Download 96 11. Use c = in vacuum, in a medium v = 12. Use λ = and 13. (i) (ii

11.
Use c =
12.
Use λ =
13.
(i)
14.
15.
16.
17.
Charge moving with constant velocity, no acceleration, can not act as a source of em wave.
Charge moving in circular orbit, has acceleration, can act as a source of em wave.
No acceleration, can not act a source of em waves.
(i) γ-rays (ii) Microwaves
18.
Use λ =
19.
20.
Extra high frequency radiowaves
Ultraviolet
in vacuum, in a medium v =
and
(ii) its electric and magnetic field vectors are along z-axis and x-axis respectively.
, microwave—in radar & microwave oven
96
6
Nature of Light
According to wave theory of light;Huygens suggested that light travels in form of longitudinal wave (as
sound wave) but later on Young showed that light is transverse in nature.This theory successfully
explained the phenomenon of Reflection,Refraction,Interference,Diffraction and Polarisation.
Wavefront
Wavefront is defined as the locus of all the particles of a medium vibrating in the same phase at a given
instant.
Types of wavefronts
(a) Spherical wavefront- If the source of distrubance is a point source,then in homogeneous
medium spherical wavefront is formed.Speed of wavefront is known as phase velocity.
(b) Cylindrical wavefront- If the source of distrubance is a slit (line source) then the wavefront
is cylindrical.
O
Point Source
Line Source
Source
at infinity
(A)
(B)
(C)
(c) Plane wavefront-At a very large distance from the source, a small part of the spherical or
cylinderical wavefront appears as plane and called plane wavefront.
A line drawn perpendicular to the wavefront gives the direction of propagation of a wave and
is called ray of light. Energy of light propagates along rays. Time taken by light to travel from any
point of one wavefront to any point on next wavefront is the same along any ray.
Huygens’ Principle
According to Huygens’ principle
(i) Each source of light produce wavefront.
(ii) Every point on wavefront(called primary wavefront) behaves as a fresh source of light called
secondary wavelet.
(iii) These secondary wavelets produce wavefronts.
(iv) The surface touching the wavefronts produced by secondary wavelets in forward direction is
called secondary wavefront.
(v) There is no backward flow of energy when a wave travels in the forward direction.
97
Laws of Reflection on the Basis of Huygens’ Principle
Consider a reflecting surface XY on which a plane wavefront AB is incident at A. CD is the reflected
plane wavefront.
Incident
wavefront
N
D
B
i i
X
h N1
r
r
A
Reflected
wavefront
C
Y
According to Huygens’ wave theory time taken by a ray to reach from B to C and from A to D
is same and is given by
t = (BC/v) = (AD/v),
where v is velocity of light
∴
BC = AD
...(i)
(In ∆ACB)
BC = AC sin i
...(ii)
(In ∆ACD)
AD = AC sin r
...(iii)
Substituting the values of BC and AD in equation (i)
⇒
sin i = sin r
⇒
Hence angle of incidence is equal to angle of reflection which is the law of reflection.
...(i)
Laws of Refraction on the basis of Huygens’ Principle
Let AB be the plane wavefront incident on the interface XY separating two media of refractive indices
µ 1 and µ2 (µ2 > µ1). Let v1 and v2 be the speeds of light in the two media (I and II) respectively (v1 > v2).
According to Huygens’ principle.
Time taken by wavelets to reach from B to C and from A to D should be same and given by
t = BC/v1 = AD/v2
i.e.
=
...(i)
In ∆ABC,
BC = AC sin i
in ∆ACD,
AD = AC sin r
Substituting values of BC and AD in equation (i), we get
...(ii)
...(iii)
=
Which is Snell's law of refraction established from Huygens’ principle
98
Behaviour of a Plane Wavefront in Prism, Lens and Concave Mirror
A
C
B
(a) Prism
X
P1
D
P2
Y E
(b) Lens(Convex)
(c) Concave Mirror
Superposition Principle
According to superposition principle,when two or more waves travelling through any medium superimpose
on one another then the resultant displacement at any time is equal to the vector sum of the individual
displacements at that time.
Coherent Sources
Two sources of light are said to be coherent if the phase difference between the waves emitted by
sources does not change with time.
Two independent sources can emit light of same frequency but their phase difference may not be
stable. The sources having same frequency but no stable phase difference are known as incoherent
sources.
Interference
A A X When two light waves Dof same frequency and constant phase difference travelling in same direction
yP =
y1 + yD
2 + y 3 + y 4 + ............
C′
P
superimpose
on each other
B then redistribution of energy takes place in the resultant wave.This phenomenon
P
C
D
Eis known as interference of light.
QBS
S1
B
Y
O
C
Young's
Double Slit
Experiment
Sz2
D
B
The phenomenon of interference
in light was demonstrated by Thomas Young.The monochromatic light
D
waves from source S made
to fall on two very fine slits S1 and S2 which were equidistant from S. These
slits S1 and S2 give waves
Screenof same frequency and same amplitude. These waves superimposed on each
other, give rise to interference. Alternate dark and bright bands (fringes) were obtained on a screen
placed at a certain distance from the plane of slits S1 and S2.
Bright and dark fringes taken together form an interference pattern on the screen.
If crest of one wave falls on the crest of the other wave and trough of one wave falls on the trough
of the other waves. Bright fringe is formed such as at print on the screen.
99
When the crest of one wave falls on the trough of the other and vice versa. Dark fringe is formed.
Through this experiment, Young conclusively established the interference and wave-nature of light.
Special cases :(i) If one of the two slits is closed, the interference pattern disappears.
(ii) If white light is incident from S then coloured fringes of unequal width are obtained.The central
bright fringe is white and sourrounded by a few coloured fringes ordering from violet to red
(iii) If two independent sources such sodium vapour lamp or bulb are used then interference pattern
is not permanent.
Mathematical Analysis of Interference
Consider a monochromatic light of
d=
D=
a,b =
φ=
Equations of waves are
wave length λ.
Distance between S1 and S2
Distance between slit and screen
amplitudes of the waves from S1 and S2
phase difference between two waves at screen.
y1 =
Using principle of superposition, resultant wave at P is represented by
y = y1 + y 2 = a sin ωt + b sin(ωt + φ)
Let
=
=
...(i)
=
...(ii)
=
...(iii)
Hence equation (i) becomes,
y=
=
or
y=
...(iv)
Thus, resultant wave is a simple harmonic wave having amplitude R and phase difference θ. This
wave is represented on screen.
Amplitude of Resultant Wave
Squaring and adding equation (ii) and (iii), we get
or
R2 =
...(v)
100
or
R=
Now, For Resultant Intensity (IR) at P- Intensity α (Amplitude)2
∴
IR =
If intensities of individual waves are given as
Ia α a2 and Ibα b2
⇒
Ia = Ka2 and Ib= Kb2
Putting equation (vii) in relation (vi), we get
IR =
...(vi)
...(vii)
)
...(viii)
Condition for constructive Interference or bright fringe or Maxima
From eqn. (vi)
IR = Imax if cosφ = maximum = + 1
or
φ = 0, 2π, 4π,....
⇒ Phase difference is given by
φ = 2nπ with n = 0, 1, 2, 3,..
∴ Corresponding path difference,
∆x =
or
φx = nγ = 0, 1, 2, 3,....
These points are called points of constructive interference or maxima.
At maxima
Rmax = (a + b)
λλ 22 =a 22λ+b 22+ 2ab cosφ
KR
(I
+
I
+
2
I
I
cos
φ
×
2n
π
a
+
b
+
2ab
cos
φ
a φ
b + 1)
I,bπ a b and
(2n
1)
a+ +
2 π I(2n
Imax = Ia + Ib + 2 Ia I b
2
2π
(
)
=
Condition for Destructive Interference or dark fringe or Minima
From eqn. (vi),
IR = Imin if cos φ = Minimum = –1
or
φ = π, 3π, 5π,.....
⇒ Phase difference is given by
φ = (2n + 1)π, n = 0, 1, 2,...
or Path difference,
∆x =
or
∆x =
=
n = 0, 1, 2, 3.....
At minima, these points are called the points of destructive interference or minima.
Rmin = (a – b)
or
Imin = Ia + I b – 2 Ia Ib =
101
(
Ia – Ib
)
2
Graphical Representation of Intensity distribution in an Interference
Imax = ( Ia + Ib )2
Intensity IR
Imin = ( Ia – Ib)2
φ
0
∆x 0
π
2π
3π
λ/2
λ
3/2λ
4π
5π (Phase difference)
2λ 5/2λ (path difference)
As Intensity of light (I) is directly proportional to the width (w) of the source.
Iα w
since I α a 2 so
aα
where a is the amplitude of light wave.
