1/2(196)(0.08)=0.627J
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Physics Grade 12
Example 9
A slingshot consists of a light leather cup, containing a stone that is pulled back
against 2 rubber bands. It takes a force of 30 N to stretch the bands 1.0 cm (a) what is
the potential energy stored in the bands when a 50.0 g stone is placed in the cup and
pulled back 0.20 m from the equilibrium position? (b) With what speed does it leave
the slingshot?
Solution
a.
P.E =
Kx2 = 1/2(3000N/m)(0.02m)2=60J
b. E=1/2mv2
V=
= 49 m/s
Example 10
A 2-kg mass hangs at the end of a spring whose constant is k = 800 N/m. The mass is
displaced a distance of 10 cm and released. What is the velocity at the instant the
displacement is x = +6 cm?
Solution
v
k
m
A2  x 2
=
= 1.6m/s
1.3 Equations of SHM and the Reference Circle
DISPLACEMENT (SHM)
The displacement of SHM is the distance from equilibrium position in either direction.
The amplitude is the maximum displacement from mean position in either direction.
x  A cos t 
or
x  A sin  t 
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Physics Grade 12
Where A is the amplitude,  (omega) is the angular frequency, and the quantity t is
called the phase.
VELOCITY (SHM)
Velocity of SHM is determined by the following equations:
(i) v = -A sin (t)
(ii) v = Acost
(iii)
vmax = A
Velocity is positive when moving to the right and negative when moving to the left
and velocity is zero at the end points and a maximum at the midpoint in either
direction.
ACCELERATION (SHM)
Acceleration of SHM is expressed as following:
(i) a = -A2cos(t)
(ii) a = -A2sin(t)
(iii)
amax = A2
Acceleration is a maximum at the end points and it is zero at the center of oscillation.
The Relationship between Kinetic Energy, Potential Energy, Displacement, Velocity and
Acceleration
X
v
A
K
U
0
A
0
-w2A
0
½kA2
T/4
0
-wA
0
½kA2
0
T/2
-A
0
w2A
0
½kA2
3T/4
0
wA
0
½kA2
0
T
A
0
-w2A
0
½kA2
T
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Physics Grade 12
1.4 Simple Pendulum
Simple pendulum is a mass on the end of a string which oscillates in harmonic motion.
The equation for the period of a simple
pendulum is:
T  2
f 
L
g
1
2
g
L
Where L is the length of the pendulum,
and g is the acceleration due to gravity
at the location of the pendulum.
Example 11
Find the period of simple pendulum with a length of 4000cm.
Solution
T=2
=4
sec
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Physics Grade 12
Table Summary for Simple Harmonic Motion
Physical Quantity
Expressed in terms of displacement (x)
Expressed in terms of time(t)
Displacement(x)
X=
)
Velocity(v)
V=
Acceleration
a=-(
a=
Restoring force (F)
F= -kx
F = -k
Potential energy
PE=
PE =
Kinetic energy
KE
Frequency
Period
f=
KE
k(
=
f
)
T=
12
A
k
Physics Grade 12
Questions and Problems
Question One
Differentiate Simple harmonic motion and periodic motion.
Question Two
Write short notes on
a. Period.
b. Frequency.
c. Force constant.
d. Reference of SHM.
Question Three
State Hooke’s law and list factors that do not affect period of simple harmonic motion.
Question Four
Define simple pendulum.
Multiple Choice Questions
1. The period of simple pendulum depends on:
a. The mass of the oscillating boy
c. The acceleration due to gravity
b. The length of the string
d. B and C
2. Which of the following is true for a particle in SHM?
a. The total energy is proportional to the square of the amplitude of the motion.
b. The acceleration is maximum at the equilibrium position
c. The kinetic energy is minimum at the equilibrium position
d. The speed decreases as the particle moves toward the equilibrium position
3. The period of SHM depends on:
a. The amplitude of the motion
b. The force constant of the spring
c. The mass of the spring
d. All of the above
4. A 2kg iron ball suspended from a vertical spring undergoes SHM with a frequency
of 20Hz. When a block of mass M is added, the frequency of vibration of the
spring is 10Hz. What is the mass of the block?
a. 8kg
b. 6kg
c. 10kg
d. 4kg
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Physics Grade 12
5. A 2kg body suspended from a spring of negligible mass is found to stretch the
spring by 10cm. If the body is pulled down and the released, the frequency of
oscillation is:
a. 0.1s–1
b.
s–1
c. 0.2s–1
d.
s–1
6. A body undergoes SHM of amplitude 3cm and frequency 20Hz. What is the
velocity of the body at the maximum acceleration of the object?
a. 1.2
b. 12
c. 12
d. 0.12
7. A 1kg object attached to a spring undergoes SHM of amplitude 5cm and a period
of 0.1 . What is the maximum acceleration of the object?
a. 20m/s2
b. 10m/s2
c. 200m/s2
d. None
8. What is the maximum force that acts on the object described in equation 7.
a. 10N
b. 200N
c. 20N
d. None
9. A body is undergoes SHM of amplitude 4cm. At what distance from the
equilibrium position is the speed of the body three – fourth of its maximum speed?
a. 0.01cm
b. 2cm
c. 2.63cm
d. 3.4cm
10. A body is undergoes SHM of amplitude 10cm. At a distance of 4cm from the
equilibrium position the acceleration of the body is 0.04m/s2. What is the period of
oscillation of the body?
a. 2
b. 0.5
c. 40
d. 0.25
11. A simple pendulum of the length 1m has a period of 2s at a certain place. If its
period is to become 4s, then its new length must be:
a. 2m
b. 4m
c. 0.5m
d. 0.25m
12. A 2kg mass is attached to a spring of force constant 100N/m and made to vibrate
on a horizontal frictionless surface with amplitude of 0.4m. The kinetic energy at
the equilibrium position is:
a. 8J
b. 6J
c. 4J
d. 32J
13. Which has a greater velocity at the equilibrium position?
a. A 1kg mass attached to a spring of K=100N/m in horizontal frictionless
surface after releasing it from a compression of 10cm.
b. A 1kg mass oscillating on a concave-upward smooth surface with an
effective vertical height of 5cm.
c. A pendulum bob of mass 1kg displaced through a vertical height of 10cm.
d. All of these could not be compared.
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Physics Grade 12
14. The acceleration due to gravity at a height h above the earth’s surface is 1/9 of that
on the surface of the earth. What is the ratio of the period of a simple pendulum at
a height to that on the surface of the earth?
a. 1/3
b. 9
c. 3
d. 1/9
15. A body is undergoes a SHM of amplitude 5cm. At what distance from the
equilibrium position is the kinetic energy four-fifth of the total energy?
a. 2.5cm
b. 1cm
c. cm
d. cm
16. If K is a positive constant of the following expression represents simple harmonic
motion (x is the displacement of particle from mean position)
a. Acceleration = kx
b. acceleration = kx2
c. Acceleration = kx
d. acceleration = –kx2
17. If k and a are the positive constants and x is the displacement from equilibrium
position of the following expression represents SHM.
a. Velocity = k(a2 – x2)
b. Velocity = √k(x2 – a2)
c. Velocity = √k(a2 – x2)
d. Velocity = k(x2 – a2)
18. A bob of mass m is hanging from the end of an elastic spring and executing SHM
with a period T. It this mass is replaced by another bob of mass 2cm, the new time
period of this system will be:
a. T
b. 2T
c. √2T
d. T/2
19. ____________________ of is not true for SHM
a. Motion is periodic
b. Elastic restoring force must be present
c. System may possess inertia
d. Total energy of system is conserved
20. A body is attached to the end of a spring is executing SHM at the extreme position its
a. Kinetic energy is maximum
b. Both kinetic and position are zero
c. Kinetic energy is zero
d. Its velocity is maximum
21. The motion of the simple pendulum is:
a. Always simple harmonic
b. May be simple harmonic
c. Can never be simple harmonic
d. Circular harmonic.
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Physics Grade 12
22. A body is executing SHM if:
a. Its acceleration is proportional to the displacement and directed away from
mean position
b. Its acceleration is proportional to displacement and directed towards mean
position
c. Its acceleration is zero
d. None of these
23. A body is executing SHM at the mean position if:
a. Its acceleration is maximum
b. Its velocity is maximum
c. It possesses maximum potential energy
d. None of these
24. A body is executing SHM if:
a. Its amplitude of motion remains constant
b. Its amplitude of motion may be constant
c. Its motion is not periodic
d. Its motion may be vibratory
25. A body is executing SHM with force constant k with an amplitude ‘a’, when its
displacement is ‘x’. Its instantaneous K.E is represented by:
a. (½ k (x2 – a2)
b. ½kx2
c. ½k (a2 – x2)
d. A(k/m)
26. A simple pendulum is performing SHM with period T. If its length is doubled. The
new time period will be:
a. 2T
b. 0.5T
c. 2.5T
d. 1.414T
27. If we increase the length of simple pendulum its time period will:
a. Increase b. Decrease c. Remain same
d. becomes infinite
28. A simple pendulum that behaves as a second pendulum on earth. If it is taken to
moon. Where gravitational acceleration is one sixth that on earth. Its time period
will become:
a. 4seconds
b. 12seconds
c. 3.5seconds
d. 4.9seconds
29. The trajectory of the bob of a vibrating simple pendulum after it has got suddenly
detached from the thread while passing through its mean position is:
a. Straight line
b. Circular
c. Parabolic d. Hyperbolic
30. Restoring force is always represent in:
a. Linear motion
b. Circular motion c. Simple harmonic d. None
31. A simple pendulum is transported to moon its frequency of oscillation will:
a. Decrease
b. Increase
c. Remain constant d. Become zero
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Physics Grade 12
32. A simple pendulum is performing simple harmonic motion SHM of the following
will remain constant throughout its motion:
a. Acceleration of the bob
b. Its amplitude
c. Force on the bob
d. Velocity of the bob
33. For a system to execute SHM, it is must to possess:
b. Only elasticity
b. only inertia
c. elasticity as well as inertia
d. neither elasticity nor inertia
34. A spring of force constant k in broken into two equal parts, then the force constant
of each part is:
a. k/2
b. 2k
c. k/√2
d. k
c. Nm-2
d. Nm
35. the SI unit of force constant are:
a. N
b. Nm-1
36. Time period of a simple pendulum is T. It is kept in a lift, which is accelerating
upward. The time period of the pendulum will:
a. Increase
b. Decrease
c. Remain the same
d. First increase then decrease
37. A body is executing SHM of amplitude A. Its potential energy is maximum when
its displacement is:
a. Zero
b. A/2
c. A
d. ±A
38. Mass m is suspended from an elastic spring of spring constant k. The time period
of small oscillation is:
a. 2 √m/k
b. 2 √2m/k
c. 2 √k/m
d. 2 √2k/m
39. The time period of a simple pendulum at the center of the earth is:
a. Zero
b. infinity
c. unity
d. same as that at the surface of the earth
40. In SHM the maximum acceleration is α and maximum velocity is β, its time period is:
a. 2 β/α
b. 2 α/β
c. α/2 β
d. β/2 α
41. In a simple harmonic motion we have the conservation of:
a. Kinetic energy
b. Potential energy
c. Total energy
d. Electrical energy
42. In a simple harmonic motion _______________ is constant:
a. K.E is constant
b. Amplitude is constant
c. Phase is constant
d. P.E is constant
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Physics Grade 12
TRUE AND FALSE QUESTIONS
1. (
) In SHM acceleration is proportional to displacement and is directed
towards mean position.
2. (
) SHM is a motion under elastic-restoring force.
3. (
) In SHM the system must have friction.
4. (
) The motion of the particle moving in a circle is SHM.
5. (
) The motion of the projection of a particle moving in a circle along one of
its diameter is simple harmonic.
6. (
) The motion of a simple pendulum is always SHM.
7. (
) The motion of simple pendulum may be simple harmonic.
8. (
) The time period of a simple pendulum increases with its length.
9. (
) The time period of a simple pendulum changes from place to place.
10. (
) Kinetic energy of a system executing SHM is always conserved.
11. (
) Potential energy of a system executing SHM is always conserved.
12. (
) Total energy of system executing SHM is always conserved.
13. (
) The total energy of a particle executing SHM is directly proportional to the
frequency of oscillation.
14. (
) The total energy of a particle executing SHM is directly proportional to
square of its amplitude of oscillation.
15. (
) If pendulum is transported to moon. It will go fast.
16. (
) The simple pendulum comes to rest after some time due to the pull of the earth.
PROBLEMS
1. A body of mass 1kg suspended from a spring of negligible mass is found to stretch
the spring by 10cm.
a. What is the spring constant?
b. What is the period of oscillation of the body, if the motion is SHM?
c. If the mass is doubled, what is the period of oscillation of a body?
2. A 2kg body attached to a spring undergoes a SHM of amplitude 0.4m and period
0.5πs. a) What maximum force acts on a body? b) Find the acceleration of the body
at a distance of 0.2m for m the equilibrium position.
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Physics Grade 12
3. A 40gm mass hangs at the end of a spring. When a 20gm mass is added the spring
stretches by an additional 5cm.
a) What is the force constant of the spring?
b) What is the frequency with which the 40gm mass vibrates?
4. A body of mass 10g is in SHM with a frequency of 40/π Hz and amplitude of 0.2m.
Find a) The maximum force acing on the body b) the maximum velocity and the
maximum acceleration and c) the acceleration and the velocity at the point 0.1m
from the equilibrium position.
5. A body is undergoing simple harmonic motion of amplitude 0.5m. The acceleration
of the body at a distance of 0.2m from the equilibrium position is 5m/s2. Find a)
the period of oscillation
b) the maximum velocity and
c) the velocity at a
distance 0.3m from the equilibrium position.
6. A 5kg ,ass attacked to a spring performs a SHM of amplitude 50cm. if the
maximum force exerted on the body is 100N,
spring
a) find the force constant of the
b) what is the kinetic energy, potential energy, the acceleration at a
distance of 40cm from the equilibrium position?
7. The mass of an object attached to the spring of force constant 50N/, is 2kg. If the
motion is simple harmonic, find the frequency and the period.
8. A body attached to the spring of force constant 200N/m vibrates in SHM with a
period of 0.5πs. Find the mass of the body.
9. A body of mass m attached to a spring oscillates with a frequency f 1. If the mass of
the body is halved, a) by how much will the frequency increase? b) Find the ratio
of the periods in the two cases.
10. An object of mass 10kg suspended from a spring of negligible mass found to
stretch the spring to a maximum distance of 0.2m a) find the spring constant b) if
the mass is pulled down and released, find the period of oscillation.
11. A body of mass m attached to a spring of force constant 100N/m is found to stretch
the spring to a distance of 0.4m form its equilibrium position. Find the mass of the
body.
12. A 10kg mass hangs at the end of a spring of force constant 200N/m. if the mass is
doubled, by how much will the displacement change?
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Physics Grade 12
13. A 1kg mass hangs at the end of a spring, stretches the spring by a distance X from
the equilibrium position. If the mass is increased by 9gm, the displacement
increases by 10cm. a) what is the force constant of the spring? b) If the spring
vibrates in SHM, find the ration of frequencies in the two cases.
14. A body vibrates with SHM of amplitude 0.5m and frequency 100Hz. Find the
maximum velocity and the maximum acceleration.
15. A 10kg vibrates in SHM with amplitude of 50cm and a period of 2s. Find the
speed, acceleration, kinetic energy and potential energy a) at equilibrium position
b) at the amplitude position and c) at a point 20cm from the equilibrium position.
16. A 2kg body vibrates in SHM with amplitude of 90cm. when the body is at a
distance of 10cm from equilibrium position; its acceleration is 1m/s2. Find the
period, the velocity, the kinetic energy, and the potential energy of the body when
it is at a distance of 10cm form equilibrium position.
17. A body undergoing a SHM with amplitude 50cm has a speed of 10m/s at a distance
of 10cm form the equilibrium position. Find a) the acceleration at this point and b)
the period.
18. A simple pendulum of length 1m has a period π sec. if its period is to become π/2s,
what is its new length.
19. The period of a simple pendulum at a certain place is 3s. If the length of pendulum
is 30cm, what is the value of acceleration due to gravity at this place?
20. The acceleration due to gravity on the surface of the Moon is 1/6 that on the
surface of the earth. Find the ratio of the period of a simple pendulum of the
surface of the moon to that on the surface of the earth.
21. The period of a simple pendulum of length 1m is 2s at a certain place. What is the
value of the acceleration due to gravity at this place?
22. A simple pendulum of length 1m makes 400 complete oscillations in 8s at a certain
place. If another pendulum makes 100 complete oscillation in 1s at the same place.
Find the length of the second pendulum.
23. A body of 0.5kg attached to a spring is displaced from its equilibrium position and
released. If the spring constant is 50N/m. Find
a. Time period
b. The frequency
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Physics Grade 12
24. Calculate the length of second’s pendulum at place where g=10.0m/s2.
25. Calculate the length of second’s pendulum on the surface of moon where the
acceleration due to gravity is 0.167 times that on the earth’s surface.
26. A simple pendulum completes 4 vibrations in 8 seconds on the surface of the earth.
Find its time period on the surface of the moon where the acceleration due to
gravity is one sixth that on the earth.
27. A mass of 4kg is suspended from a spring. The spring is stretched by 0.98m.
Calculate the period of oscillation of the mass when it is given a small
displacement.
28. Find the length of second’s pendulum on planet Jupiter where the value of ‘g’ is
2.63 times the value of ‘g’ on the surface of the earth.
29. A body of mass 0.5kg is attached to the end of a spring placed on a smooth
horizontal table and is performing SHM. Find the acceleration of the body when it
has displacement of 0.6cm. The spring constant of the spring is 150N/m.
30. A body hanging from a spring is set into motion and the period of oscillation is to
be 0.8sec. After the body has come to rest it is removed. How much shorter will the
restoring force be when it comes to rest?
31. The mass at the end of a spring oscillates with a simple harmonic motion with a
period of 0.40sec. Find the acceleration when the displacement is 4.0cm.
32. The period of vibration of a body of mass 25gm. Attached to a spring vibrating on
a smooth horizontal surface, when it is displaced 10cm to the right of its extreme
position, the period of vibration is 1.57 sec, and the velocity at the end of the
displacement is 0.4m/s. Determine the:
a. Spring constant
b. Total energy and
c. Amplitude
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Physics Grade 12
Wave Motion
At the end of this chapter, the learner should be able to:
2.1 Define waves.
2.2 State types of waves.
2.3 Describe wave motion.
2.4 List characteristics of waves and state their meanings.
2.5 Write down wave equation.
2.6 State the properties of waves
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Physics Grade 12
2.1 Introduction to waves
Many of us enjoy watching waves on the surface of water pushing fish float or a boat
up and down. Waves are generated by pebble thrown in a pond or still water. Hence,
waves are disturbances that spread and carry along energy. The waves may appear to
be plane waves which travel together as a front in a straight-line direction or circular
waves.
A wave can be described as a disturbance that travels through a medium
from one location to another location.
2.2 Types of waves
1. Mechanical waves are waves which propagate through a material medium.
Mechanical waves require material medium.
The followings are mechanical waves:
a. Water waves
b. Sound waves
c. Spring waves
d. String waves
e. Seismic waves
2. Electromagnetic waves are waves which propagate through empty space (vacuum).
Electromagnetic waves do not require a material medium.
a. Radio waves
b. Infra-red radiation
c. X-rays
d. Gamma-rays
e. Visible light
f. Ultraviolet radiation
g. Micro-waves
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Physics Grade 12
1. Visible light is a small part of the energy range of electromagnetic waves.
2. Radio waves are on the low-frequency end of the spectrum.
3. Microwaves range in length from approximately 30 cm to about 1 mm.
4. The infrared region (IR) lies between microwaves and visible light.
5. Ultraviolet radiation has a range of wavelengths from 400 down to about 10 nm.
6. X-rays are high-frequency waves that have great penetrating power and are used
extensively in medical and manufacturing applications.
7. Gamma rays are generated in nuclear reactions.
2.3 WAVE MOTION
One way to categorize waves is on the basis of the direction of movement of the
individual particles of the medium relative to the direction which the waves travel.
1. Transverse waves
A transverse wave is a wave in which particles of the medium move in a direction
perpendicular to the direction which the wave moves.
2. Longitudinal waves
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Physics Grade 12
A longitudinal wave is a wave in which particles of the medium move in a direction
parallel to the direction which the wave moves.
2.4 Characteristics of Wave Motion
1. Velocity (v) is the distance covered by the disturbance in one second.
2. Period (T) is the time taken to make one complete cycle of wave motion.
3. Frequency (f) is the number of cycles or waves per second and its unit is Hertz (Hz).
4. Wave length (λ) is the distance between two consecutive crests or troughs.
5. Amplitude (A) is the maximum displacement on either side of the undisturbed or
rest position.
6. Phase (
: There are two phases, the highest point is called Crest and the lowest
point called Trough.
2.5 WAVE EQUATION
The velocity of a wave is the product of the wavelengthand the frequency.
v  f
v 
T
V= Velocity
f= frequency,
λ = wavelength
Example 1
The wavelength of a wave is 0.75m and its frequency is 480Hz, what is the velocity of
the wave?
Solution
λ=0.75m
f = 480Hz
V=?
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Physics Grade 12
v  f  0.75m  480 Hz  360 m / s
2.6 Properties of Waves
1. Rectilinear propagation means waves travel in straight lines.
2. Reflection occurs when waves strike a surface and bounces off.
3. Refraction occurs when waves deviate from their original path while passing
from one medium to another of different optical media.
4. Diffraction is the change of wave path when passing through a slit or an aperture.
5. Interference is the phenomenon produced from the superposition of two waves in
a medium. Interference may be constructive interference or destructive
interference. Constructive interference occurs when reflected waves that are in
phase with the incoming waves and destructive interference occurs when waves
that are out of phase.
Superposition principle
 When more than one wave is present, the total oscillation of any point is the sum of
the oscillations from each individual wave.
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Physics Grade 12
 The sound waves and light waves you experience are the superposition of
thousands of waves with different frequencies and amplitudes.
 Your eyes, ears, and brain separate the waves in order to recognize individual
sounds and colors.
Two waves of equal amplitude but slightly different frequencies interfere
destructively and constructively. The result is an alternation of loudness called a
beat.
2.7 Standing (Stationary) waves
A wave that is confined between boundaries is called a standing wave. With all waves,
resonance and natural frequency are dependent on reflections from boundaries of the
system containing the wave. Standing waves arise from the superposition of two
identical waves having the same amplitude and frequency travelling with the same
speed in opposite direction.
1. Strings fixed at both ends
L  n