Energy and Interference
The law of conservation of energy holds good in the phenomenon of interference. It can be shown as below
Iavg =
=
2 (Ia + Ib )
= Ia + Ib
2
i.e. average intensity is sum of the intensities of both the waves.
=
Fringe Width
The distance between any two successive bright fringes is the width of dark fringe and vice versa.
Let seperation of source = d.
Distance of screen from source = D
Let the wave emitted by S1 and S2 meet at P on the screen at a distance y from the central bright
fringe at print O. The path difference between these waves at P is given by
∆x = S2P – S1P
...(i)
In ∆OCP;
θ=
y
D
...(ii)
102
In ∆S1S2M;
...(iii)
Comparing (ii) and (iii)
yd
D
For bright fringe, path difference ∆x = nλ
∆x =
i.e.,
= nλ
or
, with n = 0, 1, 2,...
This is the distance of bright fringe from centre.Since the distance between any two successive
bright fringes is the width of dark fringe so
i.e.,
β dark =
or
β dark =
For dark fringe, path difference
∆x =
i.e.
, with n = 0, 1, 2, 3...
yd
=
D
or
y=
so
=
or
β bright =
( 2n + 1) λD
2d
λD is the distance of dark fringe from centre.Since the distance between any two successive dark
λ
nyd
βbright
λ
D∆nxλλD
Dλ 32λD This
β
=
− is width of bright fringe.
=
−d+yy101)= fringes
(θy2n
12=−
2d
D
d dD
D
d 2 2d
d
d
It is clear that fringe widths of all bright and dark fringes are same.
Angular width
Angular width (θ) of a fringe is given by
∴
θ=
or
θ=
or
θn =
λ
d
Condition of Sustained Interference with Good Contrast
Interference pattern is said to be sustained if the positions of bright and dark fringe does not change
with time on the screen.
103
To
1.
2.
3.
4.
5.
6.
7.
have a sustained interference.
Two sources must be coherent.
Light from the sources should be continuously emitted.
Sources should be close to each other.
For good contrast amplitudes should be equal .
Width of sources should be very narrow.
The slit and screen seperation should not be small.
The light should be monochromatic.
Diffraction of Light
The phenomenon of bending of light around the corners of an obstacle or an aperture is called diffraction
of light.
The diffraction of light is more pronounced when the dimensions of the obstacle/aperture is
comparable with the wave length of light.
Diffraction of Light at a Single Slit
Monocromatic light from source S is incident on lens L1 which makes the beam of mono-chromatic light
parallel.Distance of screen from slit is D.
The plane wavefront WW' is incident on the slit AB of width d.
Central maxima: The secondary wavelets from AB reach the central point O in same phase and
hence O is the position of central maxima.
Condition of Secondary Minima:
Now let light be diffracted through an angle .Path difference between the secondary wavelets originating
from A and B and reaching at P is:Path difference =
(i) Let path difference
.
Then point P will be dark. [Divide the slit into two equal halves. For each wave from upper half
there is a wave from lower half such that their path difference is λ/2 at P. So they will form minima
at P, hence P is minima.]
Thus for 1st minima, d sinθ = λ
(ii) If path difference
, point P will be dark.[Slit AB can be divided into four
equal parts.For each wave from 1st part there exists other wave from 2nd part which produce
dark fringe at P because their paths difference is
at P.The same case hold for 3rd and
4th part.]
104
Thus, for secondary minima
= 2λ
In general, condition for minima
path difference =
or
=
or
θn =
...(1)
where, θn is the angle of diffraction of the nth order minima and n = 1, 2, 3, ...
Condition of Secondary Maxima:
If path difference, BN =
Then point P wil be bright.[In this case slit can be divided in to three equal parts.For each wave
from 1st part there is a wave from 2nd part such that their path difference is λ /2 at P. So they will
form minima at P. But remaining 3rd part produces illumination at P Hence point P is the position of
secondary maxima].
Thus, for 1st maxima, d sin θ1 = 3λ 2
Similar case arise for d sin θ = 5λ 2 , 7λ 2 ,.......
λλ
sin
ddnxλ
sin
sinθ nθ
θ2n==3n
2
fd
In general condition of maxima,
d sin θ n = (2n + 1)
sin θ n =
or
λ
2
(2n + 1)λ
2d
λ
, with n = 1, 2, 3
2d
where, θn is the angle of diffraction for nth order secondary maxima and n = 1, 2, 3, ...
or θ n = (2n + 1)
...(2)
Diffraction pattern due to a single slit
Intensity (I)
I0
–3λ –2λ –λ 0 λ 2λ 3λ
Path difference
Width of the Central Maxima
Let slit and lens L2 very close to each other.If f be the focal length of L2 and the distance of first minima
on either of the central maxima be x.
Then
tan θ =
(from diagram)
105
Since θ is very small
∴
θ=
(∵ Lens is close to slit, f = D)
...(i)
Also, for first minima
=
...(ii)
From eqns. (i) & (ii) we have
=
=
=
λf
d
2λ f
d
...(3)
...(4)
Fresnel Distance
Fresnel distance is defined as the distance of the screen from the slit when the spreading of light due
to diffraction from the centre of the screen is equal to the width of the slit. It is represented as ZF.
We know, the angular position of first minima on either side of the central maximum is given by
θ1 =
...(i)
This is known as the half angular width of the central maxima.
Let D be the distance between the screen and slit. The linear spread y1 is given by
y1 =
=
When
D = ZF
y1 = d
∴
d=
or
ZF =
(Fresnel distance)
...(5)
106
Difference between interference and Diffraction of Light
Interference
Diffraction
1. Intensity of bright fringes are same.
2. The width of fringes is same.
3. In interference two coherent sources are
required.
4. It occurs due to superposition of wavelets
emerging from two coherent sources.
1. Intensity of bright fringes are not same.
2. The width of fringes is different.
3. In diffraction a single source is required.
4. It occurs due to interference between secondary
wavelets emerging from different parts of a source.
Concept of Resolving power
According to wave optics, the image of a point object formed by an ideal lens is a diffraction
pattern(a group of bright and dark fringes).
Resolving power of an optical instrument is the power or ability of the instrument to produce
distinctly sepearate images of two close objects.
According to Rayleigh, two point objects A and B will be just resolved, when central maximum
of diffraction pattern of B lies on first secondary minimum of diffraction pattern of A.
The minimum distance between two objects which can just be seen as seperate by the
optical Instrument is called the limit of resolution of the instrument.
(a) Resolving Power of Microscope
The limit of resolution of the microscope (d) is given by
S
(d
11θ)λ2µ sin θ OBJECTIVE
OBJECTIVE
=
Sd
22µdθ
sin θλ OF TELESCOPE
d=
θ θ
Object
∴ R.P. of microscope =
Where
λ = wavelength of light
θ = Half of the angle subtended by object on objective lens.
µ = Refractive index of the medium between object and objective.
(b) Resolving Power of Telescope
The limit of resolution of telescope is measured by the angle
those two distant objects.
107
subtended at its objective, by
It is given by
1.22λ
D
dθ =
∴ RP of telescope =
Where λ= wavelength of light
D = diameter of the objective of the telescope.
Unpolarised Light
Electromagnetic wave consists of electric and magnetic field vectors which vibrate perpendicular to each
other and both are also perpendicular to the direction of propagation of the light wave.
Symbolic Representation of unpolarised light.
Linear Polarisation of Light
The phenomenon of restricting the vibrations of electric vectors of light wave in a particular direction
perpendicular to the direction of propagation of light is called polarisation of light. Tourmaline crystal
is used to polarize the light and hence is called polarizer.
This transmitted light is called the plane (linearly) polarized light and the phenomenon is called
polarisation of light.
Tourmallne
Crystal
PP′
z
y
z
x
Source
x
Unpolarised
light
QQ′ Polarised light
When unpolarised light is incident on the polarizer, the intensity of the transmitted polarized light
is ideally half the intensity of unpolarised light.
Polaroids
A polaroid is a material which is used for polarisation of light. Tourmaline is a natural polarising
material. Polaroids are artificially made.
Uses of plane polarized light and polaroids
1. One of the major uses of polaroids is to avoid glare of light.
2. Polaroids are useful in three dimensional motion pictures.
3. Polaroids are used to improve colour contrast in old oil paintings.
108
Polarisation of Light by Reflection
When a beam of ordinary light (i.e. un-polarised) is reflected from a transparent medium (like glass),
the reflected light may be completely plane polarised at a certain angle of incidence called the angle of
polarisation θp.
Also, the refracted light is partially polarised.
When the angle of incidence becomes equal to θp, the reflected and refracted beams of light are
perpendicular to each other.
A
B
Plane
polarised
light
Ordinary
light
r
i
X
90°
O
glass
r'
Partially Y
polarised
C
Brewster's Law
According to this law, the refractive index of the refractive medium (µ) is numerically equal to the
tangent of the angle of polarisation (θp).
i.e.
µ = tan θp
Experimentally, reflected and refracted light are mutually perpendicular when
Using snell's law
θ pp
∠i =sin
∠θ
sin θ p
=∴
sin(90° − θ p ) cos θ p
P
A
No
light
Ordinary Polariser Polarised Analyser
light
light
sin θp= µ sin r
µ=
(∵ sin (90° – θp) = cos θp)
µ = tan θp
Polarisation of Light by Scattering
Light scattered in a direction perpendicular to direction of incident light is always plane polarised.
Light in all other directions is partially polarized.
Experimental Verificationof Polarisation of Light (Transverse Nature of Light)
This experiment of polorisation shows that light vectors are normal to direction of propagation i.e.,
light is transverse in nature.
Malus Law
This law states that the intensity of the polarised light transmitted through the analyser varies as the
square of the cosine of the angle between the plane of transmission of the analyser and the plane of the
polariser.
109
Let A be the amplitude of the light transmitted by the polariser and θ be the angle between the
planes of the polarizer and the analyser.
Resolving A into two components :
(i) A cosθ along OP
(ii) A sinθ along OV
Component A cosθ will only be transmitted through the analyser.
∴ Intensity of the transmitted light through the analyser is given by
or
I=
But
kA2 = I0
(I0 is the intensity of the incident polarised light)
∴
or
which is Malus Law.
I=
I∝
Wave Optics
Electromagnetic radiation of wavelength about 400 nm to 750 nm which can be detected by the human
eye and produces a sensation of sight is called visible light.
Reflection of Light
It is the phenomenon of change in path of light symmetrically without any change in medium after
intereaction with any surface.
Laws of reflection
(i) Angle of incidence = angle of reflection
(ii) Incident ray, normal at the point of incidence and reflected ray are coplanar.
Spherical Mirrors
C
PC
P
PCP
AB
F
PF
=
=
=
=
=
=
=
centre of curvature
Radius of curvature (R)
Pole of mirror
Principal axis
Aperture of mirror
Focus
Focal length(f)
New Cartesian sign Conventions (for lens and mirror)
1.
2.
3.
4.
All distances parallel to principal axis are measured from pole of the mirror/Lens.
Direction of incident light is from lift to right as the object is placed on the left of the mirror/lens.
All distances in the direction of incident ray is taken as positive, and opposite to incident ray as
negative.
Heights measured above and perpendicular to the principal axis are taken as positive and measured
perpendicularly below as negative.
110
Relation Between Focal Length ( f ) and Radius of Curvature (R)
Let SP is a paraxial incident ray, (parallel & close to the principal axis OC)
∵
θ = i and i = r
PF = FC
So in ∆CPF ,
If SP is very close to OC, then
PF = OF
From eq. (1) & (2), we get
FC = OF
∵
OC = OF + FC
= OF + OF
= 2.OF
⇒
R = 2f
(OC = R & OF = f)
∴ Focal length of a concave mirror is half the radius of curvature for paraxial rays.
...(1)
...(2)
Mirror Formula
Mirror formula is the relation between focal length of mirror and distance of object and image.
Let AB be the object
and A'B' its image
Consider a concave mirror forming real image.
distance of object = OB = u
distance of image = OB' = v
focal length OF = f
∆
PQF
& ∆−
' B1v' F 1
− f1 OB'
1FB'
vQF
BO
AB
PQ
QF
OF
1vAOF
P
−S = or == ∆+−iABO
1
& ∆A ' B 'O are similar
fA'B'
f v FB'OF
fu uf r v
FB'
B'F
uB'O
OF
B'F
θ
C
∴
F
2θ
f
RFurther,
(Concave mirror)
=
O
...(1)
are also similar.
∴
=
∵
PQ = AB
⇒
=
...(2)
...(3)
From eq. (1) & (3), we get
=
(∵ Q is very close to O)
B'O
=
BO
⇒
A
v
=
u
P
B′
B
F
C
A′
=
111
Q
O
Concave
mirror
using sign convention (u = –ve, v = –ive, f = –ive)
1
=
f
This equation is valid for all mirrors and all types of images.While using this formula in numericals,
sign conventions of all quantities should be used again except unknown quantity.
Linear Magnification
It is the ratio of the size of the image to the size of object.
Linear magnification
m=
=
Using eq. (1) of suction 9.5
m=
For both real and virtual images using sign conventions
m=
=
Refraction of Light
When light ray travel from one medium to other medium then velocity of light changes.This is called
refraction.Due to refraction, light ray deviates from its original path.
Light ray bends towards normal when it travels from rare to denser medium and it bends away from
normal when it travels from denser to rare medium
Medium is said to be optically denser if speed of light decreases in that medium, otherwise medium
is called optically rare.
Refractive index (µ)
It is the ratio of the speed of light in medium to that in another medium.
If the first medium is vacuum, the value is called the absolute refractive index of the second medium.
µ medium =
is called "Refractive index of medium 2 with respect to medium 1" and is written as
=
Laws of refraction
(i) The incident ray, the refracted ray and the normal
all lie in same plane.
(ii) Snell's law: According to snell's law
where
are refractive indices of medium
1 and 2 respectively with respect to vacuum.
112
Refraction of Light Through a Glass Slab (Lateral Displacement)
When refraction takes place through a rectangular glass slab then emergent ray is parallel to incident
ray .This is called lateral displacement.It is measured by Perpendicular distance between incident and
emergent ray.
Incident ray
A
N'1
N'2
i
Air
Glass
O
i–
M
r
r
t
D
d
B
C
i
d
Air
N1
N2
Tramsmitted ray
Here
In ∆OBD
In
From eq. (1) and (2)
d = Lateral displacement
t = thickness of slab
d = OB sin (i – r)
...(1)
t = OB cos r
...(2)
d=
...(3)
∆
µ
t sin(i
OBC
r) / IQ
sin
2 r − PQ
=
µ1 cos
sin
i r PQ / OQ This expression shows that Lateral displacement is proportional to thickness of glass slab.Also d
1
incresesr with i µand
will be maximum when i = 90°. Maximum value of d is; dmax = t.
Apparent
depth
Real depth
P
Q
Medium 2
r
Real
and Apparent Depth
i
Medium 1
IIf the object is placed in a denser medium and observed from rare medium its apparent depth is less
µ2
i
than its real depth.
If the object is placed in rare medium and it is observed from a denser medium, then its apparent
O
height will be larger than its real value.
Let object O is placed in denser medium and I be its Image.
Using snell's law at Q
µ 1 sin r = µ 2 sin i
or
=
OQ OP
=
IQ
IP
(For normal observation, angles are small so P is near to Q)
µ =
1 2
or
µ =
1 2
Real depth
Apparent depth
113
If object O is in rare medium and observer is in rare medium then
Apparent depth
µ
=
1 2
Real depth
Shift in position of object = t
Where
t = thickness of medium
µ = Refractive Index (R.I.) of medium
Total Internal Reflection
The reflection of light in the optically denser medium at the
interface of denser and rare medium is known as total internal
reflection(TIR).In such reflection there is no significant loss of
energy.
Critical angle – It is the angle of incidence in the denser
medium for which angle of refraction is the rarer medium is 90°.
In this case
or
2
1
sin c
(if
1
1)
Conditions for total internal reflection
(i) Ray must travel from denser to rarer medium.
(ii) Angle of incidence is the denser medium must be greater than critical angle for the given pair
of media is contact.
Application of total internal reflection
A. Mirage: It is an optical illusion formed as a result of the bending of light rays in the atmosphere
due to abnormal vertical distribution of air density.
B. Brilliance of diamond: µ of diamond is 2.47 so critical angle for diamond air interface becomes
23.88°. The faces are so cut that whatever light gets in, it suffers total internal reflection and
produces brilliance.
C. Optical fibres: It is a thin strand of glass. It is coated from outside with some material whose
refractive index is less than that of glass. When ray of light enters in it from one of its end,
it suffers total internal reflection again and again until it comes out from the other end.Optical
fibres works on the principle of TIR.
Use of optical fibres
(a) For medical and optical examinations, e.g. endoscopy
(b) In telecommunication
D. Total internal reflection in prisms:
Deviation through 180°
Deviation through 90°
114
Deviation through 0°
Refraction from Spherical Surface
A refracting surface which is a part of a sphere of transparent refracting material is called a spherical
refracting surface. There are two types of such surfaces:–
(i) Convex surface – Which is convex towards rarer medium.
(ii) Concave surface – Which is concave towards rarer medium.