2
f=
Fn= nf1
n =1, 2, 3,.........
L

Ends fixed
Nodes
l = length
f=frequency
Anti-nodes
v= velocity
Node is the position at which the amplitude is zero.
Anti-node: is the position at which the amplitude is maximum.
2. Strings fixed at one end
L=
f=
Fn= nf1
N=1,3,5,7,........
Example 2
A string 120cm long resonates in three segments due to transverse waves sent down it
by a 210Hz .Find (a) the number of segments (b) the wavelength (c) the fundamental
frequency (d) next three frequencies (e) the speed of the wave.
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Physics Grade 12
Solution
Data: L=120cm=0.12m
f3=210Hz
a. N=3
b.
0.12m)/3=0.8m
c.
d. F2=2(70hz)=140hz ,
f3=3(70)=210hz,
f4=4(70)=280hz
e. V=2(0.12)(70)/1=168m/s
Example 3
An iron rod has a fundamental frequency of 420Hz and it is clamped at its center.
What are (a) the longest wave length and (b) the length of the wire? Assume
longitudinal vibration at speed of 5km/sec?
Solution
F1=420Hz, V=5km/sec=5000m/s
a.
=5000m/s/420Hz=12m
Questions and Problems
Question1
Explain the meanings of the words; diffraction, interference and superposition.
Question2
Explain the meaning of the words; period, frequency, amplitude, resonance, in phase
and out of phase.
Question3
Explain the difference between transverse and longitudinal waves.
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Physics Grade 12
Question4
State the relation between λ, V and f. Explain the standing waves
Question5
In a standing wave pattern, what is the relationship between the length of the vibrating
string and the wave length?
Problems
1. A string has a transverse wave of 35cm, wavelength travelling along it. If it
vibrates in
Ans. 63m/s
180Hz
with
what
speed
is
the
wave
propagating?
2. A transverse wave of frequency 2Hz are generated along a spring. The waves have
a wavelength of 0.5m.
(a) What is the speed of the wave?
(b) What would be the speed of the waves if their frequency is doubled?
Ans. 1m/s, 2m/s
3. A water wave with a wavelength of 3m is propagating with a speed of 12m/sec.
what is their frequency?
Ans. 4 Hz
4. Sound wave in a steel rail has a frequency of 620Hz if the wavelength of 10.5m.
What is the speed of a sound wave?
Ans. 6510m/s
5. What is the wavelength of a wave with a frequency of 50Hz if the speed of the
wave is 345m/sec?
Ans. 7m
6. A certain radio station broadcasts at 1600 KHz. The speed of the radio wave is 3 x
1010cm/s. On what meter band can you hear the broadcast? Ans. 188m
7. The distance between crest and the adjacent trough of water wave is 3m. They pass
a given point at a rate of 5m/sec. What is (a) the frequency (b) the speed of the
water waves?
Ans. 5m/s, 0.83m/s
8. The time taken to pass two successive crests of water waves is 0.5sec. If the speed
of the wave is 4m/sec, what is (a) frequency (b) the wavelength of the water
waves?
Ans. 2Hz, 2m
9. Microwaves travel though space at a rate of 3 x 1010m/sec. if the wavelength of
the waves is 200cm. what is its (a) frequency and (b) period?
Ans. 1.5 x
108Hz,6.7 x 10-9s
10. A train emits sound of wave of wave length 0.6m and frequency 550Hz. Find the
velocity.
Ans 330m/s
Standing waves on strings
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Physics Grade 12
11. A 1.5m long string vibrates in 4 segments. If the speed of the wave is 300/sec. what
would be the wave length of the string?
Ans 0.75m
12. A string vibrates in six segments with a frequency of 525Hz. (a) find its
fundamental frequency? (b) What frequency will cause it to vibrate in 4 segments?
Ans. 87.5Hz, 350Hz
13. The difference between the 3rd and 4th harmonics of a 70cm long fixed string is
30Hz. What is (a) the fundamental frequency. (b) The longest wavelength (c) the
wavelength of the 3th overtone (d) the speed in the 1st 2nd overtone?
Ans.30Hz,1.4m,0.35m,42m/s
14. A 2m long fixed string has nodes at both ends when it vibrates in its fundamental
mode. If the distance between this nodes is 12cm. what are the number of (1) nodes
(b) anti-nodes?
Ans. 7, 7
15. The 3th overtone of one end fixed string produce a frequency of 72Hz and the
length between a node and the adjacent anti-node is 5cm.waht is:
(a) the length of the string ?
(b) the wave length in this overtone and in the 1st overtone?
(c) the wave speed ?
(d) the fundamental frequency?
(e) the longest wavelength? Ans. a). 35cm b)20cm and 47 cm c)14.4m/s d)10.3
Hz e)1.4m
16. A 100cm long rod is clamped at its centre and stroked. At what other points might
be clamped in order to amid the same tone, (a) in the 1st overtone (b) in the 2nd
overtone?
Ans. a) 16.7 cm
b) 10 cm.
Sound Waves
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Physics Grade 12
Chapter Objectives
3.1 Introduction to Sound waves
a. Nature and Transmission of Sound
b. Sources of sound
3.2 Speed of sound
3.3 Reflection of Sound and Echo
3.4 Properties of Sound
a. Intensity and loudness
b. Frequency and pitch
c. Harmonic Contents and Quality
3.5 Doppler Effects
3.6 Standing Waves in Air Columns
3.7 Beats
3.8 Forced vibrations and Resonance
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Physics Grade 12
3.1 Introduction to Sound Waves
Sound is a longitudinal wave created by vibrating objects such as loud speakers,
piano, and human vocal cords. The particles carrying the sound vibrate in fixed
positions parallel to the direction in which the sound travels. It is a mechanical wave
that requires a medium to travel through.
Definition
Sound wave is a mechanical wave that produces a sensation of hearing.
Sound travels through liquids, gases and solids. The
nature of the medium determines the velocity of the
sound through the medium and solid is the best
transmitter of sound.
For hearing a sound, the following conditions must be
fulfilled:
1. There must be a vibrating body.
2. There must be material medium.
3. There must be receiver.
3.1.1 Sources of Sound Waves
Sound waves are produced by:
1. Vibrating strings, such as guitar.
2. Vibrating surfaces, such as drum.
3. Tubes in which air column vibrates in musical instruments such as trumpets and
saxophones.
3.2 Speed of sound
The velocity of sound in the air is 330m/s at 00C, for every 10C rise of temperature,
the velocity is increased by 0.6m/s.
 0.6m / s 
v  330m / s  
T
 1C 
Example 1
What is the velocity of sound in air at
20˚C?
Solution
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Physics Grade 12
 0.6m / s 
v  330 m / s  
T
 1C 
 0.6m / s 
0
v  330 m / s  
  20 C  342 m / s
 1C 
V  330m/s
T=200C
V ?
Example 2
The speed of sound in air is 348m/s, what is the temperature?
Solution
V = 384m/s
T =?
=
=30
3.3 Reflection of Sound and Echo
A sound persists for 1/10 seconds in our ear after exciting sound dies off. This time is
called persistence of audibility.
Definition: The reflection of sound wave is called echo.
Conditions for formation of Echoes
1.
The minimum distance between the source of sound and the reflecting body
should be 17m.
2.
The wave length of sound should be less than the height of reflecting body.
3.
The intensity of sound should be sufficient so that it can be heard after reflection.
Echoes travel twice the distance between source and reflecting surface.
d
Vt
2
d = distance
t= time
v= Speed
Reverberation is series of reflections that fall on ear from various reflectors one
after another in a closed room forming a continuous rolling sound.
Sound waves are reflected off hard surfaces like glass and wood. They are not
reflected by soft surfaces like clothes and rubber. These surfaces absorb sound. That
is why the walls of auditorium and cinema halls are covered by soft material like
curtain, to decrease reverberation.
Example 3
An echo is heard after 5sec at 25
. Find the distance between the source of sound and
reflecting surface?
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Physics Grade 12
Solution
Given: v= 330m/s + 0.6m/s
=
(25 ) = 345m/s
t= 5sec
d=?
= 862.5m
Example 4
A boy stands 66.4m in front of high wall and then blows a whistle. Find time of echo if
the speed of sound is 332m/s?
Solution
Given: v= 332m/s d= 66.4m t=?
=
= 0.4s
Example 5
A man stands between two parallel cliffs and fires a gun. He hears two successive
echoes after 3sec and 5sec what is the distance between cliffs (v= 330m/s).
Solution
Given: v= 330m/s t1= 3sec t2= 5sec d=?
Step 1: Finding d1
=
= 495m
Step2: Finding d2
=
= 825m
Step 3: Finding d
d = d1 + d2 = 495m + 825m = 1320m
Uses of Echo
1.
Determination of the depth of ocean (sea).
2.
Echoes guide insect eating bats and protect them avoid colliding with objects
or with one another.
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Physics Grade 12
3.
Dolphins communicate with each other.
4.
Echoes are used by army to locate gun positions of enemy.
5.
Echoes are used by geologists for mineral prospecting.
3.4 Properties of Sound
We shall consider three physical properties of sound:
1. Intensity
2. Frequency
3. Harmonic contents
The effects of these properties on the ear are called loudness, pitch
and quality
respectively.
Intensity and Loudness
The intensity of sound is rate at which the sound energy flows through a unit area
normal to the direction of sound waves.
or
I= intensity of sound P = power or rate of energy, r= distance or radius
Loudness describes how loud or soft a sound is perceived to be.
Example 6
The power of sound is 0.02watt. Find the intensity of the sound wave with radius of
(i) 2m and (ii) 3m.
Solution
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Physics Grade 12
a)
=
= 4× 10-4 w/m2
b)
=
= 1.8 × 10-4 w/m2
Intensity Level
Intensity level (
of sound can be
measured by the equation below and its
unit is decibel (db).
Example 7
Sound has an intensity of 4×10-6 w/m2. What is the intensity level?
Solution
Example 8
What is the intensity of 60db sound?
Solution
I = 10-6w/m2
Frequency and Pitch
Pitch is the change of frequency. Therefore, sound with long wavelength has a low
pitch, and the sound with short wavelength has a high pitch. The pitch of a note
depends on the frequency of the wave reaching the ear which in turn depends on the
frequency of the source of the sound. Musical notes have higher pitch if their frequency
is higher.
Harmonic Content and Quality
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Physics Grade 12
The number of harmonic contents determines the quality of the sound. The simplest
harmonic content is called fundamental tone. The other ones are overtones of the
fundamental tone. These frequencies are called harmonics of the fundamental. The
fundamental is the first harmonic. Each of the higher harmonics is named by the
integer by which the frequency of the fundamental must be multiplied to give its
frequencies. The modes of vibration with the frequency 2f is the 2nd harmonic or
first overtone, that with 3f is the 3rd harmonic or second overtone and so on. In all
modes of the vibration, there must be a node at each end of the string.
Range of Audibility
The range of audibility as a frequency is from 20Hz to 20,000Hz and this vibration is
called sonic vibration, the frequency which exceeds 20,000Hz, such vibrations are
called ultrasonic sound and the frequency is below 20Hz, such vibrations are called
infrasonic sound.
Some animals, like dogs and some fish, can hear frequencies that are higher than what
humans can hear (ultrasound). Bats and dolphins use ultrasound to locate prey
(echolocation). Doctors make use of ultrasound for imaging fetuses and breaking up
kidney stones. Elephants and some whales can communicate over vast distances with
sound waves too low in pitch for us to hear (infrasound). As people get older, the
upper frequency which their ear can hear drops considerably.
Frequency
Intensity
20HZ
20,000HZ
10-12w/m2
1W/M2
0db
120db
Intensity level
USES OF ULTRASONIC SOUND
1. Echo sounders
As radio waves do not travel far in sea water, ultrasonic pressure pulses are used to
measure the sea depth at a particular point. The shorter wavelengths used help stop the
pulse spreading out as it travels, ensuring that enough of the transmitted energy is
received so that a time delay can be measured.
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Physics Grade 12
2. Ultrasound scanning
Pictures of unborn babies are obtained using an ultrasound scanner. This relies on the fact
that the pressure waves will pass through the soft tissues surrounding the baby but will
reflect off the baby’s bones. The short wavelengths used allow a reasonably detailed
picture to be obtained.
3. Ultrasonic Cleaning
If a cleaning in which an object has been placed is agitated ultrasonic frequencies, the
pressure waves produce a scrapping action. The result is that the object is cleaned far
more thoroughly than would be otherwise impossible.
4. Homogenizing Milk And Cosmetics
Milk is agitated with ultrasonic vibrations so that larger fat particles break to form
smaller particles.
5. Sonar (Sound Navigation and Ranging)
In addition to locating prey, bats and dolphins use sound waves for navigational purposes.
Submarines do this too. The principle is to send out sound waves and listen for echoes.
The longer it takes an echo to return, the farther away the object that reflected those
waves. Sonar is used in commercial fishing boats to find schools of fish. Scientists use it
to map the ocean floor. Special glasses that make use of sonar can help blind people by
producing sounds of different pitches depending on how close an obstacle is.
3.5 Doppler Effect
The change in pitch produced by relative motion of source and observer is called
Doppler Effect.
CASE 1: SOURCE MOVING
A source moving (Vs) towards or away from stationary observer is determined by:
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Physics Grade 12
f'= frequency heard by stationary observer
f = Source frequency
v = Speed of sound
vs = Source velocity
CASE 2: OBSERVER MOVING
The frequency (
heard by an observer moving with a velocity (Vo) towards or away
a stationary is determined by:
fo  f
V  vo
V
CASE 3: OBSERVER AND SOURCE MOVING
When observer and source moving towards each other and away from each other is
determined by:
f 'o  f
V  vo
V  vs
Example 9
A train moving at speed of 40m/s blows its whistle which has frequency of 500Hz.
Find the frequency heard by stationary observer as the train a) Approaches
b) Recedes (v= 345m/s)
Solution
Given: v= 345m/s f= 500Hz
a)
b)
Vs= 40m/s f'=?
f ' f
V
345m / s
 500
 566 Hz
V  Vs
345  40m / s
f ' f
V
345m / s
 500 Hz
 448 Hz
V  Vs
345  40m / s
Example 10
A boy moving 6m/s towards or a way from stationary observer whistling a frequency
of 650Hz. Find the frequency when he approaches
or recedes away from the source
(Assume v= 342m/s )
Solution
39
Physics Grade 12
Given: v = 342m/s f = 650Hz
Vo = 6m/s
fo=?
Solution:
a)
=
= 661Hz
b)
=
= 639Hz
Example 11
A car travels at speed of 33.5m/s, its siren emitting sound at a frequency of 400Hz.
What frequency is heard by a passenger in a taxi travelling 24.6m/s in opposite
direction as the taxi approaches or recedes
(v= 345m/s).
Solution
Given: vs = 33.5m/s f = 400Hz Vo = 24.6m/s
f’o = ?
a)
f 'o  f
V  vo
345  24.6m / s
400 Hz
 475 Hz
V  vs
345  33.5m / s
b)
f 'o  f
V  vo
345  24.6m / s
400 Hz
 339 Hz
V  vs
345  33.5m / s
3.6 Standing Waves in Pipes
A longitudinal standing wave is set up in wind instruments which produce sound by
vibrating air column in a pipe. The pipe is closed when it is open at one end and closed
at the other end or open when it is open at both ends.
Closed Pipe
If a source of sound is held over an open end of a closed pipe, a longitudinal wave is
generated which travels through the air column to the closed end of the pipe. There are
two identical waves traveling in opposite directions give rise to longitudinal standing
(stationary) waves in the pipe.
Since the closed end acts like a rigid barrier, a node (N) is formed there but at the open
end an antinodes of air (A) is formed.
40
Physics Grade 12
=
=
Example 12
A closed pipe has a length of 50cm. What are three lowest frequencies to which it will
resonate?
[v=340m/s]
Solution
L=50cm= 0.5m
v=340m/s
=
f=?
=
=340Hz
=2(340Hz) =680Hz
=3(340Hz) =1020Hz
The lowest frequencies are 340Hz, 680Hz and 1320Hz.
Vibrating Strings
The properties of vibrating strings can be studied by instrument called sonometer. The
frequency of vibrating string is determined by its length, diameter, tension, and
density.
Law of Length
Frequency of vibrating string is inversely proportional to its length.
f1L1  f 2 L2
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Physics Grade 12
Law of Diameter
Frequency of vibrating string is inversely proportional to its diameter.
=
Law of Tension
Frequency of vibrating string is directly proportional to square root of its tension.
=
Law of Density
Frequency of vibrating string is inversely proportional to the square root of its density.
=
Combining the above four laws, we get
L = length
T = Tension
=
d = diameter
f = Frequency
= density
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Physics Grade 12
3.7 Beats
The effect of two different frequencies interfering creates a very interesting effect.
That is, adding two waves which are slightly different in frequency (f1 and f2) creates
another wave with a frequency equal to the average of the original sounds. But the
amplitude of the resulting wave varies with a frequency equal to the difference of the
two frequencies - called the beats.
3.8 Forced vibrations and Resonance
The setting up of vibrations in an object by a vibrating force is called forced vibration.
When the frequency of an applied force matches the natural frequency of an object,
energy is transferred very efficiently. The condition is called resonance.
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Physics Grade 12
Questions
Question 1
Describe in detail how sound waves are produced and how they travel through air.
Question 2
Light can be reflected from a small plane mirror, but sound cannot why?
Question 3
Explain properties of sound.
Question 4
What factors determine the frequency of vibrations of stretched strings?
Question 5
Draw a diagram of the simple stationary wave formed in a closed and open pipe when
they are sounded.
Question 5
What is Doppler Effect? Deduce an expression for the apparent frequency when a
source of sound is approaching a stationary observer.
Question 6
What are beats? How are they produced?
Question 7
On what factor does a) The loudness b) The pitch of a note depend?
Question 8
Explain sound wave and state uses of echo.
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Physics Grade 12
Problems
SPEED OF SOUND (Take v = 330m/sec, 00C)
1. What time is required for sound to travel 5km if the temperature is 150C?
[14.7m/s]
2. What is the distance between two adjacent rarefactions in a sound wave whose
velocity is 336m/s and whose frequency is 408Hz?
[ 0.82m]
3. What is the shortest wavelength of sound that can be heard by a human if the
frequency of audible sound lie between 20Hz and 20 KHz? ( take v= 344m/sec)
[1.72cm]
4. The coldest and hottest temperature ever recorded in a certain region are-800F and
1340F. What is the speed of sound at each temperature?
[293m/s, 264m/s]
5. The waves from sound a sound source of frequency 244Hz travel through air at
340m/s. Calculate the wavelength of the waves and deduce the distance from a
compression to the nearest rarefaction?
[1.4m, 0.7m]
6. What frequency of sound travelling in air at 250C has a wavelength equal to 1.7m?
[203Hz]
7. A shell fired at a target 800m distant was heard to strike it 5sec after leaving the
gun. Compute the average horizontal velocity of the shell.
[320m/s]
8. The sound of a thunderclap was heard 3sec after the flash was seen. If the
temperature at that time was 500c, find the distance of the cloud? [1.08km]
9. A sonar signal (a high frequency sound wave) sent vertically downwards from a
ship is reflected from the ocean floor and detected 3sec later. If the speed of sound
in water is 1500m/s, what is the depth of the ocean in mater?
[2.25km]
10. A person stands at a point 300m in front of the face of a sheer cliff. If the person
shows shout, how much time will elapse before echo is heard? The temperature of
air is 200c.
[1.75sec]
11. A geologist is camped 6000m from a volcano as it erupts. (a) How much time
elapse before the geologist hears the sound from the eruption? The temperature of
air is 200c (b) How much time does it take for the seismic the waves travel through
granite with a speed of 4000m/s.
[17sec, 1.5sec]
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Physics Grade 12
12. Ali and Abdi go for a walk along an abandoned railroad truck, Ali puts one ear
next to a rail, while Abdi 300m away taps in the rail with a stone. How much
sooner does Ali hear the sound (speed of sound in steel is 6500m/s).
[0.05sec]
13. A man drops a stone into a mine shaft 250m deep the temperature is 100c. How
many seconds pass before he hears the stone strike the bottom?
[7.9sec]
14. A man throws a stone over a cliff and hears it strike at the bottom in 8sec. The
temperature is 250C. What are the two times; the time that the stone travels and the
time that sound travels? What is the height of the cliff in meters?
[7.25sec, 0.75sec, 258m]
15. A rifle is fired in a valley with parallel walls. The echo from one wall is heard in
2sec; the echo from the other wall is heard 2sec later the temperature is 200C, what
is the width of the valley?
[ 1026m]
16. Thunder was heard 2.3sec after the lighting if the temperature was 250C, how far
away was the lighting?
[793.5m]
17. An object is dropped into a deep well which has water. If the well is 20m deep,
how long after the object is dropped will the splash is heard if temperature of air is
200C?
[2.08sec]
18. A sound wave is sent from a ship a point just below the water surface and the
sound is received 1.75sec later. Taking the speed of sound in water 1500m/s. how
deep is the sea under the ship?
[1312.5m]
INTENSITY OF SOUND
19. A loud speaker has a circular opening with an area of 55 cm2. Assume that the
sound it emits is uniform and outward through this entire opening. If the sound
intensity at the opening is 5×10-5w/m2. Find the power radiated as sound.
[2.75×10-7w]
20. A tiny sound source sends sound equally in all directions if the intensity is 4.5×10 4
w/m2, 4m from the source (a) how much does sound energy the source emit each
second? (b) What is the intensity 3m from the source?
[2.26×10-2w, 8×10-4w]
21. A sound source emits sound uniformly in all directions with power 2w (a) find the
sound intensity at a distance of 5m from the source (b) find the intensity level in
decibels at this distance.
[ 2.55×10-2w/m2, 104db]
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Physics Grade 12
22. What is the intensity in watt/cm2 of a sound which has an intensity level of 30db?
[10-9w/m2]
23. What is the intensity level in Bel (B) of a sound which has an intensity of 10-11
w/cm2?
[5B]
24. The intensity of a sound is determined to be 10-10w/m2. What is the intensity level?
[20 db]
25. A window whose area is 1.52m2, opens on a street where the street noise result in
an intensity level at the window of 80db. How much acoustic power enters the
[ 1.52×10-4w]
windows by the sound wave?
26. The total effective area of the auricles of a listener is about 60cm2 (a) how much
power (in watts) enters the ears of a person hearing a music if the intensity level is
75db? (b) How long would be required for 1.9×10-9J of the sound energy to enter
[1.9×10-7w, 0.01sec]
the listener’s ear?
27. A sound has an intensity of 7×10-11w/cm2, what is the intensity level?
[58.5db]
28. The intensity of the sound is determined to be 4.5×10-10w/cm2, what is its intensity
level?
[6.65B]
2
-2
29. Find the intensity in w/m corresponding to an intensity level of 10db. [ 10 w/m2]
30. Find the intensity level in decibels for a sound wave of intensity (a) 2.5×10-10w/cm2
(b)10-6.5w/m2.
31. A band has produced an intensity level 110db what is the sound intensity?
[ 6.4B]
[10-1w/m2]
32. One sound has intensity level of 60db, while a second has 70db what is the
difference combined intensity level?
[ 70.4db]
33. One sound is 600 times as intense as another sound what is the difference intensity
level of these two sounds in decibel?
[ 27.8db]
34. 10typist in room give rise to an average sound intensity of 58db, what is the
intensity level in the room when an additional three typist each generating the same
amount of noise being to type?
[ 59db]
35. What sound intensity 4db is louder than a sound of intensity 12×10-6w/cm2?
[3×10-5w/m2]
36. What fraction of the acoustic power of a noise would have to be eliminated to
lower its sound intensity level from 90dB to 70dB?
[100]
BEAT
37. Two sources of sound having frequency 514 and 552Hz are vibrating
simultaneously. How many beats are heard per second?
[38]
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Physics Grade 12
38. How many beats will be heard each second when a string frequency 280Hz is
plucked simultaneously with another string frequency 312Hz?
[32]
39. A turning fork has a frequency of 440Hz if another fork of slightly lower pitch is
sounded at the same time, the beats heard were 11 beats/sec, the is the frequency of
the second turning fork?
[ 429]
40. A guitar string under the tension of 200N, vibrating in its fundamental mode, gives
6 beats/sec. With a turning fork the player increase the string tension to 242N and a
gain gets 6beat/sec. Find the frequency of the turning fork?
[127Hz]
41. When two turning fork are sounded simultaneously, they produce one beat each
0.8sec. When the frequency of one is 475Hz, what is the frequency of the other?
[ 473.75Hz or 476.25]
42. A certain organ pipe is turned to emit a frequency of 200Hz, when a string of violin
are sounded together,12 beats are heard of 8 seconds, what are the two possible
frequencies of the violin string?
[ 198.5Hz and 201.5Hz]
43. When two turning forks are sounded simultaneously, they produce one beat each
0.3 sec a) by how much do their frequency differ? B) A tiny piece of chewing gum
is placed on a prong of one form. Now there is one beat each 0.45sc. Was this turn
fork the lower or the upper frequency fork?
[ 3.3Hz , upper]
DOPPLER’S EFFECT
44. A locomotive approaches a crossing at 95km/hr. Its horn has a frequency of
285/sec and the temperature is 150C.What is the frequency of the sound heard by
the watchman?
[ 309Hz]
45. What is the frequency of the sound a listener hears as a train, with is horn sounding
passes at 60km/hr.? The frequency of the horn is 320/sec. the temperature is 250C.
[305Hz]
46. A train approaches a station at a constant speed of 25m/sec sounding a 750Hz
whistle? A) What frequency is perceived by an observer on the platform of the
station? B) After the train passes the station platform, what frequency is perceived
by the observer as the train moves away (T=200C).
[809Hz & 699Hz]
47. A factory whistle near a road side emit a sound of frequency 500Hz A)What is