There may be six different cases:1. object in rare medium-convex surface-Real image
2. object in rare medium-convex surface-Virtual image
3. object in denser medium-convex surface-Real image
4. object in denser medium-convex surface-Virtual image
5. object in rare medium-concave surface-Virtual image
6. object in denser medium-concave surface-Virtual image
Here we will discuss only first case.
(i) Refraction from Rarer to Denser Medium at Convex Surface Forming Real Image
Assumptions:1. The object is a point object and rays are paraxial.
2. Aperture of the surface is small.
3.
All angles are small so that θ ≈ sin θ ≈ tan θ = θ
In
µ12121(tan
AM
1+
− 1tan
AM
1 β
γ )In
(i(tan
(sin
sin
αγAM
−irαγβ)+
µ12 +−+ 122+
−+−
−R
uuRPO
R
v
v
MO
PC
MC PC
PI
MC
MI Using snell's law at A
µ 2 (Denser)
A
N
(r∆
+IAC
β
tan
= θr )= to
γ −P)β
µ∵
µ∴
−M
µµµisθorclose
OAC
i
O
α
µ1
(Rarer)
u
h⇒ r
γ
or
P M
β
C
I
γ=
...(1)
i= α+γ
...(2)
=
= µ 2r (sin θ = θ)
=
=
⇒R
=
V
⇒
=
or
=
µ1 µ 2
µ 2 − µ1
+
=
u
v
R
Using sign conventions
u = –ive, v = + ive, R = + ive
⇒
=
115
(ii) General Formula for All Cases
R.I. of object Medium R.I. of other Medium R.I. of other Medium − R.I. of object medium
+
=
−(dis tan ce of object) (dis tan ce of Im age)
Radius of curvature
Lenses
A lens is a portion of a transparent refracting medium bound by two spherical surfaces. Mainly
there are two types of lens, convex lens and concave lens, having both surfaces as converse or
concave respectively.
(i) Thin Lens Formula
It is a relation between focal length of a lens, distance of object and distance of image.
Consider the case of convex lens forming real image–
In ∆ABO & ∆A'B'O are similar,
∴
Also,
=
...(1)
are similar,
∴
=
...(2)
(∵ OP = AB)
From equations (1) and (2)
OB
=
OB'
=
or
=
Using sign conventions
u = –ive, v = + ive, f = +ive
=
(ii) Linear Magnification
Linear magnification
m=
Using eq. (1) of section 9.12 (i)
h2 v
=
(h2 = – ve, u = – ve)
h1 u
This formula is valid for both real and virtual image.
⇒
m=
116
Power of Lens
Power of lens is defined as the ability to converge or diverge a beam of light falling on it. It is measured
as the reciprocal of focal length.
i.e.,
P=
The SI unit of power of lens is dioptre (D).
Combination of Thin Lenses in Contact
Let I' is the image of object O formed by first lens L1(focal length f1). So for first lens,
=
......(1)
This image I′ acts as virtual object for the second lens L2. Let I be the image formed of this virtual
object I' by the second lens L2(focal length f2).
L1
L2
Hence for second lens,
111 + 1P
11
P
−+− 2
fF
vν12' uvfu2'
O
C1
I
C2
I'
=
.......(2)
By adding equations (1) and (2), we get
u
v
v'
=
...(3)
Now, if we consider both the lenses as one only, whose equivalent focal length is F, then
=
...(4)
From eq. (3) and (4), we get
=
Accordingly,
P=
where P stands for power of these combination of lenses.
For convex lenses f1, f2, P1 and P2 all are (+). For concave lens, they are all (–).
Combination of lens is used to increase magnification,to make final image erect and to reduce
defect of lens.
117
Lens Maker’s Formula
Lens Maker’s Formula is a relation between focal length of a lens,its radii of curvature and refractive
index of the material of the lens.
All assumptions are same as in case of refraction from spherical surfaces.
Considering refraction from Ist surface ABC
A
P
µ1
µ2
I1
O
B
C1
BI1 = V'
C
then
µ1 µ 2
+
=
−u v '
...(1)
where R1 = Radius of curvature of surface ABC. Considering refraction from 2nd surface ADC.
then
=
...(2)
Where R2 = Radius of curvature of surface ADC.
Now adding eq. (1) and (2)
=
or
=
⇒
=
This is lens makers formula
118
Refraction Through Prism
The angle between the emergent ray and incident ray is called angle of deviation
( δ) .
δ=
In
δ=
δ=
....(1)
In quardrilateral ALOM
=
or
=
or
∠A + [180 – (r1+ r2)] =
or
A=
Using eq. (1) and (2)
δ=
Using snell’s law at L and M
A°
sin i1 =
α
∆A
LMO
+PLM
β+irrA
µ
∠
A
90
°(i
sin
iA
(i
(r
sin
+
−i−
)∠
r) +−)O
(r
i+ ∠
−
+O
r2rr21)+ 90
R
δ 1 1 2 121 2 2121
δm = minimum
and
µ sin r2 =
δm N 1
N2
deviation angle
L
i1 = i1= i2
P
δ
From
eq.M(4) and
i α
β
i2
r2eq. (2), (3)
From
r
B
)
...(2)
...(3)
(when angles are small)
...(4)
...(5)
(i1 + i2) =
...(6)
δ=
δ = (µ – 1)A
...(7)
0
O equation is used when
This
refracting angle of prism is small (<10 ).For large angles equation (3)
N
C
is used.
1
K
(5)
360°
180°
180 (From
r1 + r2
and (6)
(i) Variation of i and δ
The following graph shows the variation of angle of deviation with angle of incidence.
In case of minimum deviation position
i1 = i2, r1 = r2
and ray LM is parallel to base BC.
(ii) Prism formula (determination of µ)
In case of minimum deviation position :
i1 = i2 = i and r1 = r2 = r
119
Then from eq. (2) of section 9.16
⇒
and from eq. (3) of section 9.16
⇒
r1 + r2
r
δ
δ
=
=
=
=
⇒
i=
using
µ=
A
A/2
(i1 + i1) – A
2i – A
...(8)
...(9)
µ=
Dispersion of Light
It is the phenomenon of splitting of a beam of white light into its constituent colours on passing through
a prism.
Cause of dispersion
Refractive index of material increases as wavelength decreases so different colours deviate through
different angles. hence dispersion occurs.
(i) Angular Dispersion (δ
δ)
It is the difference in the angles of deviation of two extreme colours(violet and red).
∴ angular dispersion
δ = δv + δr
Also
δ=
(ii) Dispersive Power (ω )
It is the ratio of angular dispersion to the mean deviation produced by the prism.
i.e.
ω=
[δ mean = δ yellow = δ y ]
ω=
120
Scattering of Light
Spreading of light after interaction with small particle(atom, molecules etc) in every possible direction
is called scattering.
According to Rayleigh, Intensity of scattered light varies inversely as the fourth power of the
wavelength of incident light.
Is α
i.e.
Effects of Scattering of Light
(a) Blue Colour of the Sky:(2010) The wavelength of blue colour is much smaller than that of red
colour. Thus intensity of the blue light scattered from dust particle is much more than that of the
red colour
. Therefore, the blue becomes the dominating colour when the sky is clear.
Due to this fact, the clear sky appears blue.
(b) Sun appears reddish at sunset or sunrise:(2010) At sunrise and sunset, the light rays travel
large distances, so lot of blue colour gets scattered away and only red colour reaches on earth
so the sun appears reddish.
(c) White colour of clouds: The clouds consist of dust particles and water droplets. Their size is
very large as compared to the wavelength of the incident light from the sun. So there is very little
scattering of light. Hence the light which we receive through the clouds has all the colours. As
a result of this, the clouds appear white.
(d) Danger Signals are red in colour: The wave length of red light is maximum. Hence According
to Ray light the scattering is least, these can be seen from long distance. Thus danger signals are
red in colour.
r
t
r
is formed by the combined effect of dispersion, refraction and reflection of light by spherical
v water droplets of rain.
An observer can see a rainbow only when his back is towards the sun.
v Re
d
Primary rainbow: The rays of sunlight get internallyR
ed
reflected
once in the water droplets and refracted twice.Dispersion
v
takes place in the emergent light and the rainbow thus appears as
Vio
anletarc of a circle.
55° 51° 42° 40°
Primary rainbow is observed by looking at an angle of about
40° above the horizontal. In primary rainbow, the red colour is on
the top and violet at the bottom.
Secondary rainbow: It is formed when the light is internally
reflected twice. Secondary rainbow is fainter than primary rainbow.
The sequence of colours in the secondary rainbow is just
reverse of that in primary rainbow.
SUN LIGHT
r Rainbow
ole
Vi
Primary
Secondary
1 s α 1 λ 4 )r
(I
v
λ4
Rainbow
Human Eye
Light enters the eye through the pupil which is of variable aperture. It then gets refracted by a lens
(of variable focal length) which forms the image on the retina.