the apparent frequency heard by a passenger in a car that is approaching the factory
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Physics Grade 12
at 25m/sec,
b)what is the apparent frequency after the car has passes the factory
and is moving away from the whistle?(T=200C).
[537Hz and 694Hz]
48. A boy walks at 65m/sec towards a stationary man. If the child is whistling a note of
frequency 650Hz. What is the apparent frequency heard by the man?
(T=200C).
[ 651Hz]
49. An automobile at 25m/sec is approaching a factory whistle that has a frequency of
450Hz. What is the frequency heard by the driver a) As approaching
the factory with the same speed(V=340m/sec)
b)leaving
[483Hz and 417Hz ]
50. An object with the horn at 150m/sec approaches and passes a person standing at the
curb Its whistle emitting a note of frequency 1000Hz. What frequency and
wavelength will the person hear a) As the train approaches and b) As it recedes?
Vair=340m/sec.
[1789Hz, 0.19m b) 694Hz, 0.49m]
51. Two cars are heading straight at each other with the speed V. the horn of one (f=
2000Hz) is blowing, and is heard to have a frequency of 2500Hz by the driver in
the other car. Find the speed of the cars if Vair=340m/sec.
[ 37.8m/s]
52. At what velocity an engine, emitting a note of 240Hz must move toward a
stationary observer so that the apparent frequency is 266Hz? (Given velocity of
sound 356m/sec).
[ 34.8m/s]
VIBRATING IN STRINGS
53. What is the velocity of transverse wave on a string of length 3m that has a mass of
0.012kg and is stretched with a tension of 25N?
[79m/s]
54. A source vibrating with frequency of 360Hz sets up stationary waves on a string
the nodes are 30m apart what is the wave velocity?
[ 216m/s]
55. Compute the speed of transverse waves on a rope 10m long, stretched with a
tension of 40N. The mass of the rope is 2kg.
[14.1m/s]
56. Stationary waves are produced on a string for which the velocity of transverse
wave is 280m/sec. The frequency of vibration is 800Hz. How far apart are the node
and the adjacent anti node?
[ 8.8cm]
57. The fundamental frequency of a stretched string is 500Hz, if a person is able to
hear frequency up to 18,000Hz what is the highest a) Harmonic b) Overtone that he
can hear?
[36 and 35]
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Physics Grade 12
58. An 80cm long string is stretched b/w two stands so that the fundamental frequency
is220Hz a) Find the velocity of transverse waves on the string
three overtones of the string.
b)Find the first
[ a) 352m/s b) 440Hz , 660Hz, 880Hz]
59. A string under tension 1080N, of mass per unit length 0.003kg/m has many
resonant frequency on such frequency is 450Hz an d the next higher frequency is
600Hz How long is the string?
[ 2m]
60. A string when stretched with a tension of 45N produces a note of frequency F. Find
the tension required to produce four times this frequency.
[ 720N]
61. If the string on a violin is 26cm long and produces a fundamental tone of frequency
440Hz. By how much must it be shortened to produce a tone of 550Hz?
[5.2cm]
62. When the tension is 27N, a string 200cm long has a fundamental frequency of
150Hz a)What is the mass of the string b)With what tension must the string be
stretched so that it vibrates in 3 segments.
[ a) 0.15g b) 3N]
63. A string of mass 0.8kg is stretched between 9cm a part. It is observed that when
one pole is struck the transverse pulse reaches the other pole in 0.3sec. What is the
tension in the pole?
[ 80N]
64. A wire whose mass per unit length is 0.2kg/m is stretched with a tension of 800N.
The wires fundamental frequency is in tune with the first overtone of a pipe 2.5m
long that is closed at one end the speed of sound is 350m/sec how long is the wire?
[ 95cm]
65. A wire under tension vibrates with a frequency of 550Hz; what would be the
fundamental frequency if the wire were half as long, twice as thick and under
vibration in pipes.
[ 275Hz]
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Physics Grade 12
VIBRATING IN PIPES
66. A tuning fork, frequency 385/sec produces resource with a closed tube 20cm long
and 4cm in diameter what is the speed of sound?
[ 338m/s]
67. Two organ pipe closed at one end are sounded simultaneously to give 9beats/sec
b/w their fundamental frequency. If the shortest pipe has a length of 0.75m. Find
the length of the longer pipe?
(Take V=344m/sec).
[ 0.8m]
68. The fundamental frequency of a closed pipe is 300Hz. What are the first three
overtones?
[ 900Hz, 1500Hz]
69. The lowest frequency of a pipe closed at one end is 540Hz and the speed of sound
is 344m/sec. How long is the pipe?
[ 6cm]
70. What is the frequency of the next-to-the lowest natural frequency of a pipe 42cm
long and closed at one end on a day when the speed of sound is 336/sec?
[600Hz]
71. A wire 0.8m long of total mass 40g is stretched between two points, and a pipe
1.5m long and closed at one end is placed nearly. The speed of sound in air is
350m/sec. What should be the tension in the wire in order for the 6th harmonic of
the wire to be in resonance with the 2nd overtone of the pipe?
[ 302N]
72. An organ pipe closed at one end gives a note of 256 when the temperature of air is
at 400C. What will be the frequency of the note when the temperature falls to 150C?
[245Hz]
73. Find the length of an organ pipe open at both ends that will give a fundamental
note of frequency 256Hz. In air at a temperature of 200C.
[67cm]
74. An open pipe is suddenly closed with the result that the second over tone at the
closed pipe is found to be higher in frequency buy 100Hz than the first overtone of
the original pipe. What is the fundamental frequency of the open pipe?
V=340m/sec.
[200Hz]
75. A) What is the length of a column of air in a closed pipe if it vibrates with a
fundamental mode of vibration of 160Hz?
B) What would be the next shortest
length? Assuming the speed of sound to be 340m/sec.
[a) 53cm b) next node 1.6m]
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Physics Grade 12
76. If the length of an open pipe is to be 33.2cm, what would be the minimum
frequency of the standing wave, assuming the speed of sound is 340m/sec. [512Hz]
77. A whistle closed at one end has a fundamental frequency of 220Hz. What is the
frequency of the first possible overtone?
78.
[ 660Hz]
The fundamental frequency of an open pipe has the same frequency as the first
overtone of a closed pipe 2.5m in length. What is the length of the open pipe?
[1.7m]
79.
A one end closed organ pipe is 50cm long. What are the 1st three overtone
frequencies if the speed of sound is 345m/sec.?
[517.5Hz, 862.5Hz, 1207.5Hz]
80. One meter end closed pipe is sounded with an open pipe to produce 4 beats per
second. What is the length of the open pipe at a temperature of 250C.?
[2.1m]
Further Problems
81. The speed of sound is 320m/s at 0 . What is the speed of sound at temperature of
a) 20
b) 40
c) 60
[ a. 308m/s b. 344 m/s
c. 356 m/s ]
82. The speed of sound at 10
is 326m/s. find the speed of sound at 34
? [340.4m/s]
83. The speed of sound is 345m/s. Find its temperature.
[25 ]
84. The speed of sound is 320m/s at 0 . Find the temperature when its speed is a)
increased by
of speed 0
b) decreased by
of speed at 0
[a. 53.3
b -66.67
85. In a boy the speed of sound in air changes from 336m/s to 340.8m/s. What is
change of temperature?
[8 ]
86. What is the intensity level of a) 2.5×10-6w/m2 b) 5×10-6w/m2? [a. 64db b. 67db]
87. What is the intensity level of sound with intensity level of a) 30db
b)75db
[10-9w/m2 and 3.2×10-5w/m2 ]
88. If the intensity level of sound is a) 10-10w/m2 b) 7×10-7w/m2. Find the intensity
level.
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Physics Grade 12
89. Four typists in a room guess rise to an average intensity level of 50db. What is the
intensity level in the room when an additional two men and two women generate
the same amount of noise?
[ 53db]
90. An echo is heard after 5sec at 5 , what is the distance between source and surface?
[832.5m]
91. A person fires a gun in front of building 167m away. Find the time if the speed of
sound is 334m/s?
[ 1sec]
A man stands between two high buildings and blows a whistle. He hears two
successive echoes after 0.4sec and 1.6sec. Find the distance between the buildings
(v= 332m/s)
[332m]
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Physics Grade 12
Properties of Light
Chapter Objectives
4.0 Introduction
4.1 Sources of light
4.2 Properties of light
4.3 Rectilinear propagation of light
4.4Shadow, Eclipse and pin hole Camera
4.0 INTRODUCTION
Light is the invisible energy which causes sensation of sight (vision). The properties of
light are: Rectilinear propagation
 Reflection
 Refraction
 Diffraction
 Interference
4.1 SOURCES OF LIGHT
Sun is the primary source of light for the mankind. It is a huge ball of burning gases
which emits tremendous amount of light.
Luminous substances: are substances which produce light energy by themselves.
They include such substances as sun, stars, burning candle and electric bulb.
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Physics Grade 12
Non-luminous substance: are substances which do not produce light energy by
themselves. They include such substances as the moon, and most of the objects.
Definitions pertaining to Light
1.
Luminous substances: - are substances which produce light by themselves.
They include such substances as sun, stars, burning candle and electric bulb.
2.
Non-luminous substance: - are substances which do not produce light by
themselves. They include such substances as the moon, and most of the objects.
3.
Transparent substances: - are substances which allow most of light to pass
through them. They include such substance as vacuum, clear air, glass and
diamond.
4.
Translucent substances: - are substances which partially allow light energy to
pass through them. They include such substance as oiled paper, mist, and deep
water.
5.
Opaque substances: - are substances which do not allow light energy to pass
through them. They include such substances as wood, bricks and stone.
6.
Rays of light: - is the path along which light energy travels in a given direction.
7.
Beam of light: - is a collection of a number rays.
8.
Parallel beam: - consists of parallel rays.
9.
Convergent beam: - consists of rays that meet at a point.
10. Divergent beam: - consists of rays that come from one point.
4.2 RECTILINEAR PROPAGATION OF LIGHT
Light rays travel in a straight line. Shadows, eclipses and pin-hole camera are the
evidence of this property.
SHADOWS
If an opaque body is placed in the path of light rays, a shadow is formed. There are two
kinds of shadows.
a.
Umbra is the region of total darkness formed behind an opaque body.
b.
Penumbra is a region of partial darkness formed behind an opaque body.
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Physics Grade 12
Definition
Shadow is a dark patch formed behind an opaque body.
1. POINT SOURCE OF LIGHT
If the source of light is a point, umbra is only formed
Characteristics of the shadow
a. Umbra is formed only.
b. The size of the umbra increases if the distance of the screen increases and vice
versa.
c. The size of the umbra increases if the distance of the point decreases and vice
versa.
2. EXTENDED SOURCE OF LIGHT
a) When extended source is smaller than the opaque body.
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Physics Grade 12
Characteristics of the shadow
a. Umbra and penumbra are formed.
b. Sizes of umbra and penumbra increase as the distance of the screen increases and
vice versa.
c. Sizes of umbra and penumbra increase as the distance of the source increases and
vice versa.
Real life applications
Bulbs are preferred as a source of light as compared to tube lights because they form
large umbra shadows.
b) When the extended source is bigger than the opaque body.
Characteristics of the shadow
a. Umbra and penumbra are formed.
b. The size of the umbra is smaller than the size of the penumbra.
c. If the distance of the screen is increased, umbra decreases and penumbra increases.
REAL LIFE
Birds flying high in the air do not cast their shadows on the ground.
4.3 ECLIPSES
The sun is 400 times larger than the moon, yet the moon and the sun appear to be the
same size. It is because the sun is 400 times away from the earth as compared to the
moon. They subtend the same angle on the eyes and hence appear to be the same size.
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Physics Grade 12
The earth revolves around the sun in one year (3651/4 days) whereas the moon revolves
around the earth in 29 1/2 days. The sun, the moon and the earth may happen to be on
the same straight line twice in 29 1/2 days but not in the same plane because the moon
is inclined at 50 to the orbit of the earth. There are two types of eclipses. These are:-
LUNAR ECLIPSE
Lunar eclipse is always formed on a new moon day. Linear eclipse is also caused on a
full moon provided the following conditions exist:
1. The sun, the earth and the moon should be the same straight line and same plane.
2. Earth must be in the middle.
The lunar eclipse is classified as total lunar eclipse and partial lunar eclipse.
SOLAR ECLIPSE
When the sun, the moon and the earth are in the same straight line and the same plane
and the moon is in the middle solar eclipse is formed.
PIN-HOLE CAMERA
It is a box with a very small hole. The screen is opposite to the hole. When an object is
placed in front of the hole an inverted image is formed on the screen.
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Physics Grade 12
M 
hi = Image height
distance
hi  di
ho do
ho = Object height
do = Object distance
di = Image
M= magnification
Example 1
An object 4.5m high is placed in front of a pin-hole camera. If the width of the camera
is 36cm and the size of the image is 1.8cm, find the distance of the object.
Solution
di = 36cm
M 
ho =4.5m
hi  di
ho do
do 
hi= 1.8cm
ho  di
hi
do 
do = ?
4.5m  36cm
 90m
1.8cm
PROBLEMS
1. Find the height of tree 240m a way that produces an image 2cm high in a pin-hole
camera of width 20cm.
Ans: 24m
2. The magnification of an image formed by a pin-hole camera is 0.0005. (a) If the is
at a distance of 500m find the image distance. (b) If the image size is 2.4cm, find
the height of the object.
Ans: a. 0.25m b. 48m
3. Define solar eclipse and state why solar eclipse takes small time.
4. Does everyone in the earth experience solar eclipse?
5. Define lunar eclipse.
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Physics Grade 12
Reflection of Light
LIGHT AS INVISIBLE ENERGY
When we enter a dark room, the objects present there are not visible. However, if we
light a candle or switch on an electric lamp, everything in the room becomes visible. It
shows that for vision, the presence of light is essential.
Light: It is an invisible energy which causes in us sensation of sight (vision).
Generally, the very hot objects emit light. As heat is a kind of energy which gives rise
to light, therefore, light is a kind of energy.
Why do we say that light is invisible energy? When the light from a source falls on an
object or objects, we do not see it in the space between the source of light and objects.
Similarly, when the light bounces off from the objects and enters our eye, we do not
see it. So, light is invisible energy.
Sources of Light
Sun is the primary source of light. It is huge ball of hydrogen gas, in which hydrogen
fuses by thermo-nuclear process to liberate enormous amount of heat and light
energy.
Inside the core of sun, there is a temperature of about 300 million degree celsius and a
tremendous pressure. Under such conditions atoms of hydrogen fuse (join) to form
helium with the release of vast amount of heat and light. Such a process is called
thermo-nuclear process.
Stars are very distant suns in space. For us on earth
they are weak sources of light. On surface of the
earth, a glowing electric bulb or fluorescent tube light,
a burning candle or kerosene oil lamp are sources of
light.
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Physics Grade 12
Luminous bodies: The bodies which give out light energy on their own are called
luminous bodies.
Examples of luminous bodies: The sun, the stars, the burning candle or a kerosene oil
lamp, etc., are luminous bodies. In animal world the glow worm and certain kinds of
fishes are the examples of luminous bodies.
Non-luminous bodies: The bodies which do not give light energy of their own, but
reflect the light energy falling on them are called non-luminous bodies.
Examples of non-luminous bodies: In universe or space all planets, comets, moons,
etc; are non-luminous bodies. They shine on account of the sunlight falling on them.
On the surface of earth, wood, houses, furniture, stones, etc., are non-luminous bodies.
General terms of pertaining to light:
a. Optical medium: Anything (material or non-material) through which light energy
passes wholly or partially is called optical medium. Examples: Vacuum, air, all
kinds of gases, water, glass, diamond, etc.
b. Homogeneous optical medium: An optical medium, which has a uniform
composition throughout is called homogeneous optical medium. Examples:
Vacuum, distilled water, pure alcohol, glass, clear plastics, diamond, etc.
c. Heterogenous optical medium: An optical medium, which has different
composition at different points is called heterogeneous optical medium.
Examples: Air, muddy water, fog, mist, smoke, etc.
d. Transparent medium: An optical medium, which allows most of the light energy
to pass through it is called transparent medium. Examples: Vacuum, clear air,
thin layer of water or alcohol, diamond, certain kinds of clear plastics, etc.
e. Translucent medium: An optical medium which partially allows the light energy
to pass through it is called a translucent medium, in such a medium we cannot
see clearly. Example : Butter paper, oiled paper, tissue paper, ground glass, frosted
glass, deep water, fog, mist, dust laden air are examples of translucent medium.
f. Opaque bodies: Those bodies which do not allow the light energy to pass through
them are called opaque bodies. These bodies can either absorb light energy or
reflect light energy. Examples: Bricks, stones, trees, wood, all kind of metals,
certain kinds of plastics, etc.
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Physics Grade 12
g. Ray of light: The path along which light energy travels in a given direction is
called ray of light. Or the straight line drawn in the direction of propagation of
light is called ray of light. An arrow head on the straight line gives the direction of
propagation of a ray of light.
h. Beam of light: A collection (or bundle) of adjacent rays of light is called beam of
light.
Fig. 1.2: Types of light beams
i. Parallel beam: When the rays of light travel parallel to each other in the same
direction, such a bundle of rays is called parallel beam.
Example: Sun rays entering into a room through a ventilator or .the rays coming
out of searchlight constitute a parallel beam.
j. Convergent beam:
When the rays of light coming from different directions, get
closer and meet at a point, such a bundle is called convergent beam.
Examples: The rays of sun on reflection from a concave mirror or refraction from
a convex lens constitute a convergent beam.
k. Divergent beam: When the rays of light originating from a point travel in different
directions, such a bundle is called divergent beam.
Examples: The rays coming out of a car headlamp, or a burning candle, or a
glowing electric bulb constitute a divergent beam.
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Physics Grade 12
NATURE OF LIGHT
Sir Issac Newton believed that visible light consists of extremely small particles
(corpuscles) of different sizes moving with same speed in straight lines. He explained
the phenomenon of formation of shadows, reflection and refraction on the basis of
corpuscular theory of light.
However, in last quarter of the seventeenth century it was observed that if an opaque
object of extremely small size (such as razor edge of a shaving blade) is placed in the
path of light, then the light has tendency to bend around it, rather than stopping and
forming a shadow. The effect of light is called diffraction of light. Furthermore
effects such as interference of light and polarisation of light were discovered. To
explain these effects a Dutch physicist Christian Huygen (1629-1695) suggestd that
light has a transverse wave like character. His theory was not accepted immediately.
In 1801 Thomas Young (1773-1829) experimentally proved the wave character of
visible light and was able to measure its wavelength and frequency. In 1873 Maxwell
proposed that visible light is a kind of electromagenetic wave, Hertz confirmed
Maxwell's theory experimentally in 1887.
It seemed alright for wave theory of light, when effects of light such as photoelectric
effect was discovered. Albert Einstein used corpuscular theory to explain photoelectric
effect in 1905.
The confusion about the nature of light continued for sometime till the modem
quantum theory of light was proposed. According to quantum theory, light behaves
like a particle as well as a wave, depending upon the situation it encounters. In this
theory each particle of light is called quanta, which has a specific frequency and
wavelength. However, they all travel in vacuum at the same speed, i.e.; 3xl08 ms-1.
Points to Remember
 A narrow band of electromagnetic radiation (wavelength 4 x 10 -7 m for violet
to 7x10-7 m for red) constitutes visible light.
 Particulate nature of light was put forward by Newton. It explains reflection and
refraction of light. Albert Einstein used this theory to explain the photoelectric
effect.
 Dutch physicist Christian Huygen put forth the wave theory of light which
explained the phenomenon of diffraction, interference and polarisation of light.
This theory was confirmed by Thomas young experimentally.
 Presently, the light is considered to have dual nature, i.e., depending upon its
interaction with matter, it behaves like a particle or a wave.
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Physics Grade 12
Common Characteristics of Light
1. Light is an invisible energy which on rebounding from the surface of matter cause
sensation of vision.
2. Light travels along a straight-line path, but is an electromagnetic wave which is
transverse in nature.
3. Light produces shadows, when obstructed by opaque objects.
4. Light gets reflected back from polished or smooth surfaces, such as mirror,
polished metals, still water, etc.
5. Light gets refracted when it travels from one medium to another medium.
6. The velocity of light in vacuum is 3 x 108 m/s.
7. Medium to another transparent medium.
8. During the change in velocity of light, its frequency remains same, but the
wavelength changes.
REFLECTION OF LIGHT
When a ray of light (or beam of light) travelling through a certain medium (say air)
strikes an opaque, but a smooth polished surface, it bounces off the surface into the
original medium (say air), the phenomenon is called reflection of light.
Kinds of Reflection
There are two kinds of reflection, depending upon the nature of surface on which the
light falls.
(a) Regular reflection or specular reflection: When a parallel beam of light
travelling through a certain medium on striking some smooth polished surface is
reflected back in the same medium as a parallel beam of light, the phenomenon is
called regular reflection or specular reflection.
Fig, 1.3 (c): Regular reflection
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A surface is said to be smooth, if the size of irregularities on it is smaller than the
wavelength of light incident on it.
The reflection taking place from surface of mirror, highly polished objects, still water,
etc; is regular reflection. Regular reflection helps us in seeing our images as well as in
solar cookers.
(b) Irregular reflection or diffused reflection:
When a parallel beam of light travelling through a certain medium on striking some
rough surface is reflected back in the same medium in different directions, such that it
does not constitute a parallel beam, the phenomenon is called irregular reflection or
diffused reflection.
Fig. 1.3 (d): Irregular reflection or diffused reflection
In nature generally diffused reflection takes place as majority of objects have rough
surface.
 A surface is said to be rough, if the size of irregularities on it are larger than the
wavelength of light incident on it.
 Irregular reflection cuts off the glare and is soothing to rays.
 It helps the light to spread over vast area, thereby helping general illumination.
Terms Associated with Reflection of Light
(i) Mirror: A smooth and highly polished reflecting surface is called a mirror.
Commonly used mirror (or looking glass) is made by depositing very thin layer of
silver metal on one side of plane glass sheet. The thin layer of silver is protected by
coating over it a thin layer of red lead oxide [Pb3O4]. Silver is the best reflector and is
used as a reflecting surface for all kinds of mirrors.
Kinds of mirrors: There are two kinds of mirrors.
1. Plane mirror: A highly polished plane surface is called plane mirror.
2. Curved mirror: A highly polished curved surface is called curved mirror. The
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curved mirror is called spherical mirror, if its polished surface is a part of hollow
sphere of glass. The curved mirror is called parabolic mirror, if its surface is a part of
hollow parabolic piece of glass. Fig 1.4 shows different kinds of mirrors.
Fig 1.4: Kinds of mirrors
All mirrors have two surfaces. The surface P is
called polished surface and has a deposit of silver
metal, Surface R is reflecting surface.
(ii) Incident ray: A ray of light which falls on a
mirror or any polished surface is called incident
ray. PO is the incident ray (Fig 1.5). It is always
shown with an arrow head pointing towards the
mirror surface.
Fig. 1.5: Diagram showing
general
terms used in reflection
(iii) Reflected ray: A ray of light which bounces off
the mirror surface after reflection is called reflected ray.
OR is the reflected ray (Fig. 1.5). It is always shown with an arrow head pointing away
from the mirror surface, (iv) Point of incidence: The point on the mirror surface where
an incident ray strikes is called point of incidence. 'O' is the point of incidence of the
mirror surface (Fig. 1.5).
(v) Normal: A perpendicular drawn on the mirror surface at the point of incidence is
called normal. ON is the normal (Fig. 1.5). It is always shown by a dotted line,
(vi) Angle of incidence: The angle that the incident ray makes with the normal is
called the angle of incidence. <PON is the angle of incidence (Fig. 1.5). It is denoted
as <i.
(vii) Angle of reflection: The angle that the reflected ray makes with the normal is
called angle of reflection. <RON is the angle of reflection (Fig 1.5). It is denoted as <r.