121
Accommodation: It is the ability of the eye to change the focal length of its convex lens so that
the image is always formed on the retina.
Near point (the least distance of distinct vision) It is the closest point to the eye at which an
object can be clearly seen without excessive strain. Near point distance is generally taken as 25 cm for
normal human eye.
Far point: It is the farthest point from the eye for which an image can be focused on the retina.
The eye is completely relaxed in focusing an object at its far point. For normal eye, far point is
infinity.
Main defects of human eye:
1. Myopia or Short sightedness
(i) Eye can see distinctly only the near point objects
(ii) Far point of eye comes closer to eye.
Cause:-In this case focal lens of eye lens decreases.Ray from infinity focus at a point in front of
the retina.
Remedy:-It can be corrected using a concave lens of focal length equal to distance of far point of
defective eye.
2. Hypermetropia or far sightedness
(i) Eye can see only far point objects distinctly.
(ii) Near point of eye shift away from eye.
Cause:-In this case focal lens of eye lens increases.Rays emerging from near point objects focus
behind retina.
I
O
Hypermetropic eye
Remedy:-It can be corrected using a convex lens of focal length f, where f is calculated by the relation
=
Where u = distance of object from eye lens.
v = distance of near point object of defective eye.
3. Astigmatism
In this defect, object gets focused in one direction but out of focus in perpendicular direction.
Cause:-In this case cornea is not spherical.
Remedy:-It can be corrected by using cylindrical lens of appropriate focal length.
4. Presbyopia
(i) Image of nearby objects appears blurred.
(ii) Power of accomodation of the eye decreases.
Cause:- It is due to the decreasing effectiveness of ciliary muscles and loss of flexibility of the lens
with age.
Remedy:- It can be corrected by using a converging lens for reading and other similar work.
122
Simple microscope
It consists of a converging lens of small focal length. A virtual, erect and magnified image of the object
is formed at the least distance of distinct vision from the eye. The simple microscope is also called a
magnifying glass.
Angular magnification (m)- it is also called as magnifying power of an instrument. It is the ratio
of the angles formed by "the final image"(= β) and by "the object" (= α) when both are placed at the
least distance of distinct vision, D, i.e., near point of eye.
i.e. Angular magnification: m = β/α
Magnification by a simple microscope:
Case (i): the image is formed at a distance D from the lens/
eye.
(Eye focussed at near point)
Magnifying power
Case (ii): When image is formed at infinity (Normal
adjustment).
Magnifying power
Compound Microscope
It is an optical instrument used to magnify very small or tiny objects.
A It consists of two converging lenses;
⇒
A"B"
A"B"
βv D
tanD
A'B'
β
mv00
= 11+×+ D
(i) an objective lens of very small focal length and short aperture
f fαAB
−
uu 0 tan
α
AB
fe
−A'B'
0
(ii) An eyepiece or eyelens of small focal length and small aperture.
F′
Image Object
Ray diagram
is shown in figure.
F
Magnifying power
u
v = D By definition
m=
(∵ angles are small)
Eyelens
or
Objective
A "B"/ B"E
m = A B"/ B"E
1
F0
B
=
A1
A
(∵ A1B" = AB)
B'' α
h B'
O
β
E
h
u0
A'
h
m=
f0
A''
fe
D
Compound microscope
m = me × m0
...(1)
D
Since eyelens behaves as simple microscope so me = 1 + , and for objective lens m0 =
fe
Hence by (1)
m=
...(2)
123
(i) If u0 = f0 and OB' = v0 = L = length of microscope then
m=
L
D
1+
−f0
fe
...(3)
(ii) If final image is formed at infinity then eqn. (3) reduces to
m=
Astronomical Telescope
It is a device which is used for observing distinct images of heavenly bodies.
It consists of two lenses,
(i) Objective lens of large focal length and large aperture
(ii) Eyepiece of small focal length and small aperture.
case(i):-In normal adjustment, the final image of the object AB is formed at infinity.
Magnifying power
m=
or
object
(AB)
m=
f0
m = −f
e
In this case length of telescope = f0 + fe
case(ii):-When final image is formed at least distance of distinct vision(D):
Magnifying power
m=
m=
...(1)
For eye lens
Objective
f0
=
ue
fe
α
⇒
B''
O
=
α
B'
Fe F0 β
To eye
E
A'
1
=
ue
Eyepiece
...(2)
For eq. by (1) and (2)
m=
...(3)
124
A''
D
fe
Reflecting Type Telescopes
In such a telescope objective lens is replaced by a concave parabolic mirror of large aperture.
Advantages of reflecting type
(i) No chromatic aberration.
(ii) No spherical aberration takes place in parabolic surfaces chosen.
(iii) Mechanical support system is much easier.
(iv) Less chances of damaging the objective.
(v) The brightness of the image is more.
(a) Cassegrainian model:
Eyepiece
Rays from
d
distant star
Secondary
mirror
Objective mirror
In normal adjustment, magnifying power is m =
f0
fe
Answer Yourself
Very Short Questions
Q1. An object is placed at the focus of a convex lens. Where will its image be formed?
Q2. Write the Cartesian sign convention used for deriving relevant formulae for reflection and refraction
in a spherical lenses and mirrors.
Q3. The refractive index of a medium is √3 .What is the angle of refraction when angle of incidence
is 45°.
Q4. How does the frequency of UV light change when it goes from air to glass?
Q5. Two thin lenses of power +7D and -3D are in contact. What is the focal length of the combination?
Q6. What is the focal length of plane mirror?
Q7. Define critical angle.
Q8. Give two examples in nature where total internal reflection is the principle.
Q9. Why does the sky appear blue?
Q10. The image of a candle is formed by a convex lens on a screen. The lower half of the lens is
painted black to make it completely opaque. How will the image be different from the one obtained
when the lens is not painted black?
125
Q11.
Q12.
Q13.
Q14.
Q15.
Q16.
Q17.
Q18.
Q19.
Q20.
Q21.
Q22.
Q23.
Q24.
Q25.
Q26.
Q27.
Q28.
Q29.
What is the advantage of using parabolic concave mirror as objective in a telescope?
What is the cause of refraction?
How does the resolving power of telescope change if the diameter of objective is increased?
How does the resolving power of microscope change if the light of higher wavelength is used?
A substance has a critical angle. What is its refractive index?
Among red and violet which of the two colors travels slower in a glass prism.
A glass lens of refractive index 1.47disappears when immersed in a liquid . What is the refractive
index of the liquid?
What is a wave front?
What is the shape of a wavefront from a point source and line source?
Can interference be produced by X-rays?
In Young's double slit experiment if red light is replaced by blue how does the fringe width
change?
In single slit experiment if the width of the slit be made double its initial width how does the
size of central maxima change?
Light waves can be polarized but sound waves cannot be . Why?
Does the polarizing angle depend on the refractive index of the transparent medium?
What is meant by interference of light?
What is meant by diffraction of light?
Illustrate with the help of diagrams the action of 1) convex lens 2) concave mirror when a plane
wave front is incident on it.
State Malus law and graphically represent it.
State Brewster's law.
Short Questions
Q1. Write its three advantages of reflecting telescope over refracting type telescope.
Q2. Red light is incident on a thin converging lens of focal length f. Explain briefly how the focal
length of the lens will change, if red light is replaced with blue light.
Q3. What is the ratio of velocities of light of wavelength 4000A° and 8000A° in vacuum?
Q4. Using a ray diagram give the condition for total internal reflection and derive a relation between
critical angle and refractive index.
Q5. A boy uses spectacles of focal length -50cm. Name the defect of vision he is suffering from
Compute the power of the lens.
Q6. A thin prism of 60 degree angle gives a deviation of 30 degree, what is the refractive index
of the material of prism?
Q7. Express magnifying power of telescope in terms of focal length for adjustment in (i) normal
(ii) least distance of distinct vision.
Q8. A convex lens made of material of refractive index ulis kept in medium of refractive index µ2.
Parallel rays of light are incident on the lens. Complete the path of the rays when (i) µl < µ2.
(ii) µl = µ2 (iii) µl > µ2.
126
Q9. What is myopia and hypermetropia? How are they corrected.
Q10. Give reason for the following observations on the surface of the moon(i) abrupt sunrise and sun
set(ii)A rainbow is never formed.
Q11. What changes in the focal length is observed in a (i) Concave mirror and (ii) Convex lens, when
the incident violet light on them is replaced by red light.
Q12. Show by drawing a ray diagram how a totally reflecting prism may be used to invert the image
without changing its size and direction.
Q13. What are coherent sources of light? Why two different sources producing light of same wavelength
are not coherent.
Q14. Give four points of difference between interference and diffraction pattern.
Q15. Draw intensity distribution graph for single slit diffraction pattern and Young's double slit experiment.
Q16. Draw a graph showing the variation of intensity of polarized light transmitted by an analyzer.
Q17. Two sources of intensity I1 and I2 undergo interference in Young's double slit experiment obtain
a ratio of Imax to Imin in terms of the amplitude a1 and a2.
Q18. A thin prism of 60 degree angle gives a deviation of 30 degree, what is the refractive index
of the material of prism?
Q19. What is the critical angle of light, when it passes from medium 1 to medium2 of refractive index
1.7 and 1.2 respectively
Q20. A compound microscope with objective of 1cm focal length and eyepiece of 2.0cm focal length
has a tube of length 20cm. Calculate the magnifying power of the microscope if the final image
is formed at the near point of the eye.
Q21. Double Convex lenses are to be manufactured from a glass of refractive index 1.55 with both
faces of the same radius of curvature. What is the radius of curvature?
Q22. Where should an object be placed from a converging lens of focal length 10cm so as to obtain
a virtual image of magnification 2?
Q23. An astronomical telescope consists of two thin lenses set 36cm apart and has magnifying power
Calculate the focal length of the lenses.
Q24. In a double slit experiment with monochromatic light. Fringes are observed on a screen placed
at some distance from the slits. If the screen is moved by 0.05 m towards the slits. The change
in the fringe width is 0.00003m .If the distance between the slits is 0.001m, calculate wavelength
of light used.
Long Questions
Q1. Derive mirror formula for concave and convex mirrors
Q2. Defme critical angle and write condition for total internal reflection. Obtain an expression for
refractive index in terms of critical angle
Q3. Explain dispersion and rainbow formation
Q4. Derive Lens formula for convex and concave lenses
Q5. Derive an expression for refraction at spherical surfaces.
Q6. Deduce lens maker's formula for a biconvex lens *
Q7. Obtain an expression for the refractive index of the material of a prism in terms of refracting
angle and angle of minimum deviation.
Q8. Discuss Structure of eye and its defects and rectification
127
Q9. Draw a labeled diagram and determine the magnification and resolving power of (a) simple
microscope (b) compound microscope (c) astronomical telescope and (d) reflecting type telescope*
1. How is wave front defined? Using Huygens construction prove that the angle of incidence
is equal to the angle of reflection.
2. State the Huygens postulates of wave theory. Prove laws of refraction.
3. What are coherent sources? How does the width of interference fringes in the young's double
slit experiment change when (a)distance between the slits and screen is decreased?(b)frequency
of the source is increased.
4. Define polarizing angle. Derive the relation connecting polarizing angle and refractive index
of the medium.
5. Using Huygens's principle draw a diagram to show propagation of a wave front originating
from a monochromatic point source. Describe diffraction of light due to a single slit. Explain
formation of a pattern of fringes obtained o the screen and plot showing variation of intensity
with angle θ in single slit diffraction.
6. What is interference of light? Write the essential conditions for sustained interference pattern
to be produced on the screen. Draw a graph showing the variation of intensity versus the
position on the screen in the young's experiment when (a) both the slits are opened (b) one
of the slits is closed.
7. What is the effect on the pattern when
(i) Screen is moved closer to the plane of the slits (ii)Separation between the slits is increased.
8. Define polarization of light. How can one distinguish between un polarized& linearly polarized
light using a Polaroid. Show that when a light beam is incident on a refracted surface at
the polarizing angle, the reflected and refracted beams are mutually perpendicular to each
other.
To be Learnt
Q1. A concave lens made of glass of µ1 refractive index is immersed in a medium of refractive index
µ2. How will the lens behave when µl > µ2?
Q2. Calculate the apparent depth of a body immersed in a liquid of refractive index 4/3 at 8m of
depth.
Q3. Among red and violet which of the two colors travels slower in a glass prism?
Q4. In Young's double slit experiment if the distance between the slits be less than the wavelength
of light used, what will be the effect of interference pattern?
Q5. Red light is incident on a thin converging lens of focal length f. Explain briefly how the focal
length of the lens will change, if red light is replaced with blue light.
Q6. The focal length of an equi-convex lens is equal to its radius of curvature of either face. What
is the refractive index of the material of the lens?
Q7. Three rays of light red (R), green (G) and blue (B) are incident on the face AB of a right angled
prism ABC. The refractive indexes of the material of prism for red, green, blue are 1.39, 1.44,
1.47 respectively. Trace the path of the rays through the prism. How will the situation change
if these rays are incident normally on one of the faces of an equilateral prism?
128
Pedagogical Remark
1.
2.
3.
4.
5.
6.
7.
Use lens makers formula see focal lengthis positive and negative.
Refractive index equal to real /apparent depth
Recall µv > µr
Refer NCERT solved example - Wave Optics
Reciprocal of focal length depends on (µ-1) and µv > µr
Use lens makers formula and take |f| = |R|
Use µ = (1/sin iC) and check with the values given for TIR.
1.
All the apparatus required for doing the experiment are easily available and can be set
up in the lab
Teacher may do the experiment and show the students how the observation be made.
Interpretation of the graph need to be taught with proper scale being used.
2.
3.
129
7
●
●
●
●
The phenomena like interference, diffraction, polarisation of light, Maxwell's equations of
electromagnetism and Hertz experiment on production and detection of electromagnetic waves
established the wave nature of light.
Electrons are fundamental particles and electron emission can be either thermionic emission. Field
emission or phot-electric emission.
The phenomenon of emission of electrons from metal surface by the action of light is called phot
electric effect and the electron so emitted are called photoelectrons.
On the basis of experiment Lenard's observations are
(a) There is no time bag between emission and incidence of light.
Fig. 1 Experimental arrangement for study of photoelectric effect
(b) The number of emitted electrons is directly proportional
to the intensity of light falling on the metal plate.
Fig. 2 Variation of Photoelectric
current with intensity of light
130
(c)
(d)
(e)
(f)
(g)
The kinetic energy of emitted electrons does not depend upon-the intensity of incident radiation.
The kinetic energy of emitted electrons increases as the frequency of incident radiation increases.
The kinetic energy of emitted electrons increases as frequency of incident radiation increases.
The number of emitted electrons is independent of the frequency of incident radiation.
If the frequency of incident light is below
a certain minimum value then there will
be no photo emission. This minimum
frequency (Threshold frequency) is
different for different materials.
Fig. 3 Variation of stopping potential V0 with frequency v
of incident radiation for a given photosensitive material
(h) The negative potential, V0 at which the current is just reduced to zero is called stopping potential.
Fig. 4 Variation of photocurrent with collector
plate potential for different intensity of
incident radiation
(i) The value of stopping potential V0 is independent of intensity of incident light.
(j) The stopping potential is more negative for higher frequencies of incident radiation.
Fig. 5 Variation of photoelectric current with
collector plate potential for different
frequencies of incident radiation
131
●
Einstein's explaination of photoelectric effect
(a) Photoelectric effect an electron absorbs a quantum of energy hv.
where
h is planck's constant and
v is the frequency of light.
(b) A part of this energy is used in ejecting the electron from the metal surface. This is called
the work function, φ0 of the metal (different for different metals)
(c) The rest of the energy is given to the electron in the from of kinetic energy Kmax.
∴
or
hv = φ0 + K max. This is know as Einstein's photoelectic equation
Kmax = hv – φ0 ...(1)
(d) According to rinstein's photoelectric equation
Kmax = hv – φ0
...[eq. (1)]
Kmax depends linearly on r and is independent of intensity of radiation.
(e) Since Kmax must be non-negative, equation (1) implies hr > φ0 for photoelectric emission
or
hv > hv0
or
v > v0
Also as v0 =
which shows that greater the work function φ0, the higher the minimum or
threshold frequency v0 needed to emit photoelectons.
(f) As intensity is proportional to the number of energy quanta per unit area per unit time so
for greater quanta available greater number of electrons coming out of metal. This explains
for v > v0, photoelectric current is proportional to intensity.
(g) Whatever may be the intensity, photoelectric emission is instantaneous. Low intensity does
not mean delay in emission.
●
Equation (1) and also be written as.
eV0 = hv – φ0
or
V0 =
... Eq. (2)
h
This predicts that V0 versus r curve is a straight line with slope , independent of nature of
e
material.
Eq. (1) can also be written as
Kmax= eV0 =
●
Photoelectic effect gave evidence to the fact that light in interaction with matter behaved as if
it is made of quanta or packets of energy, each of energy hv, which was confirmed by compton's
scattering of X-rays experiment.
●
The mass of moving photon is
or
132
●
The momentum of moving photon is p =
●
By increasing the intensity of light of given wavelength, there is only increase in no of photons
per second crossing a given area, with each photon having the same energy.