Angle that the incident ray makes with plane mirror is called reflex angle of
incidence. <POX is reflex angle of incidence and is equal to 90° - <.i

Angle that the reflected ray makes with plane mirror is called reflex angle of
reflection. <ROY is the reflex angle of reflection and is equal to 90° - <r,
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Laws of Reflection
First law of reflection:
The angle of incidence is always equal to angle of reflection at the point of incidence.
Angle of incidence = Angle of reflection, or <i = <r
Second law of reflection:
At the point of incidence, the incident ray, the reflected ray and the normal lie in the
same plane. From the above law, it implies that if incident ray is in horizontal plane,
i.e., the plane of paper, the normal and reflected rays are also in the plane of paper.
The laws of reflection are true for:
(i)
Any kind of surface, i.e., smooth polished surface or rough surface.
(ii)
Any kind of mirror, i.e., plane mirror; spherical mirror or parabolic mirror.
Fig. 1.6: Reflection at plane and spherical mirror.
What is normal incidence?
When a ray of light strikes a plane mirror at an angle 90°, i.e., its
path coincides with normal, the normal incidence takes place.
Characteristics of normal incidence:
(i)
As the incident ray coincides with normal, therefore, angle
of incidence <i is zero.
(ii) As the <i= <r, therefore, angle of reflection <r is zero.
(iii) The incident ray of light retraces its path in opposite
Fig. 1.7: Normal
incidence
direction,
(iv) The angle between the incidence and reflected ray is zero.
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Image
Seeing the details of an object or receiving them on screen with the help of reflection
or refraction is called an image. However, scientifically an image is defined as under.
When the rays of light diverging from a point, after reflection or refraction,
either actually meet at some other point or appear to meet at some other point,
then that point is called image of the object.
There can be two kinds of images i.e.; virtual image and real image.
Virtual Image: When the rays of light
diverging from a point, after reflection or
refraction appear to diverge from another
point, then the image so formed is called
virtual image.
The virtual image is always erect, laterally
inverted and cannot be taken on screen. In
Fig. 1.8, the image T of the object 'O' is a
virtual image. The rays diverging between OA
and OB, after reflection from the plane mirror, further diverge along AC and BD
respectively. However, when these diverging rays reach, the eye, then to the eye, they
appear to diverge from point I. This T is the virtual image of the object O.
Real Image: When the rays of light diverging from a point, after reflection or
refraction actually converge at some other point, then the image so formed is called
real image.
In Fig 1.9 rays diverging from the point A,
after, reflection from spherical mirror
(concave mirror) converge at point A.
This A1 is the, real image of A.
A real image is always inverted and can
be taken on screen.
What is an inverted Image and a Laterally inverted Image?
Refer to Fig. 1.10(a): During inversion the top side of the object appears as the
bottom side of the image and vice versa. In a way the image turns around its
horizontal axis through an angle of 180°.
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Refer to Fig. 1.10(b): During lateral
inversion the left side of the object
appears as right side of the image and
vice versa. In a way the image turns
through an angle of 180° about vertical
axis.
The phenomenon due to which the
image of an object turns through an
angle of 180° through vertical axis
rather than horizontal axis, such that
the right side of the image appears as
left or vice versa is called lateral
Differences between Virtual and Real Image
Virtual Image
Real Image
1. The rays of light after reflection or  The rays of light after reflection or
refraction appear to meet at some
refraction actually meet at some other
other point or appear to diverge from
point.
some other point.
2. It cannot be taken on a screen.
 It can always be taken on a screen.
3. It is always erect, but laterally  It is always inverted.
inverted.
Characteristics of an Image Formed in a Plane Mirror
1. Image is virtual, i.e., it cannot be taken on screen.
2. Image is erect.
3. Image is of same size as the size of object. Thus, magnification of image is
4. Here, it is important to remember that magnification is the ratio of the size of an
image to the size of an object.
5. Image is laterally inverted, i.e, the left side of object appears as right side of the
image.
6. Image is formed as far behind the plane mirror as the object is in front of it.
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EXERCISE
(I) MULTIPLE CHOICE QUESTIONS
1. We can see in a room which is not directly illuminated by sunlight due to:
(A) regular reflection
(B) refraction
(C) irregular reflection
(D) none of the above.
2. When the rays of light diverging from a point, after reflection either actually meet
or appear to meet at some other point, then that point is called :
(A) shadow
(B) reference point
(C) image
(D)- none of the above.
3. No matter how far you stand from a mirror, your image appears erect. The mirror
may be:
(A) plane
(B) concave
(C) convex
(D) both (A) and (C).
4. When you stand in front of mirror, your image is always erect, of the same size and
laterally inverted. The mirror may be:
(A) plane
(B) convex
(C) concave
5. The type of mirror used in a dressing table is:
(A) convex mirror
(C) plane mirror
(D) both (A) and (B).
(B) concave mirror
(D) cylinderical mirror.
II) VERY SHORT ANSWER TYPE QUESTIONS
1. What do you understand by the term light? Name the prime source of light for the
people living on the earth.
2. Define luminous body. Give four examples of luminous bodies.
3. What are non-luminous bodies? Give four examples of non-luminous bodies.
4. Why is moon a non-luminous body?
(III) SHORT ANSWER TYPE QUESTIONS
1. What do you understand by the term optical medium? Is optical medium is always
a material body? Explain your answer.
2. What do you understand by the term (i) transparent medium (ii) translucent
medium (iii) opaque body? Give two examples each in support of your answer.
3. Classify the following as transparent medium, translucent medium and opaque
bodies. (i) fog, (ii) wooden chair (iii) diamond (iv) carbon dioxide gas (v) oiled
paper, (vi) moon (vii) iron (viii) vacuum (ix) muddy water
4. Name and define the kind of light beam formed by :
a. sunlight (ii) sunlight passing through convex lens, (iii) sunlight falling on
convex mirror
5. Give any two uses of mirror other than looking glass.
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(IV) SHORT ANSWER TYPE QUESTIONS
1. Who put forth the corpuscular theory of light? Name three optical phenomena
which are explained by this theory.
2. Who put forth the wave theory of light? Name three optical phenomena which are
explained by this theory.
3. Name and briefly explain modern theory of light.
(V) LONG ANSWER TYPE QUESTIONS
1. What is the speed of light in vacuum? How does the (i) frequency, (ii) wavelength
of light changes while travelling from one medium to another.
2. Define reflection of light. State its kinds and show by diagrams.
3. Define mirror and state its kinds.
4. State the laws of reflection of light.
5. (i) What do you understand by the terms (a) image (b) real image and (c) virtual
image, (ii) Give three differences between real and virtual image?
6. Distinguish between the inverted image and laterally inverted image.
7. State five characteristics of image formed in a plane mirror.
SPHERICAL MIRRORS AND THEIR TECHNOLOGY
We already know that, if the reflecting surface of mirror is a part of hollow sphere, it is
called spherical mirror. The spherical mirrors are further classified as (i) concave
mirror and (ii) convex mirror.
Before we discuss the kinds of spherical mirrors, let us perform the following activity.
Activity 1
Take a steel tablespoon and polish it by rubbing a fine cotton cloth over it, for some
time. Hold the straight side of spoon close to your face and then far away from your
face. What do you observe?
When the spoon is close to your face, the
image appears bigger and distorted. When
the spoon is far away from your face the
image appears inverted. The size of image
may be smaller or bigger, depending upon
the distance of spoon from your face.
Now turn the reverse side of spoon towards
your face. What do you observe?
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It is seen that image is upright, but smaller in size. Almost similar kind of images are
seen when you see your face in curved polished objects, such as cycle bell, polished
steel or a silverware, bumpers of car, etc.
In the above activity, the curved surface of the shining spoon could be considered as
curved surface of the mirror. The reflecting surface of such a mirror can be considered
a part of surface a hollow sphere. Such a mirror, whose reflecting surface which is
spherical is called a spherical mirror.
Before we move further on spherical mirrors, let us try to understand common terms
associated with them.
General terms associated with spherical
mirrors
(a) Spherical mirror: A mirror which is made
from a part of a hollow sphere is called a
spherical mirror. The back of a spherical mirror
is always shown by shaded lines.
(b)
Concave mirror: A mirror which is
polished from the outer side of a hollow sphere,
such
that
the reflecting side is towards its
hollow side, is called concave mirror.
Fig. 1.14(a): Converging action of concave
mirror
(c) Convex mirror: A mirror which is polished on the hollow side of the sphere, such
that the reflecting surface is towards its bulging side, is called convex mirror.
(d) Pole: The midpoint of a spherical mirror is called its pole. It is denoted by the
letter P.
(e) Centre of curvature: The centre of the hollow sphere, of which the spherical
mirror is a part, is called centre of curvature. It is denoted by the letter C.
(f) Principal axis: An imaginary line passing through the pole and the centre of
curvature of a spherical mirror is called principal axis.
(g) Linear aperture: The diameter of a spherical mirror is called its linear aperture.
It is shown as A'A or BB' in Fig. 1.14(b) and 1.14 (a) respectively.
BB' is a polished surface of hollow hemisphere xy, such that surface towards left hand
side is reflecting surface. Two parallel rays of light, parallel to the principal axis strike
the mirror at BB', such that ZDBC and ZD'B'C are the angles of incidence
respectively.
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Obeying laws of reflection, BF and BT are the reflected rays, which meet at F.
Similarly, all other parallel rays meet at F. i.e; the rays converge at one point. This
point of convergence of rays is called principal focus and the concave mirror is called
converging mirror.
Fig. 1.14(b): Convex mirror as diverging mirror.
AA' is the polished surface of the hollow hemisphere XY from the hollow side, such
that reflecting surface is towards left hand side. Two parallel rays, parallel to the
principal axis strike the mirror at AA' respectively, such that <DAN and <D'A'N' are
the angles of incidence respectively.
Obeying the laws of reflection AE and A'E’ are reflected rays, which diverge outward
and hence cannot meet in front of the mirror. However, when these diverging rays,
reach the eye, then to the eye, they appear to meet at point F. This point is principal
focus of convex mirror. As convex mirror diverges rays, therefore, it is called
diverging mirror.
(h) Principal focus: It is a point on the principal axis, where a parallel beam of light
after reflection, either actually meets or appears to meet. It is denoted by the letter F.
In case of a concave mirror, the principal focus is in front of it. In case of convex
mirror, principal focus is behind it.
(i) Focal length: The linear distance between the pole and the principal focus is called
the focal length. It is denoted by the letter.
(j) Radius of curvature: The linear distance between the pole and the centre of
curvature is called the radius of curvature. It is denoted by the letter R.
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It is useful to note that focal length of a spherical mirror is half of the radius of
curvature.
(k) Focal Plane: An imaginary plane passing through principal focus, such that it is
at right angles to principal axis is called focal plane. When a parallel beam of light,
not parallel to the principal axis strikes a concave mirror or a convex mirror, it
converges or appears on converge at the focal plane, rather than on the principal focus.
Representation of Images Formed in Concave Mirror by Ray Diagrams
Before, we proceed to draw images by ray diagrams in concave mirror, let us learn the
rules for drawing images.
Rule 1:
Any ray of light, travelling parallel to principal axis
of a concave mirror, after reflection passes through
the principal focus.
Rule 2:
Any ray of light, which first passes through the
principal focus of the concave mirror, after
reflection travels parallel to the principal axis of
the concave mirror.
The figure shows incident ray AB, passing through
the principal focus of a concave mirror, such that after reflection, it moves as reflected
ray BD, which is parallel to the principal axis of the concave mirror.
Rule 3:
Any ray of light which first passes
through the centre of curvature of a
concave mirror, after reflection will
retrace its path. It is because, the ray
strikes the concave mirror surface at
right angles-Fig. 1.20 shows incident
Fig. 1.20
ray AB, passing through the centre of curvature of the concave mirror C, such that
after reflection at B, it retraces its path as reflected ray BA. It is because; the ray AB
strikes the surface of mirror normally.
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Rule 4:
A ray incident obliquely to the
principal axis towards the pole P of
the
concave
obliquely,
mirror
such
that
is
reflected
angle
of
incidence at P is equal to the angle of
reflection at P. Figure shows an
Fig. 1.21
oblique incident ray EP, reflected as oblique reflected ray PF, such that <i = <r
How to Draw Ray Diagram for Locating an Image in a Concave Mirror?
Although countless rays start from a given point of an illuminated object, it is possible
to locate the position of an image by taking two standard rays as explained below:
1. Place the object in front of the spherical mirror, such that its base is in the line with
the principal axis. From the tip of the object, draw a straight line parallel to the
principal axis, such that it meets the spherical mirror. From this point, pass the
ray through principal focus of the concave mirror.
2. From the tip of the object, draw another ray, such that after reflection, it either
actually passes through the centre of curvature or appears to pass through the
centre of curvature.
3. The point where the above mentioned reflected rays meet, or appears to meet,
gives the position of the tip of the image. From this point, draw a perpendicular on
principal axis to obtain the position and size of image.
Formation of Image in a Concave Mirror for Different Positions of the Object
(a) When the object is at infinity and the rays coming from it are parallel to the
principal axis
The rays coming from an object at infinity are always parallel. For example, rays
coming from the sun or the stars are always parallel. If these parallel rays are also
parallel to the principal axis of the concave mirror, then after reflection, they will pass
through the principal focus. Thus, as all the
rays actually meet at principal focus, therefore
an image is formed, which has following
characteristics:
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Characteristics of the image:
1. The image is real.
2. The image is inverted.
3. The image is diminished to a point.
4. The image is formed at the principal focus, in front of the concave mirror.
(b) When the object is at infinity, but the rays coming from it are not parallel to
the principal axis
When the parallel rays are coming from an object at
infinity, such that they are not parallel to the
principal axis of concave mirror, they converge at
one point in the plane of principal focus as
illustrated in Fig. 1.23. The point of convergence
can be easily located by passing one ray through the
centre of curvature, such that it intersects the vertical line passing through the focal
plane. In the Fig. 1.23, point A is the point of convergence.
Characteristics of the image:
1. The image is real.
2. The image is inverted.
3. The image is highly diminished.
4. The image is formed in the plane of the principal focus, in front of the concave
mirror.
(c) When the object is beyond the centre of curvature, but is not at infinity
AB is an object situated beyond the
centre of curvature of the concave
mirror. A ray of light starting from
point A, moving parallel to the
principal axis of the concave mirror
along AG, after reflection passes
through the point F and moves along
GD.
Fig. 1.24
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Another ray starting from the point A along AC, strikes the mirror at point E retraces
its path. Thus, the divergent beams starting from point A, on striking the mirror at
points G and E, after reflection, converge at point A1, thereby forming an image A1B1.
Characteristics of the image:
1. The image is real.
2. The image is inverted.
3. The image is between the C and F, in front of the concave mirror.
4. The image is diminished.
(d) When the object is at the centre of curvature of the concave mirror
AB is an object situated at the centre of
curvature of the concave mirror. A ray
of light starting from point A and
moving along AQ after reflection from
the mirror passes through its principal
focus F. Another ray starting from point
Fig. 1.25
A along AF, after reflection from the mirror travels parallel to the principal axis of the
mirror. Thus, a divergent beam starting from point A along AG and AD, after
reflection converges at the point A1 Thus, A1 is the real image of the point A. Similarly,
the rays starting between the points A and B meet between the points A1 and B1.
Characteristics of the image:
1. The image is real.
2. The image is inverted.
3. The image is of the same size as the object.
4. The image is formed at the centre of curvature, in front of the concave mirror.
(e) When the object is in between the centre of curvature and the principal focus
AB is an object situated in between the
centre of curvature and principal focus
of a concave mirror. A ray of light
which starts from point A along AD
parallel to the principal axis, after
reflection passes through the principal
Fig. 1.26
focus F of the mirror. Another ray starting from point A and travelling along AE,
strikes the mirror normally and hence, retraces its path along EC, Thus, a divergent
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beam starting from point A on striking the mirror at D and E, converges at point A,.
Thus, A1 is the image of the point A. Similarly, the rays starting from the points A and
B converge between the points A1 and B1, so as to form a real image.
Characteristics of the image:
1. The image is real.
2. The image is inverted.
3. The image is enlarged (magnified).
4. The image is formed beyond the centre of curvature in front of the concave mirror.
(f) When the object is at the principal focus of a concave mirror
AB is an object situated at the principal
focus of a concave mirror. A ray of
light, which starts from point A and is
moving along AD, parallel to the
principal axis, after reflection passes
through the point F along DG. Another
ray starting from the point A and
Fig. 1.27
moving along AE strikes the mirror at right angles and hence, retraces its path and
passes through the centre of curvature C. Thus, a divergent beam starting from point A
along AE and AD forms a parallel beam after reflection. This parallel beam will meet
only at infinity to form a real image.
Characteristics of the image:
1. The image is real.
2. The image is inverted.
3. The image is highly enlarged (magnified).
4. The image is formed at infinity, in front of the concave mirror.
g) When the object is in between the principal focus and the pole of a concave mirror
AB is an object situated in between the pole
and the principal focus of a concave mirror. A
ray of light starting from point A along AD
parallel to the principal axis, after reflection
passes through the principal focus F. Another
Fig. 1.28
ray of light starting from A and moving along AE, after reflection retraces its path and
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passes through the centre of curvature C. Thus, a divergent beam starting from point
A on striking the mirror gets reflected as a divergent beam. This beam does not meet
in front of the mirror. However, when this divergent beam enters the eye, then the eye
retraces the path of the rays along straight lines EA1 and DA1 and hence, it appears to
come from point Ar Thus, A1B1 is the image of the object.
Characteristics of the image:
1.
2.
3.
4.
The image is virtual.
The image is erect.
The image is magnified.
The image is formed behind the concave mirror.
Uses of Concave Mirrors
1. It is used as a shaving mirror. The
reason being that when the face of a
person is between pole and principal
focus of concave mirror, an erect;
enlarged and virtual image is formed
behind the mirror.
2. It is used as a reflector in the head lights of automobiles, such as cars; trucks;
motor bikes; etc. Even the reflector of flash light is concave in nature. The bulb is
placed close to the principal focus of the concave reflector when a powerful near
parallel beam of light is formed.
3. It is used by doctors to focus a parallel beam of light on a small area, especially for
the examination of throat; ear; nose; teeth; etc. Dentists use a special concave
mirror to see the back of the tooth by placing it behind the tooth, when a virtual,
erect and enlarged.
4. Concave mirror is used as a reflector in dish type solar cookers and solar furnace.
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Formation of an image in a convex mirror
You have already learnt that image formed in a spoon held from reverse side is
diminished and erect.
Rules for Drawing Images for Convex Mirror by Ray Diagrams
Rule 1:
Any ray of light travelling parallel to the
principal axis of a convex mirror, after
reflection appears to diverge from the
principal focus of the convex mirror.
Fig. 1.30 shows an incident ray AB,
parallel to the principal axis of the
Fig. 1.30
convex mirror, such that after reflection at point B, it moves as the reflected ray BD.
However, if ray BD is produced backward as shown by dotted line, the reflected ray
appears to diverge from point F.
Rule 2:
Any ray of light which travels along the
direction of principal focus of a convex
mirror, after reflection, will travel parallel
to the principal axis of the mirror.
Fig. 1.31 shows an incident ray AB,
travelling in the direction of principal
Fig. 1.31
focus of the convex mirror as shown by the dotted line BF. This ray after reflection
moves as reflected ray BD, which is parallel to the principal axis of the convex.
Rule 3:
Any ray of light which travels along the
centre of curvature of a convex mirror,
after reflection retraces its path. It is
because; it strikes the convex mirror
surface at right angles.
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Fig. 1.32 shows an incident ray AB, which is
moving in the direction of centre of curvature
of a convex mirror as shown by the dotted line
BC. This ray after reflection at point B,
retraces its path and hence moves as the reflected ray BA.
Rule 4:
A ray of light incident obliquely on the principal axis, towards the pole P of the convex
mirror is reflected obliquely, such that the angle of incidence at P is equal to the angle
of reflection.
Formation of Image in a Convex Mirror for Different Positions of the Object
We shall consider only two positions of the object for studying the image formed by a
convex mirror,
(i) When the object is at infinity.
(ii) When the object is at a finite distance from the pole of the convex mirror. Other
positions are not possible because the focus and the centre of curvature are behind the
reflecting surface of the convex mirror.
(a) Formation of image when the object
is at infinity
The rays coming from an object situated
at infinity are always parallel. When these
parallel rays, coming parallel to the
principal axis strike the convex mirror,
they get reflected and diverge outward. However, when the diverging reflected beam
enters the eye, these rays appear to come from point F, i.e., the principal focus of the
convex mirror.
Characteristics of the image:
1. The image is virtual.
2. The image is erect.
3. The image is diminished to a point.
4. The image is formed at principal focus behind the convex mirror.
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(b) Formation of image, when the object is at a finite distance from the pole
AB is an object situated anywhere between
the pole and infinity in front of a convex
mirror. A ray of light starting from point A
and moving along AG, parallel to the
principal axis after reflection diverges along
GD, but if produced backward, it appears to
diverge from point F. Another ray starting from point A and moving towards the
centre of curvature E of the mirror, retraces its path. Thus, a divergent beam starting
from point A and going along AG and AE, diverges as GD and EA. When this
diverging reflected beam enters the eye, it appears to come from point A1. Similarly,
the divergent rays starting in between points A and B of the object will appear to
diverge from points A1B1. Thus, A1B1 is the virtual image of the object.
Characteristics of the image:
1. The image is virtual.
2. The image is erect.
3. The image is diminished.
4. The image is always formed between pole
and principal focus, behind the convex mirror.
(c) Effect on the size of image, when the object is brought closer to the pole of
convex mirror
Initially, the object in the position A'B' forms an erect and virtual image A1B1 (Fig.
1.36). When the object is moved towards the pole of the convex mirror in position
A"B", it forms an image A2B2. The size of image A2B2 is larger than image A1B1, and
it is formed closer to the pole of the convex mirror. However, image is always smaller
than the object.
Table 2 below shows the nature, position and relative size of image formed by a
convex mirror.
Table 2 — Nature, Position and Relative Size of Image formed by Convex Mirror
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Uses of Convex Mirrors
1. It is used as a rear view mirror in automobiles, because it can cover a very wide
field behind the driver and hence enables to see the traffic behind him without
turning his head. (Fig. 1.37)
A plane mirror is not useful as a rear view
mirror, because its field of view is very small.
2. It is used as a reflector for street lights,
because, it diverges rays over a large area,
and hence, illuminates that area. (Fig.
1.38)
Fig. 1.37: Field of view of convex
mirror acting as a rear view mirror
Fig. 1.38: Convex mirror as a
reflector for street lights
EXERCISE
I) MULTIPLE CHOICE QUESTIONS
1. The image of an object in a spherical mirror appears magnified, erect and behind it.
The spherical mirror is:
(A) Convex
(B) concave
(C) plane
(D) none of the above.
2. A real, inverted and larger than object, image is formed by a concave mirror. The
object is :
(A) between infinity and C
(B) between P and F (C) at F (D) between C and F.
3. Which of the following statement is not correct for irregular reflection?
(A) it cuts off glare
(B)
it takes place from most of the surfaces
(C) it helps in general illumination (D) it helps in the formation of virtual images
4. Which of the following can produce a virtual image?
(A) convex mirror (B) concave mirror
(C)
plane mirror
(D) all the above.
5. A full length image of a distant tall building can definitely be seen by using:
(A) a concave mirror (B) a convex mirror (C) a plane mirror
(D) both concave as well as plane mirror
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Physics Grade 12
(II) VERY SHORT ANSWER TYPE QUESTIONS
1. Define the following terms:
(A) spherical mirror (B) concave mirror
(C) convex mirror
2. Define the following terms in relation to a
concave mirror.
(i) pole (ii) centre of curvature
(iii) principal axis
(iv) principal focus
(v) focal length (vi) radius of curvature.
(III) SHORT ANSWER TYPE QUESTIONS
1. Define the term 'principal focus' in case of a convex mirror. Draw a ray diagram for
a convex mirror and show clearly on it: (i) the principal focus, (ii) the focal length,
(iii) the radius of curvature, (iv) the centre of curvature.
2. State the relationship between the focal length and the radius of curvature of a
convex mirror.
3. (i) What do you understand by the term 'real image'?
(ii) What type of mirror is used to obtain a 'real image'?
(iii) Does the mirror named by you form a real image for all locations of the
object? Give a reason for your answer, (iv) Is the real image always inverted?
4.
(i) What do you understand by the term 'virtual image'?
(ii) Name two types of mirrors which always form a virtual image,
(iii) Is the virtual image always erect?
5. Prove that for a concave mirror, the radius of curvature is twice its focal length.
6. What is the minimum number of rays required for locating the image formed by a
concave mirror for an object? Draw a ray diagram to show the formation of image by a
concave mirror.
7. Draw a ray diagram to show the (i) position and (ii) nature of the image formed
when an object is placed between focus F and pole P of a concave mirror.
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Physics Grade 12
(IV) SHORT ANSWER TYPE QUESTIONS
1. Why is a convex mirror used as a rear view mirror in the automobiles? Illustrate
your answer.
2. You are provided a convex mirror, a concave mirror and a plane mirror. How will
you distinguish between them, without touching them or using any other
apparatus?
3. Identify the device used as a spherical mirror or lens in the following cases, when
the image formed is virtual and erect in each case.
(i)
Object is placed between the device and its focus, image formed is
enlarged and behind it.
(ii) Object is placed between the focus and device, image formed is enlarged
and on the same side as that of the object.
(iii) Object is placed between infinity and device, image formed is diminished
and between focus and optical centre on the same side as the object.
(iv) Object is placed between infinity and device, image formed is diminished
and between pole and focus.
4. Draw ray diagram to show the formation of images when an object is placed in
front of a concave mirror (i) between the pole and focus point (ii) between the
centre of curvature and focus point.
5. An object is placed at a distance of 24cm from the optical centre of a convex lens
of focal length 18cm. Find the distance of the image from the lens. What will be
the height of the image if the object is 3cm tall?
6. How far should an object be placed from a convex lens of focal length 20cm to
obtain its image at a distance of 30cm from the lens? What will be the height of the
image if the object is 6cm tall?
LONG ANSWER TYPE QUESTIONS
1. Describe a simple method for the estimation of focal length of a concave mirror.
2. An object placed in front of a concave mirror forms a diminished image which is
not reduced to a point. (i) What is the position of the object?
(ii) What is the
position of image?
3. Draw a ray diagram in support of your answer.
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Physics Grade 12
4. Draw two neat ray diagrams for a concave mirror when an object is (i) at C (ii)
between C and F, (iii) at F and (iv) between F and P. In each case state the
characteristics of image.
5. Draw ray diagrams showing the image formation by a concave mirror when an
object is placed.
6. (i) between pole and focus of the ml ~or (ii) between focus and centre of curvature
of the mirror (iii) at centre of curvature of the mirror (iv) a little beyond centre of
curvature of the mirror (v) at infinity.
7. Draw ray diagrams showing the image formation by a convex mirror when an
object is placed : (i) at infinity (ii) at a finite distance from the mirror
SIGN CONVENTIONS FOR REFLECTION BY SPHERICAL MIRRORS
While dealing with reflection of light by spherical mirrors, we shall follow a set of
sign conventions. These conventions are called new cartesian sign conventions. In
these conventions:
a. The pole (P) of the spherical mirror is taken as origin.
b. The principal axis of the spherical mirror is taken as X-axis (X X') of the
coordinate system. The conventions are as
follows:
(i) The object is always kept on the left side of
the mirror. Thus, the light from the object is
incident on the spherical mirror from left to right
direction.
(ii) All distances parallel to the principal axis
are measured from the pole of the spherical mirror.
In other words, the pole of the mirror is origin. The zero end of measuring scale must
coincide with pole.
(iii) All distances measured to the right of origin (along +x axis) are taken as positive,
while those measured to the left of origin (along -x axis) are taken as negative.
From this, it implies that distances measured in the direction of incident light (i.e.,
behind the mirror) are taken as positive and the distances measured against the
direction of the incident light (i.e., in front of mirror) are taken as negative.
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Physics Grade 12
(iv) Distances (height) measured perpendicular and above the principal axis, (along
+ y axis) are taken as positive.
From this, it implies that height measured upward and perpendicular to principal axis
is taken as positive.
(v) Distances (height) measured perpendicular and below the principal axis, (along y axis) are taken as negative.
From this, it implies that height measured downward and perpendicular to the
principal axis, is taken as negative.
For the sake of simplicity remember following points:
1. All horizontal distances (along x-axis) are negative, if measured in front of mirror.
2. All horizontal distances (along x-axis) are positive, if measured behind the mirror.
3. All distances (along y-axis) are positive, if measured above the principal axis.
4. AH distances (along y-axis) are negative, if measured below the principal axis.
Important Points for Measuring Distances on New Cartesian Conventions
(a) For concave mirror, when the image is real
(i)
The distance of the object from the pole
of the mirror (U) is always negative.
(ii) The distance of the image from the pole
of the mirror (v) is always negative.
(iii) The focal length (f) and radius of
curvature (R) are always negative.
Fig 1.40 : Distances in case of real
image for concave mirror.
b. For concave mirror, when the image is virtual
(i)
The distance of the object from the
pole of the mirror (u) is always
negative.
(ii) The distance of the image from the
pole of the mirror (v) is always
positive.
(iii) The focal length (f) and radius of curvature (R) are always negative.
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Physics Grade 12
(c) For convex mirror
(i)
The distance of the object from the pole of
the mirror (u) is always negative.
(ii) The distance of the image from the pole of
the mirror (v) is always positive.
(iii) The focal length (f) and radius of curvature
(R) are always positive.
Mirror Formula
The relation between the distance of the object from the pole of the spherical mirror
(U), the distance of the image from the pole of the spherical mirror (v) and its focal
length (f) is given by the mathematical formula:
Furthermore, it must be remembered that focal length (f) of a spherical mirror is half
the radius of curvature (R).
Numerical Problem 1
An object 1cm high produces a real image 1.5cm high, when placed at a distance of 15
cm from a concave mirror. Calculate: (i) the position of the image, (ii) focal length of
the concave mirror.
Solution
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Physics Grade 12
Thus, the image is formed 22.5cm in front of the concave mirror.
To calculate the focal length
Thus, the focal length of the concave mirror is - 9cm.
Numerical Problem 2
An object 2cm high when placed in front of a converging mirror produces a virtual
image 3 cm high. If the object is placed at a distance of 8cm from the pole of the
mirror, calculate: (i) the position of the image, (ii) the focal length of the converging
mirror.
Solution
(i) To calculate the position.
Thus, image is formed 12cm behind the mirror.
(ii) To calculate the focal length.
Thus, the focal length of the converging mirror is - 24cm.
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Physics Grade 12
Numerical Problem 3
An object placed in front of a diverging mirror at a distance of 30cm, forms a virtual
and erect image which is 1/5 of the size of the object. Calculate: (i) the position of the
image, (ii) the focal length of the diverging mirror.
Solution
(i) To calculate the position.
Thus, image is formed at a distance of 6cm behind the mirror.
To calculate the focal length
Thus, the focal length of the diverging mirror is 7.5cm.
Practice Problem 1
1. An object 3cm high produces a real image 4.5cm high, when placed at a distance of
20 cm from a concave mirror. Calculate: (i) the position of image (ii) focal length of
the concave mirror.
[(i) v = -30 cm, (ii) f = -12 cm]
Practice Problems 2
1. An object 1.5cm high when placed in front of a concave mirror produces a virtual
image 3cm high. If the object is placed at a distance of 6cm from the pole of the
mirror, calculate: (i) the position of the image; (ii) the focal length of the mirror.
[(i) v = 12cm (ii) f = - 12 cm]
2. A converging mirror, forms a three times magnified virtual image when an object is
placed at a distance of 8 cm from it. Calculate :(i) the position of the image (ii) the
focal length of the mirror.
[(i) v = 24cm, (U) f = - 12 cm]
Practice Problems: 3
1. 1. An object 5cm high forms a. virtual image of 1.25cm high, when placed in front
of a convex mirror at a distance of 24cm. Calculate: (i) the position of the image,
(ii) the focal length of the convex mirror.
[(i) v = 6cm, (ii) f = 8 cm]
2. 2. An object forms a virtual image which is l/8th of the size of the object. If the
object is placed at a distance of 40cm from the convex mirror, calculate: (i) the
position of the image, (ii) the focal length of the convex mirror.
[(i) v = 5cm,
(U) f = 5.71 cm]
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Physics Grade 12
Refraction of Light
REFRACTION OF LIGHT
You already know that any material through
which light passes is called medium. A medium
which allows the light to pass through it almost
completely is a called transparent medium.
Furthermore, you know that light travels in a
straight line in a given transparent medium.
Now, what happens when the light travels from
one kind of transparent medium to another kin d
of transparent medium?
Let us perform the following activities:
Activity 6
Fig. 1.43: A pencil appears bent and short
 Take a pencil and immerse it partially in water contained in a beaker, such that the
pencil is held obliquely.
 Observe the part of the pencil immersed in water.
 You will observe that the part of the pencil immersed in water appears displaced.
This suggests that light reaching the eye from the immersed part of the pencil in
water is in different direction, as compared to light coming from the part of the
pencil in the air.
In simple words, we can say that light bends from its path while travelling from air to
water or vice-versa.
Activity
 Take an empty china mug and place a coin at its bottom.
 Move backward slowly, such that the coin is not visible and the line of sight is
obstructed by the edge of the mug.