●
Photons are electrically neutral and are not deflected by electric and magnetic fields.
●
In photon electron collision the total energy and total momentum are conserved.
●
According to De Broglie the wavelength, λ associated with a particle of momentum, p is given
whatever the intensity of radiation may be.
by λ =
From this equation dual aspect is evident as L.H.S. λ is the attribute of a wave while on R.H.S.
p is a typical attribute of a particle. Planck's constant, h relates the two.
hv
∴
c
De Broglie wavelength, λ of electron is
For a photon p =
●
●
λ =
=
12.3 0
A
V
where V is accelerating potential in volts.
●
According to Heisenberg's uncertainty principle it is not possible to measure both the position
and momentum of an electron (or any other particle) at the same time exactly. If ∆x is the uncertainty
in the specification of position & ∆p is the uncertainty in the specification of momentum
h
1.227
=
nm
h
2then
meV
V
=
∆x.∆p =
2π
The wave nature of electrons was verified by Davisson and Germer experimentally.
●
The De Broglie hypothesis of matter waves supports the Bohr's concept of stationary orbits.
●
hv
h ch h
== = λ =
λ2mk
pc vmv
Answer Yourself
Very Short Questions
Q1. In photoelectric effect for a metal the K.E. of the photoelectron depends upon which factor?
Q2. The threshold wavelength for photoelectric emission from a material is 5200 A°. Will the
photoelectrons be emitted, when this material is illuminated wiht monochromatic radiation from
1 watt uttravioletlamp?
Q3. How will the photoelectric current change on decreasing the wavelength of incident radiation
for a given photo sensitive material?
Q4. What is the effect on the velocity of the emitted photoelectrons. If wavelength of incident lights
is decreased?
133
Q5. On what factor the retarding potential of photocell depends?
Q6. If an electron behaves like a wave, what should determine the wavelength and frequency of the
wave.
Q7. Are matter waves electromagnetic? Write de-Broglic wave equation.
Q8. What is the main aim of Davisson-Germer experiment?
Q9. Calculate energy of a photon of freq. 6 × 1014 Hz in electron volt
Q10. How does maximum K.E. of an electron emitted vary with work function of the metal?
Q11. If green light ejects photoelectrons from a given photo-sensitive surface where as yellow light
does not. What will happen in case of violet and red light?
Q12. What is the momentum of an electron beam of wavelength 4A°?
Q13. The work function for aluminium is 4.2 eV. Find the threshold wavelength for photoelectric
emission.
Q14. What is de-Brogglie wavelength of an atom at absolute temp. T.K.?
Q15. How does the stopping potential applied to a photocell change if distance between the light source
and the cathode of cell is doubled?
[Hint : Stopping potential is independent of intensity]
Short Questions
Q1. For a photosensitive surface, threshold wavelength is λ0. Does photoemission occur if the wavelength,
λ of incident radiation is (i) more than λ0 (ii) less than λ0? Justify your answer.
Q2. De-Broglie wavelength associated with an electron accelerated through a potential differen V is
λ. What will be its wavelength when the accelerating potential is increased to 4V?
Q3. Number of ejected photoelectrons increases with an increase in intensity of light but not with
the increase in the frequency of light. Why?
Q4. An increase in the intensity of the incident light does not change the maximum velocity of photoelectrons. Why?
Q5. An electron, an λ-particle and a proton have the same K.E. which of these particles has the largest
De-Broglie wavelength?
Q6. Draw a graph to show the variation of stopping potential with frequency of radiation incident
on a metalplate. How can the value of Planck's constant be determined from this graph?
Q7. If energy of a photon is 25eV and work function of material is 7eV, then find the value of stopping
potential.
Q8. The threshold frequency of a metal is f0. When light of frequency 2f0 is incident on the metal
plate, the maximum velocity of emitted electron is V1. When frequency of incident radiation is
increased to 5f0, the maximum velocity of emitted electron is V2. Find the ratio of V1 & V2.
Q9. Calculate the maximum kinetic energy of electrons emitted from a photosensitive surface of work
function 3.2 eV for the incident radiation of wavelength 300 nm.
134
Q10. Obtain the De-Broglie wavelength of an electron of K.E. 100 eV.
Q11. A 100 W sodium lamp radiates energy uniformaly in all directions. The lamp is located at the
centre of a large sphere that absorbs all the sodium light which is incident on it. The wavelength
of sodium light is 589 nm. (i) what is the energy associated per photon with the sodium light?
Q12. The work function for a certain metal is 4.2 eV Will this metal give photoelectric emission for
incident radiation of wavelength 330 nm?
Q13. For what kinetic energy of a neutron will the associated De-Broglie wavelength be 1.40×10–10m?
Q14. On what three factors photoelectric current dependes?
Q15. On what two factors stopping potential depends?
To be Learnt
Short Questions
Q1. Sketch the graph showing the variation of stopping potential with frequency of incident radiations
for two photosensitive materials A & B having threshold frequency v0 > v01 respectively.
(i)Which of two metals A or B has higher work function?
(ii)What information do you get from the slop of the graph?
(iii)What does the value of the intercept of graph A on potential axis represents?
Q2. In a plot of photo electic current versus anodepotential how does (i) Saturation current vary with anode
potential for incident radiations of different frequencies but same intensity. (ii) the stopping potential
vary for incident radiations of different intensities but same frequency? (iii) Photo electric current vary
for different intensities but same frequency of incident radiations? Justify your answer in each case.
Q3. In a photoelectric effect experiment for radiation with hv0 = 8 eV, electrons are emitted with
2 eV. What is the energy of electrons emitted for incoming radiation of frequency 1.25 v0?
Q4. Calculate the energy of a photon in electron volt (i) whose frequency is 1000 KHz (ii) wavelength
is 6600A°.
Q5. A neutron is an uncharged particle of mass 1.67 × 10–27 kg calculate the de-Broglie wavelength
of the neutron moving with a velocity such that K.E. = 0.04 eV.
Q6. The maximum K.E. of photoelectrons emitted from a certain metallic surface is 30 eV when
monochromatic radiation of wavelength λ falls on it. When same source is illuminated with light
of wavelength 2λ, the maximum K.E. of photoelectrons is observed to be 10 eV. Calculate the
wavelength λ and determine the maximum wavelength of incident radiation for which photoelectrons
can be emitted.
Q7. The number of photons entering the pupil of our eye per second corresponding to the minimum
intensity of white light that we humans can perceive
10–10 Wm–2. If area of pupil is about
0.4 cm2 and frequency of light 6 × 1014 Hz, then why our eye can never count photons even
in barely deterctable light.
135
Pedagogical Remark
1.
(i) Metal A has higher work function as φ0 = hv0 and v0 > v0
(ii) slope =
(iii) Intercept =
2.
3.
4.
(i) Saturation current will be same.
(ii) Stopping potential will be same.
(iii) Photoelectric current depends on intensity.
Kmax = hv – φ0
2eV = 8eV – φ0 ⇒ φ0 = 6eV.
∴ Kmax = hr – φ0
= h × 1.25 v0 – 6
= 1.25 × 8 – 6
= 4 eV.
(i)
E = hv
= 6.6 × 10–34 × 1000 × 103 J
=
(ii)
5.
λ =
6.
For case I
6.6 × 10−2 g
ev = 4.125 × 10−9 ev
−19
1.6 × 10
E =
hc
= φ0 + 30 × 1.6 × 10 −19
λ
For case II
From these two λ = 310.3 A°.
∴ φ0 = 17 × 10–19 J
and threshold wavelength, λ =
7.
Light energy falling on pupil per second = 10–10 × 0.4 × 10–4
= 0.4 × 10–14 W
As energy = hv = 6.6 × 10–34 × 6 × 1014 = 39.6 × 10–20 J
∴ no. of photons entering the pupil of our eye per second
=
0.4 × 10−14
10 4 so it is large enough never to count photons
39.6 × 10−20
136
8
●
An atom has a nucleus which is positively charged. The radius of the nucleus is smaller than
the radius of an atom by a factor of 104. More than 99.9% mass of the atom is concentrated in
the nucleus.
●
On the atomic scale, mass is measured in atomic mass units (u). By definition, 1 atomic mass
unit (1 u) is 1/12th mass of one atom of 12C; 1 u = 1.660563 × 10–27 kg
●
A nucleus contains a neutral particle called neutron. Its mass is almost the same as that of proton
●
The atomic number Z is the number of protons in the atomic nucleus of an element. The mass
number A is the total number of protons and neutrons in the atomic nucleus; A = Z + N; Here
N denotes the number of neutrons in the nucleus
A nuclear species or a nuclide is represented as AZX where X is the chemical symbol of the species.