Now, ask your friend to slowly pour water in the mug without disturbing the coin, till the
mug is filled.
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Physics Grade 12
You
will
observe
that
coin
becomes visible. This suggests
that rays coming from the coin and
travelling through water, on
emerging out in air changed their
path in such a way that they are in
your line of sight. Thus, you could
see the coin. Now the coin, which
was at position A, appears at
position B. In other words, B is the
Fig. 1.44: Coin initially invisible, becomes visible on
pouring water in cup
position of image of the coin A, and hence, to you coin appears raised. Thus, to
conclude, we can say that light bends from its path while travelling from water to air.
The phenomenon responsible for activities 6 and 7 is called refraction of light. In this
phenomenon, the rays travelling from one optical medium to another optical medium
change their path at the surface of separation of the two given media.
What is the Cause of Refraction of Light?
Let us consider this analogy (Fig. 1.45).
Imagine a column of soldiers marching in
step, four in a row, on the beat of a drum.
In this situation the soldiers are marching
at a constant speed. They are taking
certain fixed number of steps in one
minute, which can be likened to
frequency. In each step they are covering
certain distance, which can be likened to
wavelength.
Fig. 1.45 : Soldiers marching in steps change
their
path while moving from hard ground to sand
Thus, Speed of soldiers per minute = No.
with the change in direction.
of steps per minute x distance covered by
each step.
Now, imagine the soldiers are marching on a hard ground, such that they enter a patch
of sand at some angle. In such a situation the soldier No 1 will slow down first, but not
the soldier number 2, 3, and 4 (Fig. 1.45).
In the next step, the soldier No. 1 and 2 slow down, but not soldier number 3 and 4.
This results in asymmetry, i.e.; the soldiers 1 and 2 appear to deviate from their path
compared to soldier number 3 and 4. This continues till all the soldiers enter patch of
sand. At this moment all will move in symmetry, but with a changed path.
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Physics Grade 12
We already know that light has a dual nature, i.e., it acts as particle as well as wave. If
we consider light energy striking another optical medium obliquely, the light photons
are likely to slow down one after another. This in turn changes their path. During the
change in path the frequency of light remains the same. However, as the speed of light
decreases, therefore, its wavelength decreases, much the same way as the soldiers
cover less distance in each step, while walking through sandy patch.
Fig. 1.46: Soldiers marching in step do not change their path, if they enter another
surface at right angles. However, if we consider the soldiers entering the sandy patch
at right angles to the surface of separation, all of them will slow down at the same
time. Their speed and distance covered in each step will decrease, but they will not
deviate from their path.
In same way, if the light energy strikes another optical medium at right angles to the
surface of separation, it will slow down, its wavelength will decrease, but it will not
change its path. So, when light energy strikes at right angles to the surface of
separation; no refraction take place.
 Refraction can take place only, if a beam of light enters from one optical medium
to another optical medium at same angle.
 No refraction of light takes place, if a beam of light strikes another optical medium
at right angles.
 During refraction of light, the frequency remains constant, but speed and
wavelength of light change.
Refraction of Light through a Rectangular Glass Slab
The phenomenon due to which a ray of light deviates from its path, at the surface of
separation of two media, when the ray of light is travelling from one optical medium to
another optical medium, is called refraction of light.
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Physics Grade 12
Terms Associated with Refraction of Light
1. Incident ray: A ray of light travelling towards another optical medium is called
incident ray.
2. Point of incidence: The point where an incident ray strikes another optical
medium is called point of incidence.
3. Normal: A perpendicular drawn at the point of incidence is called normal.
4. Angle of incidence: The angle made by the incident ray with the normal at the
point of incidence is called angle of incidence.
5. Refracted ray: A ray of light which deviates from its path on entering another
optical medium is called refracted ray.
6. Angle of refraction: The angle that the refracted ray makes with the normal is
called angle of refraction.
7. Emergent ray: A ray of light which emerges out from another optical medium,
into the original optical medium, is called emergent ray.
8. Angle of emergence: The angle that the emergent ray makes with the normal is
called angle of emergence.
Refraction through Glass Slab
Consider the incident ray PO travelling in air (rarer medium). Had there been no glass
slab (denser medium), the ray would have continued along its straight line path along
OT. However, because the presence of glass slab (denser medium), the ray deviates
along 00', i.e., it bends toward normal, such that ^N'OO' is the angle of refraction,
which is smaller than the angle of incidence ZPON. From the above observation we
can conclude:
Any ray of light travelling obliquely in an optically rarer medium, on passing
through optically denser medium, bends towards the normal, drawn at the point of
incidence.
Furthermore, the angle of incidence in optically rarer medium is always greater
than the angle of refraction in optically denser medium.
Consider the ray OO' as incident ray in glass (denser medium) (Fig. 1.48). When this
ray emerges out in air (rarer medium), at point 0', it travels along O'S, Le., it bends
away from the normal MM', such that ^VI'O'S is larger than the angle of incidence
^OO'M. From the above observation, we can conclude:
Any ray of light travelling obliquely in an optically denser medium, on passing
through optically rarer medium bends away from the normal drawn at the point of
incidence. Furthermore, the angle of incidence in an optically denser medium is
always smaller than the angle of refraction in an optically rarer medium.
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Physics Grade 12
Now imagine a ray of light travelling
along NN' or MM', such that it meets
the surface of separation of two media
at an angle of 90° (Fig. 1.48) In such a
situation the angle of incidence is zero,
because, the angle of incidence is the
angle between the incident ray and the
normal. Thus, the angle of refraction
will be also zero. In other words, the
ray will not deviate from its path, i.e., it
does not suffer refraction. From the
above observation, we can conclude:
When a ray of light strikes the surface
of separation of two optically different
Fig. 1.48
media normally, then it does not suffer
any refraction, i.e., the ray continues in a straight line path. Furthermore, the angle
of incidence as well as angle of refraction is zero.
Lateral Displacement
The incident ray PO, which should have continued along straight line path OT, gets
displaced along O'S due to the refraction produced by the glass slab. This lateral shift
of the incident ray on emerging out of the glass slab is represented by perpendicular
distance O'L, which is called lateral displacement.
The perpendicular shift in the path of incident ray, while emerging out of a
rectangular optical slab is called lateral displacement.
It has been found that lateral displacement:
(i) Increases with the increase in the angle of incidence, (ii) Increases with the increase
in the thickness of the optical slab. (iii) Increases with the increase in the optical
density of the slab.
Laws of Refraction
Following are the laws of refraction:
(i)
The incident ray, the refracted ray and the normal to the surface of the
separation of two media at the point of incidence, all lie in the same plane.
(ii) The ratio of the sine of angle of incidence to the sine of angle of refraction is a
constant, for the light of given colour, for the given pair of media.
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Physics Grade 12
This law is also known as Snell's law. It was stated by Willibrod Snell in 1621.
Verification of Laws of Refraction
Perform the activity 8 by making angle of incidence 30°, and hence, find the angle of
refraction. Repeat the activity for the angle of incidence as 35°, 40°, 45° and 50°. In
each case, find the angle of refraction. Tabulate the angles of incidence, angles of
refraction and their sine values. Find the ratio between the sine values as shown in the
table below:
Table 3 — Verification of Laws of Refraction
In the above table, you can see that the ratio between the sine i and sin r is a constant
quantity. This verifies the Snell's law of refraction.
Furthermore, as the incident ray PO, the refracted ray
OO' and the normal to the surface of separation of
two media at the point of incidence O, lie in the
horizontal plane paper, therefore, we can say that all
of them lie in the same plane. This verifies the first
law of refraction.
Alternatively if a graph is plotted between sin / and
sin r, it is a straight line. From this it implies: sin i
Fig. 1.49 : Graph between sin / and
sin r
oc sin r
= constant. This verifies Snell's law of refraction.
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Physics Grade 12
Refractive Index
The above discussion you have seen that when a ray of light travels obliquely from
one transparent medium (air) to another transparent medium (glass), it changes its
direction in the second medium. This extent of change of direction that takes place in a
given pair of media is expressed in terms of refractive index and is defined as under:
The ratio between the sine of angle of incidence in one optical medium to the sine of
angle of refraction in another optical medium, is called refractive index of the second
medium with respect to the first medium,
In the above expression 1 stands for the first medium, in which angle of incidence is
formed and 2 stands for the second medium, in which angle of refraction is formed.
Thus, if the angle of incidence is formed in air and angle of refraction is formed in
glass, then refractive index of glass with respect to air is:
However, if ray is travelling from glass to air, such that angle of incidence is formed in
glass and angle of refraction is formed in air, then the refractive index of air with
respect to glass is:
Refractive Index and The Speed Of Light
Light has a constant speed of 3 x 10s ms-1 for all the colours or the wavelengths in
vacuum. However, if the light travels through any other optical medium, it is slowed
down. The extent of slowing down depends upon:
(i) The optical density of the medium. (ii) The colour or wavelength of the light.
It is this slowing down of light, which is responsible for the phenomenon of refraction.
It has been found experimentally that the refractive index of a given optical material is
the ratio between the speed of light in vacuum divided by the speed of light in a given
optical medium.
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Physics Grade 12
To show n
=
Consider an incident ray of light AB on passing through a glass slab moves along BC,
such that it strikes a plane glass at 90° and retraces its path.
When the ray travels from air to glass
Fig: 1.50
The value of refractive index so obtained is called absolute refractive index.
We also know that:
Speed of light = Wavelength x Frequency
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Physics Grade 12
Note:
It has been found that speed of light changes very slightly while travelling
from vacuum to air. For all practical purposes the speed of light in air is taken as same
as is in vacuum.
Thus, absolute refractive index (n) - 
Speed of light in vacuum or air
Speed of light in a given optical medium
Table - 4: Refractive index w.r.t. Vacuum
Material
n
Material
n
Diamond
2.42
Benzene
1.50
Sapphire
1.77
Turpentine oil
1.47
Ruby
1.71
Fused quartz glass
1.46
Flint glass
1.65
Kerosene oil
1.44
Carbon disulphide
1.63
Alcohol
1.36
Rock salt
1.54
Water
1.33
Canada Balsam
1.53
Ice
1.31
Crown glass
1.52
Air
1.0003
What is Optical Density?
We have been using the word 'optical density' without explanation. Let us try to make
sense out of it. The ability of a medium to refract light is expressed in terms of optical
density.
From the above statement it implies, more a ray of light bends in a given medium,
lesser will be the angle of refraction and lesser will be value of sin r. Thus, refractive
index of the medium n - sin i! sin r will increase. Thus, the optical density will
increase. Remember that optical density in a way is another name for refractive index.
How to Distinguish between a Rare and Dense Medium?
We have been also using the term 'rare medium' and 'dense medium.' How do we
know, which amongst the two given media is rare or dense? Look for the refractive
indices of the given two media. The media with lesser refractive index is rare and the
medium with more refractive index is dense.
Furthermore, optical density should not be confused with mass density. For example,
mass density of kerosene oil is less than water; however, the optical density of
kerosene oil is more than water.
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Physics Grade 12
IMPORTANT POINTS
 Speed of light in air or vacuum is 3 x 108 m/s.
 Speed of light decreases when light travels through some denser medium.
 Speed of light again becomes 3 x 108 ms^1 when it emerges out in air from denser
material without the loss of energy. Why?
Simple Effects of Refraction of light
1. A stick appears bent and short, when
immersed obliquely in water.
Consider a stick AC, immersed obliquely in
water, such that part AB of stick is within
Fig: 1.51
the water. Consider a point A on the tip of
stick. A divergent beam starting from A,
after refraction will bend away from the normal.
When this refracted beam reaches the eye, the eye retraces a straight line path. Thus, to
the eye the rays appear to come from point
A1, which is higher than A.
This is true for any other point between A
and B. Thus, on the whole, the stick within
the water appears bent and short. However,
it does appear magnified because the image
Fig 1.52 : Coin appears raised
is formed closer to the eye.
2. The bottom of a beaker filled with water appears raised. Take an empty beaker
and place in it a fifty paisa coin. Cover the sides of the beaker with paper and move
away from it, till the coin is just out of the line of sight. Ask your friend to pour water
gently in the beaker, till it is completely filled. It is observed that coin visible.
The reason for the above observation is that rays of light diverging from the coin, on
emerging out of water, suffer refraction and hence, bends away from the normal.
When these refracted rays reach the eye, then to the eye, they appear to come from the
point I, which is higher than O. Thus, the coin and hence, the bottom of the beaker
appears raised when filled with water. It is for the same reason:
(i) A stamp placed under a thick glass block appears raised.
(ii) A swimming pool, when filled with water appears shallow.
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Physics Grade 12
Numerical Problems on Refractive Index
Numerical Problem 1
It has been found that a ray of light while travelling from air to glass has angle of
incident 45° and angle of refraction 28°. What is refractive index of glass correct to
one decimal place.
Practice Problems 1
1.
A ray of light strikes a glass slab of refractive index 1.5 at an angle of 37°. What
is the angle of refraction? [24°]
2.
The ray of light strikes the surface of water at an angle of 30°, such that angle of
refraction is 22.1°. What is refractive index of water?
Numerical Problem 2
The refractive index of a material is 1.4. If velocity of light in vacuum is 3 x 10 8 m/s,
find the velocity of light in the material.
Practice Problems 2
1. The velocity of light in air is 3 x 10s ms"1 and in glass is 2 x 108 ms"1. Find the
refractive index of glass.
[1.5]
2. The velocity of light in air is 3 x 108 ms"1. Calculate the velocity of light in
diamond of refractive index 2.5.
[1.2 x 108 ms/s]
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Physics Grade 12
Numerical Problem 3
A light of wavelength 500'nm in air enters a glass block of refractive index 1.5. Find: (a)
speed; (b) frequency; (c) wavelength of light in glass. Velocity of light in air is 3 x 108 m/s.
Practice Problems 3
4
1. A light of wavelength 600 nm in air enters in water of refractive index . Find: (a)
3
speed; (b) frequency; (c) wavelength of light in glass. Velocity of light in air is 3
x 108 m/s.
[(a) 2.25 x 108 m/s (c) 450 nm]
2. Calculate: (i) speed; (ii) wavelength of light in diamond for a ray of light of
wavelength 800 nm in air, when refractive index of diamond is 2.42. Speed of light
in air is 3 x 108 m/s.
[(i) 1.24 x 108 m/s, (ii) 330.6 nm]
Exercise
VERY SHORT ANSWER TYPE QUESTIONS
1. What do you understand by the term refraction of light?
2. Fill in the blank spaces:
(a) When a ray of light is travelling obliquely in_______________medium, enters a
denser medium, it always bends __________ normal.
(b) When a ray of light is travelling obliquely in denser medium, enters ------medium,
it always bends away from the____________.
(c) When a ray of light strikes the surface of separation of two optical media at right
angles, it ______________ suffer any _________________.
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3. State Snell's law of refraction.
4. What do you understand by the term refractive index?
5. The refractive index of glass is 1.5. What do you understand by the statement?
6. What do you understand by the term lateral displacement?
7. How does frequency and wavelength of right change when light passes from rarer
(air) to denser (glass) medium?
8. For the same angle of incidence, the angle of refractive in three media A, B and C
are 15°, 25° and 35°. In which medium the velocity of right is minimum?
9. Refractive index of glass for light of yellow, green and red colours are n y , ng , and
nr respectively. Rearrange these symbols in increasing order of values.
10. When light undegoes refraction at the surface of separation of two media, what
happens to its wavelength?
11. When light undergoes refraction, what happens to its frequency?
SHORT ANSWER TYPE QUESTIONS
1. Why does a fisherman aim at the tail of a fish, during spear fishing? Explain.
2. Why does light ray incident on a rectangular glass slab (immersed in any medium
emerges parallel to itself? Explain using a diagram.
3. Draw a ray diagram to show the passage of two rays of light through rectangular
glass slab, when the angle of incidence is zero in one case and little less than 90° in
the other case.
4. Why does a ray of light bend when it travels from one medium to another medium?
5. With respect to air, refractive index of ice is 1.31 and that of rock salt is 1.54.
Calculate the refractive index of rock salt with respect to ice.
6. Light enters from air into glass plate which has a refractive index of 1.50.
Calculate the speed of light in glass. The speed of light in air is 3 x 10s m/s.
7. In an experiment with a rectangular glass slab, a student observed that a ray of light
incident at an angle of 55° with the normal on one face of the slab, after refraction
strikes the opposite face of the slab before emerging out into air making an angle of
40° with the normal. Draw a labelled diagram to show the path of this ray. What
value would you assign to the angle of refraction and angle of emergence?
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SHORT ANSWER TYPE QUESTIONS
1.
How lateral displacement produced by an optical block is related to :
(a) Thickness of the block?
(b) Refractive index of the block?
(c) Angle of incidence?
2.
By drawing a neat diagram show:
3.
Why does a stick immersed obliquely in water appear bent and short?
4.
Why does a coin placed in a water tank appear raised?
5.
Under what condition in an arrangement of two plane mirrors, incident ray and
reflected ray will always be parallel to each other, whatever may be angle of
incidence. Show the same with the help of diagram.
6.
Draw a ray diagram showing the path of rays of light when it enters with oblique
incidence
(i)
from
air
into
water;
(ii)
from
water
into
air.
A pencil when dipped in water in a glass tumbler appears to be bent at the
interface of air and water. Will the pencil appear to be bent to the same extent, if
instead of water we use liquids like, kerosene or turpentine. Support your answer
with reason.
7.
How is the refractive index of a medium related to the speed of light? Obtain an
expression for refractive index of a medium with respect to another in terms of
speed of light in these two media?
8.
Refractive index of diamond with respect to glass is 1.6 and absolute refractive
index of glass is 1.5. Find out the absolute refractive index of diamond.
9.
An object is placed at a distance of 24cm from the optical centre of a convex lens
of focal length 18cm. Find the distance of the image from the lens. What will be
the height of the image if the object is 3.0cm tall?
10. How far should an object be placed from a convex lens of focal length 20 cm to
obtain its image at a distance of 30 cm from the lens? What will be the height of
the image if the object is 6cm tall?
11. An object 20 cm in size is placed 20.0cm in front of a concave mirror of focal
length 10.0cm. Find the distance from the mirror at which a screen should be
placed in order to obtain sharp image. What will be the size and nature of the
image formed?
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12. (i) Draw a ray diagram to show the passage of two rays of light through a
rectangular slab of glass, when the angle of incidence is zero in one case and a
little less than 90° in other case.
13. Prove that if a ray enters a rectangular glass slab obliquely and emerges from the
opposite face, the emerged ray will be parallel to the incident ray.
14. An object 5cm high is placed 20cm in front of a concave mirror of focal length
15 cm. At what distance from the mirror, should a screen be placed to obtain a
sharp image? Calculate the size of the image formed.
15. (i) State Snell’s law of refraction of light.
(ii) A transparent medium A floats on another transparent medium B. When a ray
of light travels obliquely from A into B, the refracted ray bends away from the normal.
Which of the two media A and B is optically denser and why?
LONG ANSWER TYPE QUESTIONS
1. Write the laws of refraction. Explain the same with the help of ray diagram, when a
ray of light passes through a rectangular glass slab.
2. A ray of light incident obliquely on the face of rectangular glass block placed in air
emerges from the opposite face parallel to the incident ray. State two factors on
which lateral displacement of emergent ray depends on emergent ray.
REFRACTION BY SPHERICAL LENSES
You might have seen a magnifying glass in your biology laboratory. It is used for
magnifying small biological specimens. Some people use spectacles for reading or
seeing far-off objects. Try to touch the surface of magnifying glass or spectacle lens
and find out whether its surface is plane or curved. Also try to find out whether it is
thicker in the middle or at the edges. You will notice that surface of magnifying glass
is not plane, but thicker in the middle and tapering at the edges. Similar will be your
observation in case of spectacle glasses. The curved glass used in the magnifying glass
or the spectacle is called lens.
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Lens
A lens is defined as a portion of a transparent optical material, having one or two
spherical surfaces. From the above statement, it is clear that a lens can either have one
spherical surface and one plane surface or it can have two spherical surfaces.
Lenses are divided into two broad classes:
(a) Converging lens or convex lens.
(b) Diverging lens or concave lens.
(a) Converging lens or Convex lens
A piece of transparent optical material,
having one or two spherical surfaces,
such that it is thicker in the middle and
tapering (thinner) at the edges is called
convex lens or converging lens.
Fig: 1.54
Depending upon their shape, these lenses are further classified as : (i) double convex
lens, (ii) plano-convex lens, (iii) concavo-convex lens as shown in Fig. 1.54.
If you carefully study their overall shape, you will notice that all of them are thicker
in the middle and tapering at their edges.
(b) Diverging lens or Concave lens
A piece of transparent optical material,
having one or two spherical surfaces,
such that it is tapering (thinner) in the
middle and thicker at its edges, is called
concave lens or diverging lens.
Depending upon their shapes, these
lenses are further classified as: (i)
Fig: 1.55
double concave lens, (ii) plano-concave lens, (iii)
convexo-concave lens, as shown in the Fig. L55.
If you carefully study their overall shape, you will notice that all of them are thicker
at their edges and tapering in the middle.
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A convex lens can be differentiated from a concave lens by moving your fingers in the
middle and along its circumference. If it appears tapering (thinner) at the
circumference and thick in the middle, then it is convex lens. If it appears tapering
(thinner) at centre and thick at the circumference, then it is concave lens.
Simple Terms Used in Lenses
(a) Principal Axis
An imaginary line joining the
centre of curvatures C1 and C2 of
two spherical surfaces of the lens
is called the principal axis.
XY is the imaginary line passing
through C1and C2 and hence is the
principal axis (Fig. 1.56).
(b) Optical Centre
A point within the lens, where a line drawn through the aperture (diameter of the lens),
meets the principal axis is called optical centre. It is denoted by the letter 'O'.
The property of optical centre is that any ray of light passing through this point,
emerges parallel to its direction of incidence,
as shown in Fig. 1.56. The central portion of
the lens may be regarded as a small slab with
parallel sides and therefore, the ray passing
through it, is slightly displaced parallel to its
original direction.
However, if the lens is very thin, the small
lateral displacement t is generally ignored.
Fig: 1.57
Thus, the ray is assumed to travel straight
through the optical centre and hence, does not
suffer any refraction as shown in the Fig, 1.57.
(c) Principal focus of a lens
It is a point on the principal axis of the lens,
such that a beam of light parallel to the
principal axis, after refraction through the lens,
Fig: 1.58
either actually meet or appear to come from this point.
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(d) (i) First Principal focus of a convex lens(F1)
It is a point on the principal axis of a convex lens, such that the rays originating from
it, after refraction travel parallel to the principal axis, F1 is the first principal focus of
the convex lens, as shown in Fig. 1.58.
(ii) Second Principal focus of a convex lens (F2)
It is a point on the principal axis, where the rays coming parallel to the principal axis,
after refraction through the lens actually meet.
In Fig. 1.59, F2 is the second
principal focus of a convex lens.
(e) Focal length
The
distance
between
the
optical centre and the principal
focus (first or second) of a lens
is called the focal length.
In Fig. 1.58, is the focal length. In Fig. 1.59, OF2 is the focal length. It is denoted by
the letter f.
(f) Focal plane
A vertical plane passing
through the principal focus
(first or second) of the lens
is called a focal plane.
(g) Second Principal focus of concave lens (F2)
It is the point on the principal axis at which incident
ray would have converged before refraction, but
after refraction through lens becomes parallel to the
principal axis.
Iri Fig. 1.60, F2 is the second principal focus of the
concave lens. PQ is the second focal plane of the
concave lens and OF2 = / is its focal length.
Fig: 1.60
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(h) First Principal focus of a concave lens (F1)
It is a point on the principal axis, where the rays travelling parallel to the principal
axis, after refraction through the lens appear to come.
Image Formation by Lenses using Ray Diagrams in convex lens
1. A ray of light travelling parallel to the principal axis, after refraction through the
ray of light passes through the second principal focus of the lens.
2. A ray of light which first passes through the first principal focus of the convex lens,
after refraction the ray of light travels parallel to the principal axis.
3. A ray of light which passes through the optical centre of a convex lens does not
suffer any refraction, i-e., it passes through the lens without
Rules for drawing images in concave lens
1. A ray of light travelling parallel to the principal axis of a concave lens, after
refraction, appears to come from its first principal focus.
2. A ray of light initially travelling along the direction of the second principal focus,
after refraction, travels parallel to the principal axis of the concave lens
3. A ray of light which passes through the optical centre of a concave lens does not
suffer any refraction, i.e., it passes through the lens without any deviation.
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Construction of Images in Convex Lens by Ray Diagrams
(a) When the object is at infinity, such that the rays coming from it, are parallel to
the principal axis of the convex lens.
When the object is at infinity the rays coming from it are parallel. These rays, on
passing through the convex lens, meet at the second principal focus as illustrated in
Fig. 1.73.
Following are the characteristics of the image formed:
(1) The image is real.
(2) The image is inverted.
(3) The image is diminished to a point.
(4) The image is formed at F2 on the other
side of the lens.
Uses: Because of the above property, a
Fig: 1.73
convex lens can be used as a burning glass. When the rays of the sun are passed
through it and a piece of paper is placed at F2, on the other side of the lens, the paper
catches fire. It is because, the heat radiations focus on the paper. Thus, the temperature
of paper rises above the ignition point and hence, it catches fire.
(b)
When the object is at infinity,
such that the rays coming from it are
not parallel to the principal axis of
the convex lens.
If the parallel rays coming from the
distant object are not parallel to the
Fig: 1.74
principal axis of the convex lens, then the image is formed at the second focal plane of
the lens.
In order to locate the position of the image draw a line parallel to the rays of light,
such that it passes through the optical centre. The point where this line intersects the
focal plane is the position of the image. In Fig. 1.74, the line passing through optical
centre intersects the focal plane at I. Thus, I is the image of the distant object. All other
rays will meet at point I. Following are the characteristics of the image formed:
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Physics Grade 12
(1) The image is real.
(2) The image is inverted.
(3) The image is highly diminished.
(4) The image is formed in the second focal plane on the other side of the lens.
Uses: The above properties of the convex lens is used as an objective lens of the
telescopes. It forms small inverted image of a far off object in the second focal plane
of the lens.
(c) When the object is beyond 2Fp but not at infinity.
Consider an object AB situated on the
principal axis of the convex lens, beyond
2Fj, but object- not at infinity. A ray of
light which starts from point A, and
moves parallel to the principal axis, after
Fig: 1.75
refraction passes through F2.
Another ray of light which starts from point A, on passing
through the optical centre, passes on un-deviated. These
refracted rays meet at the point A1, thereby forming an
image A1B1. Following are the characteristics of the image
formed:
Fig: 1.76
(1) The image is real.
(2) The image is inverted.
(3) The image is diminished.
(4) The image is formed between F2 and 2F2 on the other side of the lens.
Uses: The above property of the convex lens is
used in photographic camera, when a small
inverted image of an object is formed on the
film placed between F2 and 2F2 of the lens.
(d) When the object is located at 2Ff
Fig: 1.77
AB is an object located at 2F1 of the convex lens. A ray of light which starts from
point A and moves parallel to the principal axis, after refraction through the lens,
passes through its second principal focus F2. Another ray of light starting from point
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Physics Grade 12
A, on passing through the optical centre O, passes on undeviated. These two refracted
rays, meet at A1, there by forming an image A1B1.
Following are the characteristics of the image formed:
(1) The image is real.
(2) The image is inverted.
(3) The image is of the same size as the object.
(4) The image is formed at 2F2, on the other side of the lens.
Uses: Above case is used in terrestial telescope for erecting the inverted image formed
by the objective lens of the telescope.
(e) When the object is located between 2F1 and F1.
AB is a small object located in between
Fig: 1.78
2F1 and F1 of a convex lens. A ray of
light which starts from the point A,
moves parallel to the principal axis,
after refraction through the lens, passes
through its second principal focus F2. Another ray of light
starting from point A, on passing through optical centre O,
passes on undeviated. These two refracted rays meet at A1
thereby forming an image A1B1.Following are the characteristics
of the image formed:
(1) The image is real.
(2) The image is inverted.
(3) The image is enlarged, i.e., bigger than the size of the object.
Fig. 1.79: Cine
projector
(4) The image is formed beyond 2F2, but not at infinity on the
other side of the lens.
Uses: The above case is used in an optical projector (slide projector), when enlarged
image of a small slide is formed on the screen.
(f) When the object is located at F1
AB is an object located at the principal
focus of a convex lens. A ray of light
which starts from point A, and moves
parallel to the principal axis, after
refraction through the lens, passes through
its second principal focus F2Fig: 1.80
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Physics Grade 12
Another ray of light starting from point A, on passing through the optical centre O,
passes on undeviated. These two refracted rays produce a near parallel beam, which
meets only at infinity to form an image.
Following are the characteristics of the
image formed:
(1) The image is real.
(2) The image is inverted.
(3)
The image is formed at infinity on the other side of
the lens.
(4) The image is highly enlarged.
Uses: The above case is used in a searchlight. A powerful
lamp is placed at the principal focus of a convex lens,
Fig. 1.81: Searchlight
when a powerful parallel beam of light is produced.
(g) When the object is located between the optical centre and first principal focus.
AB is an object located in between the optical centre and the first principal focus of the
convex lens. A ray of light which starts from point A, and moves parallel to the
principal axis, after refraction through the
lens, passes through its second principal
focus F2. Another ray of light starting from
point A on passing through the optical centre
Fig. 1.82
O, passes on undeviated. These refracted rays give rise to a divergent beam, which
does not meet on producing forward.
However, when this divergent beam enters the eye, it appears to come from point A1,
thereby forming an image A1B1.
Following are the characteristics of the image formed:
(1) The image is virtual.
(2) The image is erect.
(3) The image is enlarged.
(4) The image is formed on the same side as the
object.
Fig. 1.83: Hand Lens
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Physics Grade 12
Uses: In the above case, the lens acts as a simple microscope and is put to the
following uses:
(1) It is used in seeing clearly the weave patterns of the cloth.
(2) It is used by the palmists to study lines of a hand.
(3) It is used to read fine print.
(4) It is used to study biological specimens, such as parts of a flower, etc.
Construction of Images in Concave Lens by Ray Diagrams
(a) When the object is located at infinity.
The rays coming from an object it infinity are always parallel. When these parallel
rays pass through a concave lens, they get refracted and form a diverging beam. This
beam of light on reaching the eye appears to come from F1 of the concave lens,
thereby forming a virtual image.
Following are the characteristics of the
image formed:
1. The image is virtual.
2. The image is erect.
3. The image is diminished to a point.
4.
The image is formed at F1 on the same
Fig. 1.84
side as the object.
Uses: This case is used in a Galilean telescope, when the concave lens acts as an eye lens.
(b) When the object is anywhere between the optical centre and infinity.
AB is an object located anywhere
between the optical centre O and infinity.
A ray of light which starts from point A,
and moves parallel ;a the principal axis,
after refraction through the lens diverges
outward, but on producing backward, appears to come from F1. Another ray of light
starting from point A, on passing through the optical centre, passes on undeviated.
These refracted rays give rise to a divergent beam which does not meet on being
produced forward. However, when these diverging rays enter the eye, they appear to
come from point A1, thereby forming a virtual image A1B1.
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Physics Grade 12
Following are the characteristics of the image formed:
(1) The image is virtual.
(2) The image is erect.
(3) The image is diminished.
(4) The image is formed between the optical centre (O) and the second principal focus
(F2) on the same side as the object.
Uses: It is used for correcting short-sightedness in human the eye.
POWER OF A LENS
Power of a lens is a measure of degree of convergence or divergence of light rays
incident on it. In other words, it is the ability of a lens to bend the rays of light incident
on it. Mathematically, it is defined as follows:
The reciprocal of focal length in metres is called power of a lens.
Thus, power of a lens =
Focal length of the lens (m metres)
However, if the focal length is given in centimetres, then the above formula can be
modified as under
As the, power of a lens is inversely proportional to its focal length in metres, therefore,
higher the power of a lens, lesser is its focal length.
Units of Power of a Lens
The SI unit of power of a lens is called dioptre It is denoted by the letter D.
A lens is said to have a power of one dioptre, if its focal length is one metre.
As the convex lens has a positive focal length, therefore, its power is said to be
positive and is denoted by the sign + D.
As the concave lens has a negative focal length, therefore, its power is said to be
negative and is denoted by the sign - D.
In optical instruments, a number of lenses are combined, so as to increase the
magnification or the sharpness of the image.
The net power (P) of these lenses placed in contact with one another, is the algebraic
sum of the individual powers Pj, P2, ...., etc., of the lenses. Thus, P = P! + P2 + P3 + …..
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Physics Grade 12
Advantages in Using Power of Lenses
Using power, instead of focal length for the lenses avoids a good deal of arithmetical
calculations. Thus, it makes the calculations very convenient and straight forward.
For example, an optician while testing eye sight, places a convex lens of power + 2D
in front of the eye of the patient. However, vision is not clear. He, then places another
convex lens of power + 0.5 D in contact with first lens, when the vision becomes clear.
Thus, optician immediately knows that power of the lens suitable for his patient, is
+2.50 D. Such simple calculations are not possible, if the optician uses focal length of
lenses.
Similarly, using power of the lens is very convenient in designing lenses for optical
instruments, such as cameras, microscopes, telescopes, optical projectors and variety
of other instruments.
Points to Remember

A lens whose focal length is 1m has a power of one dioptre.

The power of convex lens is positive and that of concave lens negative.

The power of lens is a measure of its degree of convergence or divergence.

With the increase power of lens its focal length decreases and vice versa.

Opticians can design a lens of particular power by combining two or more lens of
appropriate power.

The spectacle lens made by combining two or more lenses are free from defects
of astigmatism which cause distortion.
NEW CARTESIAN SIGN CONVENTION FOR LENSES
Following are the sign conventions for measuring distances in ray diagrams for lenses:
1. All distances are measured from the optical centre of the lens.
2. The distances measured from the optical centre in the direction of the incident
light are taken as positive.
3. The distances measured from the optical centre against the direction of the
incident light are taken as negative.
4. The distances measured upward and perpendicular to the principal axis are taken
as positive.
5. The distances measured downward and perpendicular to the principal axis are
taken as negative.
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Physics Grade 12
Important points for Measuring Distances for Convex/Concave Lens
(a) A convex lens when a real image is formed.
(i) The distance of the object from the optical centre (u) is always negative. (ii) The
distance of the image from the optical centre (v) is always -positive. (in) The focal
length of the lens is always positive.
(b) A convex lens when a virtual image is formed.
(i) The distance of the object from the optical centre (u) is always negative. (ii) The
distance of the image from the optical centre (v) is always negative. (iii) The focal
length of the lens (/) is always positive.
(c) Concave lens
(i) The distance of the object from the optical centre (u) is always negative. (ii) The
distance of the image from the optical centre (v) is always negative. (iii) The distance
of the focal length of the lens (/) is always negative.
Lens formula
A formula which gives a relationship between the distance of object from the optical
centre (u), distance of image from optical centre (v) and the focal length of the lens (f)
is called lens formula. The lens formula for convex and concave lenses is same as
stated below.
Derivation of the above formula is deleted from the syllabus.
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Physics Grade 12
Important Points in using Lens Formula
(i) Put the correct signs of known variables according to the sign conventions.
(ii) Do not put the sign of unknown variable. The sign will automatically show up
during calculations.
(iii) If the calculated sign of a variable turns out positive, then the variable calculated is
on the other side of the lens, i.e., on the opposite side to the object. However, if
calculated variable is of negative sign, then it is on the same side as the object.
Linear Magnification Produced by Lenses
The ratio between the height of the image produced by a lens to the height of the
object is called the linear magnification.
While using above formulae, use conventional signs for v and u.
Points to Remember
In case of convex lens, the magnification m is positive, when image is virtual, and it is
negative when image is real. In case of concave lens, m is always positive, as the
image formed is always virtual.
Numerical Problems on Lenses
Numerical Problem 1
A convex lens produces a real and inverted image 2.5 times magnified at a distance of
25 cm from the lens. Calculate focal length of the lens.
Solution:
Magnification (m) - -2.5 (real image)
Distance of the image from the lens (v) = 25cm.
Distance of the object from the lens (u) =? (to be calculated)
Focal length of the lens (/) =? (to be calculated)
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Physics Grade 12
Thus, the focal length of the lens is 7.14 cm.
Practice Problems: 1
1. An object when placed in front of a convex lens forms a real image of 0.5
magnification. If the distance of the image from the lens is 24 cm, calculate the
focal length of the lens.
[f = 16 cm]
2. A convex lens forms a real image magnified four times, when placed at a distance
of 6cm from the lens. Calculate the focal length of the lens.
[f = 1.2 cm]
Numerical Problem 2
An eye specialist prescribes a number + 4-5 D to a person for his glasses. What is the
nature of the lens? What is the focal length of the lens?
Practice Problems: 2
1. State the nature of the lens and the focal length if its power is + 4D.
[(i) convex, (ii) 25 cm]
2. The number of the glasses of a person is + 0-75 D. What is the nature of the lens
and what is its focal length?
[(i) convex, (ii) 133-33 cm]
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Physics Grade 12
Numerical Problem 3
The focal length of the glasses of a short-sighted person is 37.5 cm. Calculate the
power of the glasses and their nature.
Solution: As the person is short-sighted, the lens of his glasses is concave.
Minus sign indicates that the lens is concave in nature.
Practice Problems 3
1. The focal length of a concave lens is 10cm. Calculate its power.
[10 D]
2. The focal length of the lens of a myopic person is 40cm. What is the power of the
lens? [- 2.5 D]
Numerical Problem 4
The power of the lens of a myopic person is – 3.75D. Calculate the focal length of the lens.
Solution:
Power - 3.75 D indicates that lens is concave.
Practice Problems 4
1. Calculate the focal length of a lens of power - 2.75 D.
[- 36.36 cm]
2. The power of a concave lens is - 12.5 D. What is the focal length of the lens?[- 8 cm]
Numerical Problem 5
Two thin lenses of power + 2.5 D and -1.5 D are placed in contact with each other.
Calculate: (i) power of the combination, (ii) focal length of the combination.
Solution
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Physics Grade 12
Practice Problems 5
1. Two thin lenses of power + 4.5 D and -5D are placed in contact with each other.
Calculate: (i) power of the combination, (ii) focal length of the combination.
[(i) - 0.5 D (ii) 2 m]
2. The power of a thin convex lens is + 3.5 D. It is placed in contact with another thin
lens, such that the combined power is + 1.0 D. Calculate: (i) power of second lens, (ii)
focal length of the second lens.
[(i) -2.5 D (ii) 0.4 m]
Numerical Problem 6
A convex lens of focal length 40 cm and a concave lens of focal length 50 cm are
placed in contact with each other. Calculate: (i) the power of the combination, (ii)
focal length of the combination.
 Focal length of the combination
F 

1 (m)
Power of combination
1 (m)
 2 m.
0.5
Practice Problems: 6
1. Calculate the power and focal length of a combination of convex lens of focal
length 25 cm and concave lens of focal length 10cm.
[(i) -
10D, (ii) 0.10 m]
2. A convex lens of focal length 40 cm is placed in contact with concave lens of focal
length 25cm. Calculate (i) power of the combination (ii) focal length of the
combination.
[(i) - 1.5D, (ii) 0.666m]
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EXERCISE
(I) MULTIPLE CHOICE QUESTIONS
1- You have blindfolded your friend and then gave him a lens for recognition. He
moves his fingers on the lens and declares it convex lens, because:
(A) it has tapering edges
(B) it is thicker in the middle
(C) both (A) and (B)
(D) none of the above.
2. When a newspaper is seen through a lens, its print appears smaller. The nature of
the lens is:
(A) convex
(B) concave
(C) double convex
(D) concavo-convex
3. A converging beam of light on striking a concave lens, appears to converge at a
point on the principal axis. The point is called:
(A) optical centre of lens
(B) first principal focus of lens
(C) second principal focus of lens
(D) none of the above.
4. An object is at infinity with respect to the optical centre of a converging lens. The
image formed by it is:
(A) diminished and erect
(B) diminished
(C) diminished to a point and inverted
(D) magnified and erect.
5. Which of the following statements is true?
(A) A convex lens has 4 dioptre power having a focal length 0.25m
(B) A convex lens has -4 dioptre power having a focal length 0.25m
(C) A concave lens has 4 dioptre power having a focal length 0.25m
(D) A concave lens has -4 dioptre power having a focal length 0.25m
VERY SHORT ANSWER TYPE QUESTIONS
1. (i) What do you understand by the term lens?
(ii) Define a convex lens. What are its kinds? Show each kind by a diagram.
(iii) Define concave lens. What are its kinds? Show each kind by a diagram.
2. Define the following with reference to a convex lens:
(i) the principal axis,
(ii) the optical centre,
(iii) the principal focus,
(iv) the focal length.
3. Define the following with reference to a concave lens:
(i) the principal focus,
(ii) the optical centre.
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4. How does a focal length of convex lens change if monochromatic red light is
replaced by monochromatic blue light?
5. Draw the given diagram in your answer book and complete it for the path of light
beyond the lens.
6. Take down this diagram in your answer book and complete the path of the ray.
SHORT ANSWER TYPE QUESTIONS
1. One half of convex lens of focal length 20cm is covered with black paper
(i)
Will the lens produces a complete image of object?
(ii) Show the formation of an image of an object placed at 2F1 of such a covered
lens, with the help of a key diagram.
(iii) How will the intensity of image formed by a half covered lens compare with
non-covered lens?
2.
Mr. Ali finds out that the sharp image of the window pane of her science
laboratory is formed at a distance of 15 cm from the lens. She now tries to focus
the building visible to her outside the window instead of the window pane without
disturbing the lens. In which direction will she move the screen to obtain a sharp
image of the building? What is the approximate focal length of this lens?
3. How are power and focal length of a lens related? You are provided with two
lenses of focal length 20cm and 40cm respectively. Which lens will you use to
obtain more convergent light?
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4. A concave lens has focal length of 20cm. At what distance from the lens is a 5 cm
tall object be placed so that it forms an image at 15cm from the lens? Also, calculate
the size of the image formed.
5. A convex lens has a focal length of 25cm. Calculate the distance of the object from
the lens if the image is to be formed on the opposite side of the lens at a distance of
75cm from the lens. What will be the nature of the image?
6. A convex lens has a focal length of 30cm. Calculate at what distance should the
object be placed from the lens so that in forms an image at 60cm on the other side of
the lens. Find the magnification produced by the lens in this case.
7.An object 50cm tall is placed on the principal axis of a convex lens. Its 20cm tall
image is formed on the screen placed at a distance of 10cm from the lens. Calculate
the focal length of the lens.
8. An object 3cm high is placed at a distance of 20cm in front of a convex lens of
focal length 12cm. Find the position, nature and size of the image formed.
9. Where should an object be placed from a converging lens of focal length 20 cm, so
as to obtained a real image of magnification 2?
10. An object 3.0cm high is placed perpendicular to the principal axis of a concave
lens of focal length 15.0cm. The image is formed at a distance of 10.0cm from the
lens. Calculate:
(i) distance at which the object is placed and
(ii) size and nature of the image formed.
11. An object 20 cm tall is placed on the principal axis of a convex lens. Its 30 cm tall
image is formed on the screen placed at a distance of 10cm from the lens. Calculate
the focal length of the lens.
12. An object 30cm tall is placed on the principal axis of a convex lens. Its 10cm tall
inverted image is formed on the screen placed at a distance of 15cm from the lens.
Calculate the focal length of the lens.
13. A 5.0cm tall object is placed perpendicular to the principal axis of a convex lens
of focal length 20cm. By calculation determine:
(i) the position and
(ii) the size of the image formed.
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14. A convex lens of focal length 20cm produce a magnified virtual as well as real
image. Is this a correct statement? If yes, where shall the object be placed in each case
for obtain these images?
15. Under what condition in an arrangement of two plane mirrors, incident ray and
reflected ray will always be parallel to each other, whatever may be angle of incidence.
Show the same with the help of diagram.
LONG ANSWER TYPE QUESTIONS
1. You are required to form an upright image which is: (i) enlarged, (ii) diminished by
using a lens. State the nature of the lens in each case and draw a neat ray diagram in
support of your answer.
2.
(i) State the lens formula.
(ii) State whether «, v and/are positive or negative, when:
(a) The lens is convex and the image formed is real.
(b) The lens is convex and the image formed is virtual.
(c) The lens is concave.
3. What do you understand by the term linear magnification of the lens? State two
formulae for the linear magnification.
4.
(i) What do you understand by the term power of a lens?
(ii) State and define the unit of the power of a lens.
(iii) When is the power of the lens (a) positive, (b) negative?
5. Draw a neat diagram to illustrate the use of convex lens as magnifying glass.
6. Draw ray diagrams showing the image formation by a concave lens when an object
is placed:
(i) at the focus of the lens
(ii) between focus and twice the focal length of the lens (iii) beyond twice the
focal length of the lens
7. The image of a candle flame formed by a lens is obtained on a screen placed on the
other side of the lens. If the image is three times the size of the flame and the distance
between lens and image is 80cm, at what distance should the candle be placed from the
lens? What is the nature of the image at a distance of 80cm and the lens?
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8. Define power of a lens. What is its unit? One student uses a lens of focal length 50
cm and another of-50cm. What is the nature of the lens and its power used by each of
them?
9. A student focussed the image of a candle flame on a white screen using a convex
lens. He noted down the position of the candle screen and the lens as under:
Position of candle = 12.0cm
Position of convex lens = 50.0cm
Position of the screen = 88.0cm
(i) What is the focal length of the convex lens?
(ii) Where will the image be formed if he shifts the candle towards the lens at a
position of 31.0cm?
(iii) What will be the nature of the image formed if he further shifts the candle
towards the lens?
(iv) Draw a ray diagram to show the formation of the image in case.
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Human Eye and Colourful World
INTRODUCTION
In the previous chapter we have learnt about the phenomena of reflection and
refraction of light. In this chapter, we will learn how, the basic ideas developed in the
formation of images in the lenses help us to understand the working of human eye and
dispersion of light. In the human eye, we shall study its structure and various optical
defects. In dispersion we shall learn, how and why the light splits up into seven
colours and its natural consequences.
HUMAN EYE
The human eye is almost a spherical ball with a slight bulge in the front part. The
structure and function of each part of the eye is discussed below:
The eye ball is approximately spherical in shape. Its diameter in a grown up person is
2.3cm to 2.5cm.
1. Sclerotic: It is the outermost covering of the
eye. It consists of white tough fibrous tissue.
Its function is to protect and contain the vital
internal parts of the eye.
2. Cornea: It is the front bulging portion of the
eye. It is made of transparent tissues and
contains no blood vessels. Its function is to act
Fig. 2.1(a): Human eye.
as a window to the world, i.e., to allow the light to enter the eye ball.
Most of the refraction of light takes place while passing through cornea.
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3. Choroid: It is a grey membrane attached to the sclerotic from the inner side. Its
function is to darken the eye from inside, and hence, prevent any internal
reflection.
4. Optic nerve: It is a bundle of approximately 70,000 nerves originating from the
brain and entering the eye ball from the posterior side. The left eye receives optic
nerves from the right side of the brain and the right eye from the left side of the
brain. Its function is to carry the optical messages to the brain.
5. Retina: The optic nerve on entering the eye ball spreads like a canopy and each
nerve attaches itself to the choroid. The nerve endings form a kind of hemispherical
screen called retina. The nerve endings are sensitive to light. Thus, any image
formed on the retina is converted into optical impulses, which are then sent to the
brain. The retina has two very important regions, which are called the yellow spot
and the blind spot. The function of the retina is to receive the optical image of an
object and then convert it into electrical pulses which are finally sent to the brain
through the optic nerve.
6. Yellow spot : It is situated at the centre of the retina and is slightly raised. It has a
little depression called Fovea-Centralis, which is extremely sensitive to light. Its
function is to form an extremely clear image, i.e., when we want to examine an
object very minutely, its image is brought to focus at this point.
7. Blind spot : The region on the retina, where the optic nerve enters the eye ball is
called the blind spot. It has no nerve endings, and hence, is insensitive to the light.
Apparently it has no function. Any image formed at this spot is not visible.
8. Crystalline lens : It is a double convex lens, more bulging on the posterior side. It
is made of transparent, flexible tissues. It is held in position by a ring of muscles,
commonly called ciliary muscles. It divides the eye ball into two unequal parts.
The front part of the eye is called the anterior portion and the rear part is called the
posterior portion. Its function is to focus the images of the objects at different
distances, clearly on the retina. The crystalline lens merely provides the finer
adjustments hi focal length, so that objects at different distances are clearly
focussed on the retina.
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9. Ciliary muscles: It is a ring of muscles, which along with the suspensory ligament
that holds the crystalline lens in position. When these muscles contract, they
decrease the focal length of the crystalline lens. Similarly, when these muscles
relax, they increase the focal length of the crystalline lens.
Its function is to alter the focal length of the crystalline lens, so that the image of the
objects at various distances is clearly focussed on the retina.
10. Iris: It is a circular contractile diaphragm, suspended in front of the crystalline
lens. It has a tiny hole in the middle, commonly called the pupil. The diaphragm
has tiny muscles arranged radially and is heavily pigmented. The colour of the eye
depends upon the colour of the pigment.
Its function is to control the amount of light entering the eye. This is achieved by the
muscles present in the diaphragm. When the muscles contract, they increase the size of
the pupil, thus allowing more light to enter the eye and vice-versa.
11. Vitreous humour: It is a dense, jelly like fluid, slightly grey in colour, filling the
posterior part of the eye ball. It has the following functions:
a.
It prevents the eye ball from collapsing, due to the changes in the atmospheric
pressure.
b.
It partially helps in focussing the image clearly on the retina.
12. Aqueous humour: It is a watery, saline fluid, filling the anterior portion of the
eye. It has the following functions:
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a. It prevents the anterior portion of the eye from collapsing, due to the changes in
the atmospheric pressure.
b. When we wink our eyes, a tiny drop of the aqueous humour flows out from the
side of the eye. Then it washes the eye and keeps the cornea moist. Otherwise, the
cornea will shrivel and become opaque.
How Do We See Objects?
When the light coming from an object is incident on the cornea, it gets refracted and
then passes through the pupil. The partial refracted light on passing through the
crystalline lens of the eye suffers further refraction and hence converges on to the
retina to form an extremely small, real and inverted image on the retina. The nerve
ending on the portion of retina on receiving the light impulses get activated and change
into electrical impulses. These electrical impulses are then sent to the brain. It is the
brain which converts these electrical impulses into visual signals and hence we see an
erect image of an object in its actual size.
Accommodation of the Eye
The process by which the ciliary muscles alter the focal length of the crystalline
lens, so as to focus nearer or far-off objects clearly on the retina is called the
accommodation of the eye.
In order to focus at far-off objects, the ciliary muscles relax. In doing so, they make
the crystalline lens thin. Since the crystalline lens is thin, its focal length increases.
Thus, the images of far-off objects are clearly focussed on the retina, which at the
moment is at the principal focus of the crystalline lens.
Defects of Vision and Their Correction
A normal eye can see any object, which is situated between a distance of 25 cm and
infinity, using its power of accommodation.
However, sometimes due to some disease or age, some abnormalities occur hi the eye,
with the result; the eye is unable to accommodate itself to various distances. Following
are the most common defects of vision:
1. Short-sightedness or myopia
2. Long-sightedness or hypermetropia.
3. Presbyopia.
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1. Short-sightedness or Myopia
A person suffering from this defect can see the nearer objects clearly, but cannot see
the far-off objects clearly. It must be remembered that the person is not blind as far
as far-off objects are concerned. Only the images of far-off objects are blurred.
Causes :( 1) Due to some disease or age, the ciliary muscles, attached to the lens get
weak. They are unable to relax, and hence, focal length of the crystalline lens does not
increase. Thus, the image of the far-off object is formed in front of the retina, and
hence, appears blurred. (2) Sometimes the eye ball gets elongated and therefore the
images of far-off objects are formed in front of the retina.
Correction of short-sighted eye: A short-sighted person can see clearly to some
distance (say 4 m). Beyond this distance, the images get blurred. The farthest point
from which a short-sighted person can see clearly is called far-off point of clear
vision. To enable such a person see the objects situated at infinity, we must use some
lens, so that the image of the object is formed, at the far-off' point of clear vision.
Generally, the lens used is a concave lens and its focal length depends upon the degree
of abnormality in the eye, Figures given below show how shortsightedness can be
corrected with the help of a concave lens.
Fig. 2.2(a) shows a defective shortsighted
eye when the image of a far-off object is
formed in front of the retina.
Fig. 2.2(b) shows a defective shortsighted eye when the object is situated at
the far-off point of clear vision.
Fig. 2.3, shows a corrected shortsighted
eye when image of a distant object is
formed at the far-off point of clear vision.
The focal length of a concave lens in
case of myopic eye is always equal to the
distance of the far point from the eye,
because, the virtual image is always formed at far point.
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Points to Remember
 A short-sighted (myopic) person can see objects at the point of distinct vision
clearly.
 A short-sighted (myopic) person cannot see objects which are far-off.
 Short-signtedness (myopia) occurs (a) due to elongation of eyeball (b) weakening a
ciliary muscles.
 Short-sightedness can be corrected by using a concave lens of appropriate focal
length or power.
2. Long-sightedness or Hypermetropia
Fig. 2.4: A short-sighted person
cannot see far-off objects clearly
A person suffering from this defect can
see far-off objects clearly, but cannot
see clearly the objects situated at a
distance of 25cm or at the distance of
distinct vision.
It must be remembered that the person
is not blind at the distance of 25 cm.
Only the images formed at this point
are blurred.
Causes : (1) Due to some disease or
age, the ciliary muscles become stiff,
and hence, they do not contract. Thus,
they are unable to make the crystalline
lens thicker, with the result that the focal length of the crystalline lens does not
decrease as desired. Thus, the image of the object situated at the point of distinct
vision is formed behind the retina.
(2) Due to some disease, the eye ball becomes smaller in size, and hence, the image of
the object, situated at the point of distinct vision, is formed behind the retina.
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Correction of long-sightedness:
A long-sighted person can see a near object clearly, if it is held at some distance away
from the point of distinct vision (say 75cm). This minimum distance from which a
person can see clearly is called the near distance of clear vision.
To enable such a person to see from a distance of 25cm, a lens must be used, such that
it forms the image of the object at the near distance of clear vision. Generally, the lens
used is convex lens. The focal length of the lens used, depends upon the abnormality
in the eye length. Fig. 2.5 shows a defective long-sighted eye when the object is
situated at the point of distinct vision and its image is formed behind the retina. Fig.
2.6 shows a defective long-sighted eye when the object is situated at the near point of
clear vision and its image is clearly formed on the retina. Fig. 2.7(a) shows a corrected
long-sighted eye when the convex lens forms the image at the point of clear vision.
Points to Remember
 A hypermetropic (long-sighted) person can see far-off objects clearly.
 A hypermetropic (long-sighted) person cannot see
objects at the point of distinct vision clearly.
 Hypermetropic (long-sighted) person is not totally
blind for near objects. He can see these clearly, by
moving objects away from his eyes.
 Hypermetropia occurs:
a. due to shortening of the eye ball.
b. stiffness of ciliary muscles.
Fig. 2.7
(b) : A hypermetropic
person reading by moving book
away from him.
 Hypermetropia can be corrected by using convex lens of appropriate focal length or
power.
3. Presbyopia
In this defect, the eye can see neither the nearer nor the far-off objects clearly. This
defect can be overcome by using two glasses. For long-sightedness, use glasses fitted
with convex lenses and for short-sightedness, use glasses fitted with concave lenses.
Alternately, a bifocal lens (Fig. 2.8(a)) can be used. Its lower half is made of convex
lens and is used for reading and the upper half is made of concave lens and is used for
seeing far-off objects.
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\
Advantages of having Two Eyes in Humans
1. A single human eye has a horizontal field of view of 150°. However, with two
eyes, separated by few centimetres, the horizontal field of view increases to 180°.
Furthermore, the ability to detect faint objects is enhanced with two eyes.
2. The two eyes separated by few centimetres, in the front portion of head, no doubt
limits the field of view to 180°. However, this positioning increases our ability to
judge the depth, a phenomenon called steropsis.
With one eye, the world is two-dimensional and flat, much the same, as you see
images or television screen or a movie. With both eyes open, the world takes on a third
dimension of depth. As our eyes are separated by few centimetres, each eye receives
an image which is slightly different. When our brain combines these images into one,
the sensation of depth is produced.
Here, it is interesting to note that most of the animals and birds have two eyes, placed
almost diametrically on their head. Thus, animals or birds have almost 360° view.
However, they do not have sensation of depth.
Numerical Problems on Human eye
Numerical Problem 1
A nearsighted person wears eye glass with power of - 5.5D for distant vision. His
doctor prescribes a correction of + 1.5D in near vision section of his bifocals, which is
measured relative to main part of the lens.
(i) What is the focal length of his distant viewing part of lens?
(ii) What is the focal length of near vision section of the lens?
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Practice Problems 1
1.
The power of a lens for distant vision for a near-sighted person is -3.5D. If a
correction of +1.0D is required in his bifocal lens, in order to see near objects,
calculate: (i) focal length for distant vision, (ii) focal length for near vision.
[(i)
-28.57 cm, (ii) - 40 cm]
2.
A near-sighted person wears eye glasses of power —4.5D for distant vision. His
doctor prescribes a correction of +0.75D in near vision section of his bifocals,
which measured relative to main part of the lens.
(i) What is the focal length of his distant viewing part of the lens?
(ii) What is the focal length of his near viewing section of the lens?
[(i) -22.22 cm, (ii) -26.67 cm]
EXERCISE
VERY SHORT ANSWER TYPE QUESTIONS
1. Describe and state the function of the following parts of eye.
(i) choroid (ii) cornea (iii) sclerotic (iv) retina (v) optic nerve (vi) yellow spot
(vii) iris (viii) pupil (ix) ciliary muscles (x) crystalline lens (xi) vitreous humour
(xii) aqueous humour.
2. What do you understand by the term accommodation of eye?
3. A person is advised to wear spectacles with convex lenses. What type of defect of
vision is he suffering from?
4. What do you understand by the following terms?
(i) Normal eye
(ii) Defective eye
(iii) Least distance of distinct vision
(iv) Distance of far-off vision.
5. What kind of lens is used in the spectacles of a person suffering from myopia (nearsightedness)?
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SHORT ANSWER TYPE QUESTIONS
1.
(i) Name two common defects of the human eye.
(ii) Name the kinds of lens used for correcting the above defects.
2.
(i) What do you understand by the term myopic eye? (ii) State two causes of the
above defect.
(iii) By drawing neat diagrams, explain how:
(a) myopic eye sees far-off objects (b) myopic eye is corrected by the use of a
lens.
3.
(i) What do you understand by the term hypermetropic eye?
(ii) State two causes of the above defect.
(iii) By drawing neat diagrams, explain:
(a) how hypermetropic eye can see objects at the point of distinct vision,
(b) hypermetropic eye corrected by the use of lens.
4. How are we able to see nearby and also the distant objects clearly?
5. What happens to the image distance in the eye, when we increase the distance of an
object from the eye?
SHORT ANSWER TYPE QUESTIONS
1. Draw ray diagrams each showing (i) myopic eye and (ii) hypermetropic eye.
2. A student sitting at the back of the classroom cannot read clearly the letters written
on the blackboard. What advice will a doctor give to the student? Draw a ray diagram
for the correction of this defect.
3. A person needs a lens of power - 4.5D for correction of her vision. (i) What kind of
defect in vision is she suffering from?
(ii) What is the focal length of the corrective lens? (iii) What is the nature of the
corrective lens?
LONG ANSWER TYPE QUESTIONS
1. Explain the structure and functioning of Human eye. How are we able to see near
as well as distant objects?
2. When do we consider a person to be myopic or hypermetropic? Explain using
diagrams how the defects associated with myopic and hypermetropic eye can be
corrected?
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3.
(i) Draw a diagram to show the formation of image of a distant object by a
myopic eye? How can such a defect be remedied?
(ii) State two reasons due to which this eye defect may be caused.
(iii) A person with a myopic eye cannot see objects beyond a distance of 1.5 m.
What would be power of corrective lens used to restore proper vision?
4.
(i) What is hypermetropia? Draw diagrams to show image formation by an
object by (a) hypermetropic eye, (b) correction made by a suitable lens for
hypermetropic eye.
(ii) We cannot see an object clearly, if it is placed very close to the eye.
Explain, (iii) What is presbyopia? Write two causes of this defect.
5.
A 14-year old student is not able to see clearly the questions written on the
blackboard placed at a distance of 5 m from him.
(i) Name the defect of vision he is suffering from.
(ii) With the help of labelled ray diagrams show how this defect can be
corrected, (iii) Name the type of lens used to correct this defect.
6. Describe with the neat diagram how near sightedness (myopia) can be corrected by
using appropriate lenses.
7.
(i) Explain the following terms used in relation to defects in vision and
correction provided by them :
(a) Myopia
(b) Astigmatism
(c) Bifocal lenses
(d) Farsightedness
(ii) Describe with a ray diagram how a person with myopia can be helped by
spectacles.
8. What is longsightedness? List two causes for development of longsightedness.
Describe with the ray diagram, how this defect may be corrected by using spectacles.
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Dispersion of Light
DISPERSION OF WHITE LJGHT BY A GLASS PRISM
Sir Issac Newton, while working with the astronomical telescope, observed that the
images of stars were coloured near the fringes. He then got the lenses of the telescope
polished, but found that the colour still persisted. From these observations he
suggested that the fault may not be with the lenses, but there is something in the nature
of the light itself. To investigate this conclusion, he performed the following
experiment.
Experiment 1
Newton allowed sunlight to enter
through a small aperture in a
window of a darkened room. He
placed an equilateral prism in the
path of the beam of light. He
received the light emerging from
the prism on a white screen. He
found that the light obtained on a white screen was a band of seven colours,
resembling the rainbow formed after the rain. The order of colours from the base of the
prism was, violet, indigo, blue, green, yellow, orange and red. The order of colours can
be easily remembered by the word VIBGYOR. This phenomenon of splitting of white
light is called, dispersion and the band of colours, spectrum.
Dispersion: The phenomenon due to which a white light splits into its component
colours, when passed through a prism is called dispersion.
Spectrum: The band of seven colours obtained on the screen, when a white light splits
into its component colours is called the spectrum.
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It has been further observed that the bands of colours obtained on the screen have no
sharp boundaries, but merge in each other. Such a spectrum is called the impure
spectrum.
It can be argued that, a prism colours the white light and the light by itself is white. In
order to disapprove the above argument, Newton performed the following
experiments.
Experiment 2
Passing any one colour of the
spectrum through another
prism: White light passing
through a slit, is made to fall
on an equilateral prism A, in a
dark room. The dispersed light
is then received on a white
screen ST, having a small hole in it. The light passing out of the hole (say green light),
is made to pass through another prism B, placed in an inverted position. It is observed
that light on passing through prism B only-gets refracted, but does not suffer any
dispersion.
By moving the screen ST, up or down, the experiment can be conducted for any colour
of light. It is seen that in each case, only refraction takes place.
Thus, the experiment proves that white light gets dispersed at the first face of the prism
only. Afterwards it gets refracted. Hence, prism does not colour the white light.
Experiment 3
Recombination of white light
For this experiment, Newton took two
exactly similar prisms i.e., the prisms were
of the same material and had the same
refracting angle.
He passed white light through a slit and
allowed it to fall on the prism A. He
obtained a spectrum on the white screen.
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Now he interposed prism B, in an inverted position as shown in the side figure. He
found that patch of white light is formed on the white screen. This phenomenon is
called Recombination.
The reason for the above phenomenon is that the second prism B, turned the dispersed
colours towards its base through different angles, such that before emerging from it,
they were incident at the same point. Thus, they rejoined to form white light.
In fact prisms A and B, collectively act like a glass slab with parallel sides. The prism
A disperses white light and is called the dispersing prism, whereas prism B
recombines dispersed light and is called the recombination prism.
Why a glass prism causes dispersion of white light, but not a glass slab?
The glass ABCD can be regarded as
double prism ABD and CBD, joined at
the base BD, The prism ABD acts as a
dispersing prism and hence disperses the
white light. However, the prism CBD acts
as
recombination
prism
and
hence
combines the dispersed light to white
light. Thus, on the whole no dispersion
takes place in a glass slab.
What causes dispersion of white light?
White light is a mixture of several waves of electromagnetic radiations, whose
wavelengths vary from 700 nm to 400 nm. The highest wavelengths produce red
sensation in the eye, whereas the lowest wavelengths produce violet sensation. The
wavelengths between the 700 nm and 400 nm produce the effect of indigo, blue, green,
yellow and orange. These waves travel with same speed (3 x 10s m/s) in vacuum.
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However, on passing through the prism, the waves of different wavelengths slow
down, the red slowing down the least and violet the maximum. This in turn bends the
waves of different colours through different angles. Thus, the white light splits to
component colours or the dispersion of white light takes place.
Monochromatic light
Light of a single colour or single wavelength is called monochromatic light. Light
emitted by sodium vapour is golden yellow in colour and is a monochromatic light.
The word chrome means colour. So, the light of single colour is called
monochromatic. Polychromatic light
A light which made of two or more colours is called polychromatic light. Sunlight is
polychromatic as it is made of seven colours.
Rainbow
Rainbow is one of the most beautiful examples of
spectrum formed due to the dispersion of light in
nature.
The rainbow is produced due to the
dispersion of sunlight by tiny droplets of water
suspended in air, just after rain.
The suspended tiny droplets of water act as
innumerable small prisms. When the sunlight is
incident on the side A of the tiny droplet of water,
it gets refracted as well as dispersed. The dispersed rays on striking the surface B of
the tiny water drop suffer total internal reflection, and hence, moves on towards
surface A. At the surface A, the rays further suffer refraction and emerge out as the
band of colours in the form of a circular arc along the horizon. The red colour appears
on the upper arc of rainbow and violet colour on the innermost arc.
You can also see rainbow on a bright sunny day, in the mist created by a waterfall or a
water fountain.
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Rainbow is always formed in the direction opposite to that of the sun. For
example, if the sun is towards west, the rainbow is formed on the eastern horizon.
ATMOSPHERIC REFRACTION
1. When we sit around a campfire, the face of a person sitting opposite to you,
appears to shimmer. This happens due to refraction of light. The rays of light
reflected from the face of the person, sitting opposite to you, on passing through
the hot air (produced by burning wood), get refracted, Since the hot air is rapidly
moving and its optical density is continuously changing, therefore the path of
refracted rays also changes. This gives rise to shimmering effect.
2. Apparent position and twinkling of stars.
Atmosphere consists of a number of parallel layers of air of varying densities, such
that the most dense layer is near the surface of the earth and the least dense layer is
far away from the surface of the Earth. These layers of air are not stationary, but
continuously intermingle, thereby rapidly changing the density of one or more
layers of the air.
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In Fig. 2.20, A is the true position of a star. When the rays of light coming from this
star pass through the atmosphere of varying density, they bend towards the normal.
Finally, when the refracted rays of light reach the eye, the eye retraces the straight line
path. Thus, to the eye the rays appear to come from a point B, which is higher in the
horizon. This point B gives the apparent position of the star.
The twinkling can be explained by saying, that as the layers of air of different densities
mix, they change the apparent position of star. Thus, when a star is within the line of
sight, it is visible. However, when it falls out of the line of sight, it is no longer
visible. The collective effect of the above changes in apparent position of a star is, that
it appears to twinkle.
3. Why do planets not twinkle?
Planets are very close to us as
compared to stars. Their apparent
position also changes with the
change in density of different
layers of atmosphere. However,
the size of their apparent image is
still fairly large, such that it
seldom falls outside the line of
sight. Hence, they do not appear to twinkle.
4. Why does sun appear bigger during sunset or sunrise?
As already pointed out, the apparent position of a star is higher than its actual position
in horizon. Moreover, due to refraction, the apparent image of star is closer to eye
than its actual position. Since, during sunset or sunrise, the rays of light travel
through maximum length of atmosphere, therefore refraction is also maximum.
Hence, apparent image of sun is very much closer to eye. Thus, it appears bigger.
Further on, even when sun is below horizon, its rays manage to reach earth due to
refraction. It has been found that sun is visible to us about 2 minutes before the actual
sunrise and 2 minutes after the actual sunset, because of atmospheric refraction.
Furthermore, the light from the sun can be seen in the atmosphere, because of the
atmospheric refraction before the actual sunset or sunrise. This, in a way gives rise to
twilight. Thus, refraction helps in increasing the length of the day.
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SCATTERING OF LIGHT
The interplay of white light (sunlight) with the molecules of the air and tiny droplets of
water, suspended particles in air (such as dust and carbon particles), etc., gives rise to
several spectacular phenomena in nature. The blue colour of sky, the reddening of Sun
and horizon at sunrise or sunset, the yellowish colour of sunlight, the distant hills
appearing blue, the smoke coining out of chimney appearing blue, etc., are some
familiar phenomena.
What is scattering?
The phenomenon due to which a particular wave of light is absorbed by a particle,
which is greater in diameter than the wavelength of light and then transmits it in all
possible directions is called scattering of light.
It has been established that when an incident ray of light strikes a particle which
has a diameter greater than the wavelength of incident light, then incident light is
first absorbed by the particle and then transmitted in all possible directions. This is
called scattering of light.
Fig. 2.22 shows a large particle whose diameter is greater than the wavelength of the
blue light, but smaller than the wavelength of red light. Now, when a mixture of red
and blue light is made incident on such a particle, then the blue light is absorbed by the
particles and then transmitted in all possible directions, i,e., the blue light is scattered.
The red light, however, continues moving straight as it is not absorbed or scattered.
Now, if your eyes receive this scattered blue light, then to you the particle will appear
blue in colour.
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When the white sunlight enters the atmosphere of earth, the particle size is smallest.
Thus, the violet light which has the smallest wavelength in white light scatters.
From the above statement it does not imply, that all the violet waves present in
sunlight scatter. Instead it means that if only a particular ray, which strikes a particle,
whose diameter is greater than the wavelength of violet light gets scattered. From this
it follows that quite a large number of rays of white sunlight will reach us without
scattering.
As the white light continues its journey through the atmosphere, the number and
particle size present in the air increase. This results in the scattering of indigo and blue
wavelengths. If the white light travels through a longer distance through the
atmosphere, as during sunset or sunrise than even yellow, orange and red scatter. The
red light has wavelength 1.8 times the wavelength of violet light, and hence, scatters
last of all.
Some Common Phenomena of Atmospheric Scattering
1. Why does sky appear blue?
When the white sunlight passes through the atmosphere, the violet, indigo and blue
wavelengths, encounter suspended particles in air whose diameter is more than their
wavelengths. Thus, these waves are absorbed and then scattered in all possible
directions. The scattered light from these particles, suspended all around in the sky
reach our eyes, then the sky appears blue.
2. Why is the sunlight reaching the earth yellowish?
When the white sunlight passes through upper atmosphere, the violet, indigo and blue
colours scatter which makes the sky to appear blue. However, the white light gets
deficient in violet, indigo and blue colour on account of scattering. Thus, it appears
yellowish instead of white.
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3. Why does the sun appears yellowish?
When the violet, indigo and blue colours scatter in the upper atmosphere, the resultant
sunlight is yellowish in colour. When this light enters our eyes, then to us sun appears
yellowish instead of white hot.
4. Why does the sky appears dark instead of blue to an astronaut?
In space no particles are present. Thus, no scattering of light takes place. Hence, the
sky appears dark as light by its own nature is invisible, but produces in us the
sensation of vision.
5. Why do the sun and the horizon appear reddish during sunset or sunrise?
During sunset and sunrise, the sun light travels the maximum distance through the
atmosphere. With the increase in distance, the size and number of particles suspended
in air increases. Thus, not only the violet, indigo or blue, but yellow, orange and red
wavelengths of white light scatter. As the red light scatters last of all and is nearest to
the eye, therefore, the sun and the horizon appear reddish.
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However, as the sun rises up the horizon, the distance travelled by sunlight in the
atmosphere decreases, and hence, yellow, orange or red light
A
" colours are not
scattered. Thus, sky appears blue and the sunlight yellowish. Most of the colours of
white light scatter. The red colour scatters close to the eye and hence the sun and sky
appear red.
6. Why do the distant hills appear blue?
In the thick growth of trees on the hills there are always present some amount of tiny
droplets of water in the air. When the white light passes through this moisture laden
air, the blue light is scattered. When this scattered light reaches our eyes, the hills
appear blue.
7. Why does smoke coming out of coal fired chimney appear blue on a misty day?
The tiny particles of smoke and moisture
scatter blue colour of the white light
passing through it. When this scattered
blue light reaches our eyes, the smoke
appears blue.
It is for the same reason that on a misty
day, the smoke coming from an incense
stick appears blue.
8.Why do motorist use orange lights, rather than normal white light on a foggy day?
If a motorist uses white light while driving in fog, then the tiny droplets of water
scatter large amount of blue light. This scattered blue light on reaching the eyes,
decreases visibility, and hence, driving becomes extremely difficult. However, when
orange lights are used, they don't get scattered on account of their longer wavelengths,
and hence, the driver can see ahead clearly.
It is for the same reason that rescue workers wear orange coloured suits. Similarly,
children's buses are painted orange, so that they are easily visible in the foggy
conditions, smoke or rain.
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9. Why is red light used as a universal danger signal?
The red light has largest wavelength among the spectral colours, and hence, is least
scattered. Thus, red light can easily pass through fog or mist or smoke without getting
scattered, and hence, is visible from a long distance. Thus, it is used as universal
danger signal.
Let us perform the following experiment to show scattering of white light.
Experiment
Dissolve 100 g of sodium thiosulphate (hypo) crystals in 1 litre of distilled water. Fill a
rectangular glass container with the hypo-solution.

On the left-hand side of the recwigular glass container arrange a source of light
(flash light or torch light), a cardboard screen with a hole Cj and a convex lens
Lj, such that a parallel beam of light is incident on the hypo-solution.

On the right-hand side of rectangular glass container arrange another cardboard
screen with a hole C2, a convex lens L2 and a white screen, such that the light
coming out of hypo-solution is focussed on the white screen.

Notice that the path of the light is not visible in the hypo-solution. It is because
the particle size in the hypo solution is so small that, the scattering of light does
not take place.

Now add 1 ml of concentrated sulphuric acid in the hypo-solution and stir its
contents with a glass rod.
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Physics Grade 12
You wUl observe
1. In 2-3 minutes, the three sides of the rectangular glass container start giving out blue light.
2. The light coming out of the rectangular glass container is initially orange-red. This
light gradually changes-to crimson-red.
Reason: When sulphuric acid reacts with the hypo-solution, it forms colloidal sulphur.
Initially, sulphur particles are very small, and hence, they scatter blue light. Thus, the
light passing out of rectangular glass container is or'^^e m colour. However, as the
reaction proceeds, more and more colloidal sulphur particles are formed. These
sulphur particles then join to form bigger colloidal particles. These bigger particles
scatter most of the colours of white light except red colour, and hence, the light
coming out of rectangular container is crimson-red.
EXERCISE
MULTIPLE CHOICE QUESTIONS
1. Rainbow is formed due to:
(A) refraction of light
(B) diffraction of light
(C) reflection of light
(D) dispersion of light.
2. Stars appear to twinkle because:
(A) of atmospheric refraction
(B) movement of air
(C) both (A) and (B)
(D) none of the above.
3. The sky generally appears blue, because the colour which scatters closest to the eye
is:
(A) violet
(B) indigo
(C) blue
(D) violet and indigo.
4. During sunset or sunrise the sun appears reddish because:
(A) at this time sun is not very hot.
(B) sun produces red light at this time
(C) due to longer passage in atmosphere, even red light in the sunlight scatters
(D) none of the above.
5. The danger signals installed at the top of tall buildings are red in colour. These can
be easily seen from a distance because among all other colours, the red light:
(A) is scattered the most by smoke or fog
(B) is scattered the least by smoke or fog
(C) is absorbed the most by smoke or fog
(D) moves fastest in air
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VERY SHORT ANSWER TYPE QUESTIONS
1. Why do different components of white light deviate through different angles when
passing through a triangular glass prism?
2. Name the components of white light deviate through different angles when passing
through a triangular glass prism?
3.
What will be the observed colour of the sky on a planet where there is no
atmosphere? Why?
SHORT ANSWER TYPE QUESTIONS
1. Draw a neat diagram to show the refraction of light through an equilateral glass
prism. On the diagram clearly show:
(i) angle of incidence
(ii) angle of refraction
(iii) angle of emergence
(iii) angle of deviation
(iv) angle of prism.
2. What do you understand by the terms:
(i) dispersion
(ii) spectrum?
3. In an experiment with a rectangular glass slab, a student observed that a ray of light
incident at an angle of 60° with the normal on one face of the slab, after refraction
strikes the opposite face of the slab before emerging out into air making an angle of
42° with the normal. Draw a labelled diagram to show the path of this ray. What value
would you assign to the angle of refraction and angle of emergence?
4. State the reason for the following observations recorded from the surface of moon.
(i) Sky appears dark
(ii) Rainbow is never formed
SHORT ANSWER TYPE QUESTIONS
1.
(a) What is a rainbow? How is a rainbow formed?
(b) Which spectral colour is on:
(i) outer arc of a rainbow
(ii) innermost arc of a rainbow?
2. Why do the faces sitting on the opposite side of campfire appear to shimmer?
3. A star appears on the horizon. What is the true position of star? Explain with the
help of a diagram.
4. Why do the stars twinkle, but not the planets?
5. Why do sun or moon appear bigger in size, when they are just going to rise or set?
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6. Why the sun during sunrise becomes visible, when actually, it is below the horizon?
7. Why does the sky appear blue? Explain.
8. Why does the sky appear dark instead of blue to an astronaut? Explain.
9. Why is the sunlight reaching the earth yellowish? Explain.
10. Why do the distant hills appear blue? Explain.
11. Why does the sun appear red during sunrise or sunset? Explain.
12. Why are orange coloured lights used by the motorists as fog lights? Explain.
13. Why is red coloured light used as a universal signal for danger? Explain.
14. How will you use two identical prisms so that a narrow beam of white light
incident on one prism emerges out of the second prism as white light? Draw the
diagram.
15. Draw a ray diagram showing the dispersion through a prism when a narrow beam
of white light is incident on one of its refracting surfaces. Also indicate the order of the
colours of the spectrum obtained.
16. Is the position of a star as seen by us its true position? Justify your answer.
17. Why do we see a rainbow in the sky only after rainfall?
18. Why is the colour of the clear sky blue?
19. What is the difference in colours of the Sun observed during sunrise/sunset and
noon? Give explanation for each.
20. Give reasons for the following:
(i) Colour of clear sky is blue.
(ii) The sun can be seen about two minutes before actual sunrise.
21. To an astronaut, why does the sky appears dark instead of blue?
LONG ANSWER TYPE QUESTIONS
1.
Explain the refraction of light through a triangular glass prism using a labelled
ray diagram. Hence define the angle of deviation.
2.
How can we explain the reddish appearance of the sun at sunrise or sunset? Why
does it not appear red at noon?
3.
Explain the phenomenon of dispersion of white light through a glass prism, using
suitable ray diagram.
4.
How does refraction take place in the atmosphere? Why do stars twinkle but not
the planets?
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5.
(i) What is dispersion of white light? What is the cause of such dispersion?
Draw a diagram to show dispersion of white light by a glass prism.
(ii) A glass prism is able to produce a spectrum when white light passes through
it, but a glass slab does not produce any spectrum. Explain, why it is so?
6.
Explain how the ray of white light is dispersed. Why does this take place? Which
colour deviates more? Why?
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Modern Physics
ELETROMANAGNETIC SPECTRUM
The properties common to all types of electromagnetic waves include:
(i)
They consist of varying electric and magnetic fields which are perpendicular to
each other.
(ii) They travel at the speed of light, i.e., 3 x 108m/s in vacuum.
(iii) They are transverse waves.
(iv) They are not affected by electric and magnetic fields. Carry no charge.
(v)
They undergo reflection, refraction, polarisation and difraction..
(vi) They travel in straight lines.
(vii) They transfer energy (E = hf)
Generation of Electromagnetic Waves
Type
Gammarays
Example of
Detection
generation
 Energy changes  G-M tube.
inside the
 Lonisation
nucleus.
chambers.
 Photographic
films.
 Energy changes
outside the
nucleus.
Ultra- violet  Carbon arc.
 Mercury
vapour lamp.
 The sun.
X-rays
 Photographic
film.
 Photocells.
 Photographic
film.
 Fluorescent
materials.
Uses
 Detecting faults in welded
materials.
 Controlling thickness of
materials in industries.
 Treatment of malignant
growth.
 Locate bone fructures.
 destroy cancer cells, etc.
 Produces ionisation and
fluorescence in photography.
 photoelectricity.
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Visible
light
Infrared
 Discharge tube.
 Incandescent
lamps.
 Lasers, flames.
 Molecular
vibrations.
Microwaves  Magnetrons,
lasers.
 Electrical
oscillations.
Radio
waves
 Theeye.
 Photocell.
 Films.
 Thermopile.
 Photographic
films.
 Crystal
detectors.
 In photosythesis.
 Photography.
 The eye (vision).






 Radio receivers. 
Produces heating effect.
Photgraphy
Monitoring (security)
Radar.
Electron spin resonace.
Heating effect.
Radio comunication.
Worked Example
Determine the frequency of yellow light given that the wavelength is 6 x 10
-7
and
speed of light c = 3x 108m/s.
Solution
3  10 8 m / s
f  
 6  10 7 Hz
c
= 5 x 1014 Hz
Revision Exercise
1. A source emitted a radiation of energy 1.29eV. Find the wavelength of the
radiation.
2. Given the range of frequency for ultraviolet rays is 7.9 x 1014 Hz to 5 x 1Q17 Hz.
What is the corresponding range of energies of the photons for ultraviolet light?
(Planks constant h = 6.62 x 10- 34 Js)
3. A radio station broadcasts at a frequency of 5 x 107 Hz, Calculate the wavelength
of it's broadcasts.
Cathode rays
 Cathode rays are streams of electrons moving at high speed from a heated
cathode. The properties of cathode rays are as follows:
(i)
They travel in straight lines from the cathode
(ii)
They cause certain substances to fmoresce.
(iii)
They possess kinetic energy
(iv)
They are deflected by magnetic and electric fields
(v)
They produce X-rays on striking a metal target.
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Physics Grade 12

The speed of the cathode rays in the electric field is expressed as eV =
1
mv2,
2
where m is the mass of the electron, V the voltage, e the charge and v the speed.
v
2eV
m
 The cathode ray oscilloscope consists of three parts, namely;
(i)
The electron gun, which is composed of the cathode, the grid and the anode.
It controls the brightness and focuses the beam on screen.
(ii) The deflecting plates, to allow horizontal and vertical deflections of the
beam.
(iii) A fluorescent screen, on which the beam produces a spot of light. The screen
is coated with a phosphor such as zinc sulphide, which fluoresces under the
impact of the electrons.
Fig. 30.1
Note:
There must be a return path for the electrons that hit the screen. This is provided by
coating the inner wails of the tube with graphite, which is electrically connected to A,
and is therefore at earth potential. When the electrons hit the zinc sulphide, they knock
electrons from it. The secondary electrons are collected by the graphite coating and
flow to earth.
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The cathode ray oscilloscope is used for:
(i)
measuring a.c. and d.c. voltages.
(ii)
(iii)
(iv)
(v)
measuring frequencies.
measuring phase differences.
measuring small time intervals,
study wave pattern (forms).
X-Rays
X-rays are electromagnetic radiations of
short wavelengths of the order 10 -10m.
An
X-ray
tube
is
usually
highly
evacuated and contains an anode made of
copper block and on which tungsten or
molybdenum (high-melting point metals)
is embedded. The purpose of copper
block is to conduct heat away from the
target see the figure 30.2.
Fig. 30.2 X-ray tube
The copper block is cooled by circulating oil and the use of cooling fins. Electrons are
produced from the filament by thermionic emission and are accelerated to the anode
by a high potential difference. On collision with the target electrons decelerate rapidly
and X-rays are produced, but about 99.5% of the kinetic energy goes into producing
heat.
• If an electron has mass m and velocity v on striking the target, its energy hf is given
by; hf =1/2mv2 where h is Planck's constant and f is the frequency of the radiation.
With a potential difference V across the X-ray tube, an electron of charge, e gains
kinetic energy eV. Hence;
1
mv2 = eV
2
hf = eV ( h= 6.63 x 10-34 Js)
The intensity of the X-rays increases with the number of electrons hitting the target
and therefore depends on the temperature of the filament, which in turn is controlled
by the heating current.
The penetrating power or quality of the X-rays increases with the operating voltage of
the tube.
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Physics Grade 12

Soft X-rays have low penetrating power, and are produced by relatively low
voltage.

Hard X-rays have high penetrating power, produced fay high voltage.

Properties ofX-Rays:
(i) They travel in straight lines, at the speed of light.
(ii) They cannot be deflected by electric or magnetic fields.
(iii) They readily penetrate matter. Penetration is least with materials of high
density" and high atomic mass.
(iv) They can be reflected through very large angles of incidence.
(v) Refractive indices of all materials are very close to unity for X-rays.
(vi) There is mining boundary. They can be diffracted.
X-rays are detected by the following properties:
(i) They cause fluorescence in certain substances
(ii) They affect photographic emulsion
(iii) They ionise gases
(iv) They are not deflected by electric or magnetic fields.

X-rays are used for:
(i) locating bone fractures.
(ii) destroying cancer cells.
(iii) locating internal imperfections in welded joints and castings.
(iv) studying of crystal structures.
(v) Very soft x-rays are used for determining the authenticity of works
of art and picture restoration.
Precautions when using X-rays:
(i)
Minimise the exposure time as much as possible.
(ii)
Workers exposed to X-rays should always wear protective clothing.
Dangers of X-rays:
(i)
Kill living cells when overexposed
(ii)
Interfering with gene structure.
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Photoelectric Effect
Photoelectric effect is the process whereby electrons are emitted from matter when an
electromagnetic radiation falls on it.
Factors affecting Photoelectric Emission
(i)
The number of photoelectrons emitted per second is proportional to the intensity
of the incident radiation. The intensity of the radiation has no effect on the
kinetic energies of the emitted electrons.
(ii) The emitted electrons have various kinetic energies with velocities ranging from
zero up to some maximum value, which increases as the frequency of the
radiation increases. This energy is independent of the intensity of the radiation.
(iii) Emission of electrons occurs only if the frequency of the incident radiation is
above some minimum value called threshold frequency, which depends on the
particular metal being irradiated. If the incident radiation is a single frequency,
the number of electrons emitted per second is proportional to the intensity of the
radiation.
Work function of a metal is the minimum energy required to liberate an electron from
the surface of the metal and it is expressed as hfo, where fo is the threshold frequency.
The excess energy hf-hfo appears as kinetic energy of the emitted electron, which
escapes with a velocity (V max ).
1
Thus, hf- hfo = mv2max
2
Experiments can be set to determine the maximum K.E, for a given radiation on a
given metal surface. In this case a potential is set in the negative direction to oppose
the movement of electron.
The minimum potential required to stop the electrons is called stopping potential (Vs).
The work done against the electron is thus equal to the maximum K.E
eVs =
1
MV2
2
Therefore eVs = hf – hf0
Vs =
h
h
f  f0
e
e
A graph of Vs again f
I ev = 1.6 x 10-19 J.
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Photocell changes radiation into an electric current. If light
of a frequency which is greater than the threshold
frequency of the cathode-coating is incident on the
cathode, electrons are emitted and move to the anode.
Examples of photocells are photo -emissive cell and
photoconductive cells.
Photocells are used in counting systems, intruder alarms, sound reproduction from
film, light-meters, artificial lighting, etc.
Worked Examples
1. The figure below shows the features of a cathode ray tube.
a) Explain how the electrons are produced in the tube.
(b) What is the purpose of the anodes?
(c) Why is the tube evacuated?
(d) Explain how the grid controls the brightness of the spot on the screen.
(e) State two uses of a CRO.
Solution
(a) Produced by the cathode when heated by the filament.
(b) To accelerate and focus the electrons to a narrow beam.
(c) To avoid electrons from colliding with air molecules.
(d) Making the grid more negative repels the electrons reaching the screen.
(e)
(i) As a voltmeter
(ii) Determine frequency
(ii) Study waveforms.
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2.
X-rays are produced by a tube operating at 104 V. Calculate the wavelength of the
radiation. (Take h = 6.63 x l0-34Js, e= 1.6 x l0-19C
c = 3x x l08m/s).
Solution
KE on impact = eV
eV = 1.6 x 10-19xl04= 1.6 10-15 J
The energy of X-rays is hf, where f =
c
min
c
min
= 1.6 × 10-15
 min 
6.63  1034  3  108
1.6  1015
 min  1.243 ×10-10m
3.
X-rays are emitted when a tube operates at 2 x 104 V and a current of 0.015 A is
passing through it. Calculate:
(a) the number of electrons striking the target per second.
(b) the velocity of the electrons on hitting the target.
(c) the power input.
(d) the minimum wavelength of the X-rays emitted.
Solution
(a) I = n x e, where n is the number of electrons.
n
=
I
e
0.015
1.6 10 19
= 9.375 × 1016 electrons
(b) Velocity of the electrons is given by:
1
eV- mv2; e=l.6x 10-19C,m = 9x 10-31 kg, V= 2 x 104 V
2
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Physics Grade 12
4. Given that sodium has a work function of 3.68 x 10"19 joules, calculate:
(a) its threshold frequency.
(b) the maximum velocity of the electrons emitted by light of wavelength 4.10
x 10~7m.
(c) the stopping potential with light of this wavelength.
(Takeh-6.63x 10 34Js, c = 3x 108 ms
19
-1
and me = 9 x 10-31kg and e = 1.6 x 10-
C).
Solution
(a) hfo = W. where W is the work function.
6.63 x 10-34fo = 3.68 x 10-19J
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Physics Grade 12
(c) The stopping potential V is given by;
Electrons emitted from an electron gun are accelerated by a voltage of 500 V.
Given that e = 1.6 x 10 19C and m - 9 x 10 3i kg, calculate:
(a) the energy of the electrons.
(b) the speed of the electrons.
Solution
(i)
Energy = eV
= 1.6 x 1019x500 =
= 8 x io-17 J
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Physics Grade 12
Revision Exercise
1. Define the term photoelectric effect.
2. State three laws of photoelectric emission.
3. Explain the role of stopping potential in photoelectric emission.
4.
(a) What do you understand by the term photocell?
(b) State three applications of photocells in daily life.
5.
(a) With the aid of a well-labelled diagram, describe how an X-ray tube works.
(b) Explain how:
(i) soft X-rays are produced and controlled.
(ii) hard X-rays are produced and controlled.
6. Explain how heating is controlled in an X-ray tube.
7. List three properties of X-rays that can be used to detect them.
8. Derive a formula for minimum wavelength X,mm in the production of X-rays.
9. State four applications of X-rays in daily life.
10. Explain how the radiation from an X-ray tube is affected by changing:
(a) the potential difference between the filament and the target.
(b) the material of the target.
(c) the current from the filament.
11. State briefly how you would control electrically:
(a) the intensity of the emitted X-rays.
(b) the penetrating power of the emitted X-rays.
12. Draw and label a cathode ray oscilloscope. Explain its operation.
13. Explain how a cathode ray oscilloscope can be used to display the current-voltas
characteristic of a diode.
14. State how a circular trace can be obtained on the screen of a cathode ray
oscilloscope.
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Physics Grade 12
Nuclear Physics
SECTIONA: ATOMIC PHYSICS
Soon after the discovery of electron and proton in an atom the quest started to find the
way in which these charged particles are present in an atom. From his experiments
Ernest Rutherford developed a nuclear model of the atom. His model of the atom
consisted of a small dense, positively charged nucleus with negative electrons orbiting
about it. In 1920 Rutherford suggested that there is probably another particles with the
nucleus, neutral one, to which he gave the name neutron. James Chardwick discovered
neutron in 1932.
ATOMIC MODEL
J.J Thomson believed that the atom is a sphere with a
positive and negative charges scattered inside it. He
said that there is an attraction between the unlike
charge.
Rutherford discovered that atoms have central
positively charged part known as nucleus of the atom and the electrons revolve in an
empty space of the atom.
Bohr discovered that the electrons revolve around the nucleus in a special orbit known
as the energy levels and he proved that the angular momentum of the electron is
defined by the following equation.
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Physics Grade 12
ATOMIC STRUCTURE
At the centre of each and every atom there is an infinitesimally small nucleus. The
entire positive charges of the atom and about 99.9 percent of its mass is concentrated
in the nucleus. The tininess of the nucleus can be imaginated by comparing that the
radius of the atom is 105 times the radius of the nucleus.
A nucleus consists of nucleons comprising of protons and neutrons. A proton has a
positive charge equal to 1.6x10-19C and its mass is 1.673x10-27kg. A neutron has no
charge on it, but its mass is 1.675 x10-27kg. The mass of a neuron is almost equal to
mass of proton. To indicate the mass of atomic particles, instead of kilogram, unified
mass scale (u) is generally used. By definition 1u is exactly one twelveth the mass of
carbon 12 atom (1u =1.6606x10-27kg). In this unit the mass of a proton is 1.007276 u
and that of a neutron is 1.008665 u while that of an electron is 0.00055u.
The charge on a proton is equal in magnitude to the charge on an electron. The charge
on the proton is positive while that of an electron is negative. As an atom on the whole
is electrically neutral, therefore, we can conclude that the number of protons inside the
nucleus. The number of protons inside a nucleus is called the atomic number or the
charge number of an atom. It is denoted by Z. thus the total charge of any nucleus is
Ze, where e indicates charge on one proton.
The combined number of all the protons and neutrons in a nucleus is known as its
mass number and denoted by A.
The number of neutrons N present in a nucleus is given by we now consider different
elements of the periodic table Hydrogen atom is simplest of all the atoms. Its nucleus
is composed of only one proton; that is for hydrogen A=1, Z=1.
ISOTOPES
Isotopes are such nuclei of an element that have the same charge number Z, but have
different mass number A, that is in the nucleus of such an element the number of
protons is the same, but the number of neutrons is different. Helium, for example has
two isotopes. These are symbolically represented as
and
. As the charge
number of helium is 2, therefore, there are two protons in the helium nucleus.
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Physics Grade 12
the chemical properties of all the isotopes of an element are alike, as the chemical
properties of an element depend only upon the number of electrons around the nucleus,
that is upon the charge number Z, which for all the isotopes of an element is the same.
It is, therefore, not possible to separate the isotopes of an element by chemical
methods physical methods are found to be successful for this purpose. A device with
the help of which not only the isotopes of any element can be separated from one
another but their masses can also be determined quite accurately is called massspectrograph.
RADIOACTIVITY
Radioactivity is the explosion of the nucleus of the atom. The key feature of
radioactivity that makes it so fascinating is that the energy released is enormous at
least when compared to typical chemical energies. The typical energy release in the
explosion of one nucleus of one atom is about million times greater than in a chemical
explosion of a single atom.
Elements with great atomic number emit rays and particles naturally. This property is
known as the radioactivity and Henary Bacquerel discovered the radioactivity when he
put a piece of uranium and photographic plates in the same place.
NATURE OF RADIOACTIVITY
1. They affect photographic plates.
2. They ionize the air molecules.
3. They have a fluorescent effect.
4. They have physiological effect.
5. They pass through a sequence of decay.
RADIOACTIVE EMISSIONS
The radioactive elements emit:
1. Alpha-particles (α): are helium nuclei (
) that are positively charged. When the
radioactive atom emits alpha particle, the atomic weight decreases by 4 a.m.u and
the atomic number decreases by 2 a.m.u.
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Physics Grade 12
Example
a)
b)
c)
d)
2. Beta-particles (β): is similar to electrons with high velocity (
) and negatively
charged. When a negatively charged. When a beta particle is emitted, the mass
number is constant, but the atomic number increases by 1.
Example 2
a.
b.
3. Gamma (γ) rays are electromagnetic waves and are uncharged. When an atom
emits γ-rays, it is charged to more stable atom.
MASS DEFECT AND BINDING ENERGY
It is usually assumed that the whole is always equal to the sum of its parts. This is not
so in the nucleus. The results of experiments on the masses of different nuclei show
that the mass of the nucleus is always less than the total mass of all the protons and
neutrons making up the nucleus. In the nucleus the missing mass is called the mass the
defect M given by
As Z is the total number of protons in the protons in the nucleus and
a proton, then Z
as
is the total mass of all the protons. (A-Z) is the total neutrons and
is the mass of a single neutron, (A-Z)
The term
is the mass of
is the total mass of all the neutrons.
is the experimentally measured mass of the entire nucleus. Mass
defect represents the difference in mass between the sum of the masses of its
constituents and the mass of the nucleus itself.
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Physics Grade 12
The missing mass is converted to energy in the formation of the nucleus. This energy
is found from Einstein’s mass energy relation and is called the binding energy (B.E) of
the nucleus.
From equations above, the binding energy of a nucleus is
HALF LIFE AND DECAY CONSTANT
The time in which half of the atom decays is called half life (T½). The ratio of the
number of atoms that decays in unit of time to the whole number of the atoms is called
decay constant (λ).
Derivation
No= Initial Mass
when t=0
N=
N= No e-
and N = Half mass, t= T1/2
λt
= No e-λt1/2
ln( ) = - λt1/2
T1/2 =
T1/2 =
LAW OF RADIOACTIVE DISINTEGRATION
No at
t= 0
N
at
t
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Physics Grade 12
DN = - λNdt
=
Ln N – Ln No = - (t) + (0)
Ln(
)=- t
Example 3
The half life of radioactive substance is 30sec, find:
a. Decay constant
b. Time taken for the sample to decay
Solution:
a.
= 30s
of its intial value.
=?
=
b.
FISSION AND FUSION
There are two main types of nuclear reactions that can release energy:
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Physics Grade 12
Fission: The process of causing a large nucleus (A > 120) to split into multiple smaller nuclei,
releasing energy in the process.
1. It can start when the large nuclei absorbs a neutron, causing it to become unstable
to the point that it falls apart.
2. This is the reaction that we use in nuclear power plants and nuclear weapons.
3. Fission is relatively easy to do, but also leaves us with lots of nuclear waste that
must be stored for thousands of years before it is safe.
Example 4
Fusion is the process of causing small nuclei to stick together into a larger nucleus, in
the process releasing energy or union of small atoms to form a large atom.
Example 5
APPLICATION OF RADIOACTIVY
Carbon dating
The beta of
is commonly used to date organic sample.
Cosmic rays (high energy particles from outer space) in the upper atmosphere cause
nuclear reactions that create 14C from 14N.
The ratio of 14C to 12C as constant value of about 1.3 × 10-12 as determined by
measuring carbon ratios in three rings. Using carbon dating sample of wood, charcoal,
bone and shell have been identified as having lived from 1,000 to 25,000 years ago.
This knowledge has helped scientists and researchers to reconstruct the history of
living organism including humans- during this time span.
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Physics Grade 12
Example 6
A wooden coffin is found that contains a skeleton holding a gold statue which of three objects
(Coffin, Skeleton and Gold statue) can be carbon dated to find out how old it is?
Solution
Only coffin and the skeleton can be carbon dated these two items exchanged air with
the environment during their lifetimes, resulting in a fixed of carbon-14 to carbon-12.
Once the human and the tree died, the amount of carbon-14 begins to decrease due to
radioactivity decay. The gold in the statue was never to detect.
Example 7
A 50g sample of carbon is taken from the pelvis none of a skeleton and is found to
have a carbon-14 decay rate of 200decay/min. It is known that carbon from a living
organism has a decay rate of 15decay/min.g and that 14C has a half-life of 5730 years
× 109min. find the age of the skeleton.
Solution
Question2
The normal activity of living carbon containing matter is found to be about 15decays
per minute for every grams of carbon. This activity arises from the small proportions
of radioactive 14C present with the ordinary carbon isotope 12C when the organism
dead. Its interaction with the atmosphere (Which remains the above equilibrium
activity) ceases and its activity begins to drop from the known half-life (5730 years) of
14C dating used in archaeology. Suppose a specimen from Mohenjadaro gives an
activity of 9decays/minute/grams of carbon estimate the approximate age of the Indus
valley civilization.
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Physics Grade 12
Solution
Exercise
Instructions: Fill the blank spaces
1. (______)The sum of the number of protons and neutrons making the nucleus is
called mass number.
2. (_______) The electrons present inside the nucleus.
3. (_______) The energy equivalent to mass of the electron is 0.51Mev.
4. (_______) In heavy nuclei the ratio of the number of protons and the number of
neutrons is less than unity.
5. (_______) 1curie=3.7x1010decays/sec
6. (_______) Uranium series to Actinium series to Thorium series to Neptunium
series is called radioactive series.
7. (_______) The process of splitting up of a heavy nucleus line
two nearly equal fragments, when it
into
is bombarded with suitable slow
neutrons is called fission.
8. (_______) The spontaneous emission of radiation is called Decay constant.
9. (_______) The unity of decay rate is /sec.
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Physics Grade 12
Basic Electronics
10.1 Semiconductors
10.2 Logic gates
10.3 Diodes
10.4 Transistors
10.5 Integrated Circuits
10.6 Amplifiers
10.0 Introduction
Electronics essentially deals with electonic devices and their utilization.An electonic
device is that in which current flows through vacuum or gas or semiconductors.
Electronics gained much importance due to its numerous applications in our daily life.
Electronic devices are capable for performing functions:
1.
Rectification means
the conversion of alternating current to direct electric
current.The D.C supply is used for charging storage batteries,field supply of D.C
generators,electroplating, etc.
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Physics Grade 12
2.
Amplification means the process of raising the strength of weak signal.The
devices that raise strength of weak signal are amplifiers. Amplifiers are used in
radioset , public address system, television so that the weak signal can be heard
loudly.
3.
Control devices are devices that are widely applied in automatic control.Speed
of a motor,voltage across the refrigerator can be automatically contolled with the
help of such devices.
4.
Generation is an electronic device that can convert d.c power to a.c power of
any frequency.For example electronic high frequency heating is used for
annealing and hardening.
5.
Conversion
of
light
into
electricity.
Electronic devices can convert light to
electricity.This
coversion
of
light
into
electricity is known photo-electricity. photoelectric
devices
are
used
in
Buglar
alarms,sound recording on motion pictures
etc.
6.
Conversion
of
electricity
into
light.
Electronic devices can convert electricity to
light.This valuable property is ulitized in televion and radar.
10.1 Semiconductors
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Physics Grade 12
Certain substances like germanium, silicon, carbon etc. are neither good conductors
like copper nor insulators like glass. In other words, the resistivity of these materials
lies in between conductors and insulators.
semiconductors.
Such substances are classified as
Semiconductors have some useful properties and are being
extensively used in electronic circuits.
For instance, transistor—a semiconductor device is fast replacing bulky vacuum tubes
in almost all applications. Transistors are only one of the family of semiconductor
devices; many other semiconductor devices are becoming increasingly popular.
A semiconductor is a substance which has resistivity (10−4 to 0.5 Ω m) in between
conductors and insulators e.g. germanium, silicon, selenium, carbon etc.
Properties of Semiconductors
(i)
The resistivity of a semiconductor is less than an insulator but more than a
conductor.
(ii) Semiconductors have negative temperature co-efficient of resistance i.e. the
resistance of a semiconductor decreases with the increase in temperature and
vice-versa. For example, germanium is actually an insulator at low temperatures
but it becomes a good conductor at high temperatures.
(iii) When a suitable metallic impurity (e.g. arsenic, gallium etc.) is added to a
semiconductor, its current conducting properties change appreciably.
Commonly Used Semiconductors
There are many semiconductors available, but very few of them have a practical
application in electronics. The two most frequently used materials are germanium (Ge)
and silicon (Si). It is because the energy required to break their co-valent bonds (i.e.
energy required to release an electron from their valence bands) is very small; being
about 0.7 eV for germanium and about 1.1 eV for silicon.
1.
Germanium. Germanium has become the model substance among the
semiconductors; the main reason being that it can be purified relatively well and
crystallized easily. Germanium is an earth element and was discovered in 1886.
It is recovered from the ash of certain coals or from the flue dust of zinc smelters.
Generally, recovered germanium is in the form of germanium dioxide powder
which is then reduced to pure germanium.
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Physics Grade 12
The atomic number of germanium is 32. Therefore, it has 32 protons and 32
electrons. Two electrons are in the first orbit, eight electrons in the second,
eighteen electrons in the third and four electrons in the outer or valence orbit
[See Fig. (i)]. It is clear that germanium atom has four valence electrons i.e., it
is a tetravalent element. Fig (ii) shows how the various germanium atoms are
held through covalent bonds. As the atoms are arranged in an orderly pattern,
therefore, germanium has crystalline structure.
2.
Silicon. Silicon is an element in most of the common rocks. Actually, sand is
silicon dioxide. The silicon compounds are chemically reduced to silicon which
is 100% pure for use as a semiconductor.
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Physics Grade 12
The atomic number of silicon is 14. Therefore, it has 14 protons and 14 electrons. Two
electrons are in the first orbit, eight electrons in the second orbit and four electrons in
the third orbit [See Fig.(i)]. It is clear that silicon atom has four valence electrons i.e. it
is a tetravalent element. Fig. (ii) shows how various silicon atoms are held through
covalent bonds.
Like germanium, silicon atoms are also arranged in an orderly
manner.
Intrinsic Semiconductor
A semiconductor in an extremely pure form is known as an intrinsic semiconductor.
In an intrinsic semiconductor, even at room temperature, hole-electron pairs are
created. When electric field is applied across an intrinsic semiconductor, the current
conduction takes place by two processes, namely; by free electrons and holes as shown
in Figure below. The free electrons are produced due to the breaking up of some
covalent bonds by thermal energy. At the same time, holes are created in the covalent
bonds. Under the influence of electric field, conduction through the semiconductor is
by both free electrons and holes. Therefore, the total current inside the semiconductor is
the sum of currents due to free electrons and holes.
It may be noted that current in the external wires is fully electronic i.e. by electrons. What
about the holes? Referring to Figure above, holes being positively charged move towards
the negative terminal of supply. As the holes reach the negative terminal B, electrons
enter the semiconductor crystal near the terminal and combine with holes, thus cancelling
them. At the same time, the loosely held electrons near the positive terminal A are
attracted away from their atoms into the positive terminal. This creates new holes near
the positive terminal which again drift towards the negative terminal.
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Physics Grade 12
Ex tr in s ic Se mic o nduct or
The intrinsic semiconductor has little current conduction capability at room
temperature. To be useful in electronic devices, the pure semiconductor must be altered
so as to significantly increase its conducting properties. This is achieved by adding a
small amount of suitable impurity to a semiconductor. It is then called impurity or
extrinsic semiconductor. The process of adding impurities to a semiconductor is
known as doping.
The amount and type of such impurities have to be closely
controlled during the preparation of extrinsic semiconductor. Generally, for 108 atoms
of semiconductor, one impurity atom is added.
The purpose of adding impurity is to increase either the number of free electrons or
holes in the semiconductor crystal. As we shall see, if a pentavalent impurity (having 5
valence electrons) is added to the semiconductor, a large number of free electrons are
produced in the semiconductor. On the other hand, addition of trivalent impurity
(having 3 valence electrons) creates a large number of holes in the semiconductor
crystal. Depending upon the type of impurity added, extrinsic semiconductors are
classified into n-type semiconductor and p-type semiconductor.
When a small amount of pentavalent impurity is added to a pure semiconductor, it
is known as n-type semiconductor.
The addition of pentavalent impurity provides a large number of free electrons in the
semiconductor crystal. Typical examples of pentavalent impurities are arsenic (At. No.
33) and antimony (At. No. 51).
Such
impurities
which
produce
n-type
semiconductor are known as donor impurities
because they donate or provide free electrons
to the semiconductor crystal. The addition of
pentavalent impurity pro- vides a large
number of free electrons in the semiconductor
crystal.
Typical examples of pentavalent
impurities are arsenic (At.No.33) and antimony (At. No. 51). Such impurities which
produce n-type semiconductor are known as donor impurities because they donate or
provide free electrons to the semiconductor crystal
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Physics Grade 12
To explain the formation of n-type semiconductor, consider a pure germanium crystal.
We know that germanium atom has four valence electrons. When a small amount of
pentavalent impurity like arsenic is added to germanium crystal, a large number of free
electrons become available in the crystal. The reason is simple. Arsenic is pentavalent
i.e. its atom has five valence electrons. An arsenic atom fits in the germanium crystal
in such a way that its four valence electrons form covalent bonds with four germanium
atoms. The fifth valence electron of arsenic atom finds no place in covalent bonds and
is thus free as shown in Figure above. Therefore, for each arsenic atom added, one free
electron will be available in the germanium crystal. Though each arsenic atom provides
one free electron, yet an extremely small amount of arsenic impurity provides enough
atoms to supply millions of free electrons.
p-type Semiconductor
When a small amount of trivalent impurity is added to a pure semiconductor, it is
called p-type semiconductor. The addition of trivalent impurity provides a large
number of holes in the semiconductor. Typical examples of trivalent impurities are
gallium (At. No. 31) and indium (At. No. 49). Such impurities which produce p-type
semiconductor are known as acceptor impurities because the holes created can accept
the electrons.
To
explain
the
formation
of
p-type
semiconductor, consider a pure germanium
crystal. When a small amount of trivalent
impurity like gallium is added to germanium
crystal, there exists a large number of holes in
the crystal. The reason is simple. Gallium is
trivalent i.e. its atom has three valence
electrons. Each atom of gallium fits into the
germanium crystal but now only three covalent bonds can be formed.
It is because
three valence electrons of gallium atom can form only three single covalent bonds with
three germanium atoms as shown in Figure above. In the fourth covalent bond, only
germanium atom contributes one valence electron while gallium has no valence
electron to contribute as all its three valence electrons are already engaged in the
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Physics Grade 12
covalent bonds with neighbouring germanium atoms. In other words, fourth bond is
incomplete; being short of one electron. This missing electron is called a hole.
Therefore, for each gallium atom added, one hole is created. A small amount of
gallium provides millions of holes.
10.2 Logic Gates
Logic gates are very useful in dealing with and processing a combination of different
inputs. This switching logic can be applied to electrical switches and sensors,
pneumatic valves or hydraulic systems. Switching logic uses logic gates to perform
decisions.
Digital systems are said to be constructed by using logic gates. The basic gates are the
AND, OR, NOT gates. The basic operations are described below with the aid of tables
in the following, called truth tables.
AND gate
The AND gate is an electronic circuit that gives a high output (1) only if all its inputs
are high. A dot (.) is used to show the AND operation i.e. A.B. Bear in mind that this
dot is sometimes omitted i.e. AB
OR gate
The OR gate is an electronic circuit that gives a high output (1) if one or more of its
inputs are high. A plus (+) is used to show the OR operation.
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Physics Grade 12
NOT gate
The NOT gate is an electronic circuit that produces an inverted version of the input at
its output. It is also known as an inverter. If the input variable is A, the inverted
output is known as NOT A. This is also shown as A', or A with a bar over the top, as
shown at the outputs.
EXOR gate
The 'Exclusive-OR' gate is a circuit which will give a high output if either, but not
both, of its two inputs are high. An encircled plus sign ( ) is used to show the EOR
operation.
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Physics Grade 12
10.3 Transistor
Transistor was invented in 1948 by J. Bardeen and W.H. Brattain of Bell Telephone
Laboratories, U.S.A and transistor has now become the heart of most electronic
applications. Though transistor is only slightly more than 58 years old, yet it is fast
replacing vacuum tubes in almost all applications.
A transistor consists of two pn junctions formed by "sandwiching either p-type or ntype semiconductor between a pair of opposite types. Accordingly; there are two types
of transistors:
1. n-p-n transistor
An n-p-n transistor is composed of two n-type semiconductors separated by a thin
section of p-type as shown below.
2. p-n-p transistor
A p-n-p transistor is formed by two p-sections separated by a thin section of n-type as
shown below.
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