Nuclides with the same atomic number Z, but different neutron number N are called isotopes.
Nuclides with the same A are isobars and those with the same N are isotones. Most elements
are mixtures of two or more isotopes. The atomic mass of an element is a weighted average of
the masses of its isotopes. The masses are the relative abundances of the isotopes.
●
A nucleus can be considered to be spherical in shape and assigned a radius. Electron scattering
experiments allow determination of the nuclear radius; it is found that radii of nuclei fit the formula
R = R0A1/3,
Where R0= a constant = 1.2 fm. This implies that the nuclear density is independent of A. It
is of the order of 1017 kg/m3.
●
Neutrons and protons are bound in a nucleus by the short-range strong nuclear force. The nuclear
force does not distinguish between neutron and proton.
●
The nuclear mass M is always less
than the total mass, m, of its
constituents. The difference in mass
of a nucleus and its constituents is
called the mass defect,
∆M = {Zmp+{A–Z)mn)–M
Using Einstein’s mass energy
relation, we express this mass
difference in terms of energy as
∆Eb = ∆Mc2
Fig. 1
137
The energy Eb represents the binding energy of the nucleus. In the mass number range A = 30
to 170, the binding energy per nucleon is nearly constant, about 8 MeV/nucleon.
●
Energies associated with nuclear processes are about a million times larger than chemical process.
●
The Q-value of a nuclear process is
Q = final kinetic energy - initial kinetic energy.
Due to conservation of mass-energy, this is also
Q = (sum of initial masses - sum of final masses) c2
●
Radioactivity is the phenomenon in which nuclei-of a
given species transform by giving out
α or β or γ rays; α-rays are helium nuclei, β-rays are
electrons, γ-rays are electromagnetic radiation of wave
lengths shorter than X-rays
N(t) = N(0) e–λ
●
Fig. 2
Law of radioactive decay:
where λ is the decay constant or disintegration constant. The half-life T1/2 of a radio nuclide is
the time in which N has been reduced to one-half of its initial value. The mean lifex, is the time
at which Af has been reduced to e–1 of its initial value.
T1/2 =
ln 2
= τ ln 2
λ
●
Energy is released when less tightly bound nuclei are transmitted into more tightly bound nuclei.
In fission, a heavy nucleus like 235
U breaks into two smaller fragments.
92
●
The fact that more neutrons are produced in fission than are consumed gives the possibility of
a chain reaction with each neutron that is produced triggering another fission. The chain reaction
is uncontrolled and rapid in a nuclear bomb explosion. It is controlled and steady in a nuclear
reactor. In a reactor, the value of the neutron multiplication factor k is maintained at 1.
●
In fusion, lighter nuclei combine to form a larger nucleus. Fusion of hydrogen nuclei into helium
nuclei is the source of energy of all stars including our sun.
Atom
●
Atom, as a whole, is electrically neutral and therefore contains equal amount of positive and negative
charges.
●
In Thomson’s model an atom is a spherical cloud of positive charges with electrons embedded
in it.
138
●
In Rutherford’s model, most of the mass of the atom and all its positive charge are concentrated
in a tiny nucleus (typically one by ten thousand the size of an atom), and the electrons revolve
around it. Rutherford nuclear model has two main difficulties in explaining the structure of atom
(a) It predicts that atoms are unstable because the accelerated electrons revolving around the nucleus
must spiral, into the nucleus. This contradicts the stability of matter, (b) It cannot explain the
characteristic line spectra of atoms of different elements. In Rutherford’s model the electrons can
revolve in orbits of all possible radii & so they should emit continuous radiation all frequencies
and so the atom should exhibit a continuous spectrum while the experimental evidence show that
atom like H emit line spectra of radiation of only certain discrete frequencies.
●
Let ro be the distance of closest approach of α-particle to the nucleus. The positive charge on
nucleus is Ze and that on α-particle is 2e Potential Energy on closest approach is
1 Ze(2e)
4πε 0 r0
At this instant α-particle is momentarily at rest & Kinetic Energy is changed to potential Energy
Kinetic Energy
= EK =
1 2
1 2 ze 2
mv =
2
4π ∈0 r0
1 2 Ze 2
⇒ r0 =
4π ∈0 1 mv 2
2
Fig. 3
●
Atoms of each element are stable and emit characteristic spectrum. The spectrum consists of a
set of isolated parallel lines termed as line spectrum. It provides useful information about the atomic
structure.
●
The atomic hydrogen emits a line spectrum consisting of various series. The frequency of any
line in a series can be expressed as a difference of two terms
Lyman series
:
v = RC
1 1
−
; n = 2,3, 4... where Rhydberg constant
12 n 2
me4
R =
8 ∈02 h3C
= 1.03 ×107 m–1
139
:
v = RC
1 1
−
; n = 3, 4,5...
22 n 2
Paschen series :
v = RC
1 1
−
; n = 4,5, 6...
32 n 2
Brackett series :
v = RC
1 1
−
; n = 5, 6, 7...
42 n 2
Pfund series
v = RC
1 1
−
; n = 6, 7,8...
52 n 2
Balmer series
:
Fig. 4
●
Absorption Transition of H-atom. Normally all atoms are in their lowest energy level . When we
pass white light through a tube filled with hydrogen, then H-atom absorb proton of appropriate
energies and rise to various energy levels which are absorption transition. At ordinary temperature
almost all the atoms remain in their lowest energy level n = 1. Hence only Lyman series is obtained
in the absorption spectrum of H-atom.
●
To explain the line spectra emitted by atoms, as well as the stability of atoms, Niel’s Bohr proposed
a model for hydrogenic (single elctron) atoms. He introduced three postulates and laid the foundations
of quantum mechanics:
(a) In a hydrogen atom, an electron revolves in certain stable orbits (called stationary orbits) without
the emission of radiant energy.
(b) The stationary orbits are those for which the angular momentum is some integral multiple of
h/2π (Bohr’s quantisation condition.) That is L = nh/2π where n is an integer called a quantum
number.
(c) The third postulate states that an electron might make a transition from one of its specified
non-radiating orbits to another of lower energy. When it does so, a photon is emitted having
energy equal to the energy difference between the initial and final states. The frequency (v)
of the emitted photon is then given by
●
●
hv = Ei – Ef
An atom absorbs radiation of the same frequency the atom emits, in which case the electron is
transferred to an orbit with a higher value of n.
Ei + hv = Ef
As a result of the quantisation condition of angular momentum, the electron orbits the nucleus
at only specific radii. For a hydrogen atom it is given by
140
vn =
n2
m
h
2π
2
4π ∈0
e2
The total energy is also quantised:
En = −
me 4
8n 2 ∈20 h 2
= – 13.6 eV
Then n = l state is called ground state. In hydrogen atom the ground state energy is –13.6 eV.
Higher values of n correspond to excited states (n > 1). Atoms are excited to these higher states
by collisions with other atoms or electrons or by absorption of a photon of right frequency.
●
de Broglie’s hypothesis that electrons have a wavelength h/mv gave an explanation for Bohr’s
quantised orbits by bringing in the wave-particle duality. The orbits correspond to circular standing
waves in which the circumference of the orbit equals a whole number of wavelengths.
●
Bohr’s model is applicable only to hydrogenic (single electron) atoms. It cannot be extended to
even two electron atoms such as helium. This model is also unable to explain for the relative
intensities of the frequencies emitted even by hydrogenic atoms.
Answer Yourself
Very Short Questions
Q1. Name two elemenary particles which have almost infinite life time.
Q2. Name the radioactive radiations emitted from a radioactive element. Arrange them in the order
of increasing ionising power.
Q3. Why the nucleus can not have negative mass defect?
Q4. What do you mean by Q value of a nuclear reaction?
Q5. What is pair production?
Q6. Define decay constant.
Q7. What fraction of tritium will remain after 25 years when its help life is 12.5 years?
Q8. The half-life of a radioactive element A is same as mean life-time of another radioactive element
B. Initially both have same no. of atoms but B decays faster than A. Whay?
Q9. What is the order of nuclear density?
Q10. What is the ratio of Kwh to MeV?
Q11. You are given two nuclides 37 × and 43 y. Are they the isotopes of same element? Why? Which
is more stable?
Q12. Define atomic mass unit. Write its energy equivalent.
Q13. What is meant by critical mass?
Q14. Draw the graph shoming the distribution of kinetic energy of electron emitted during beta decay.
15. Does the ratio of neutrons to protons in a nucleus increase, decrease or remain the same after
the emission of an α-particle?
141
16.
What are thermal neutrons?
17.
In which model (Thomson or Rutherford), is it completely wrong to ignore multiple scattering
for the calculation of average angle of scattering of α-particles by a thin foil?
18.
The total energy of electron in first excited state of hydrogen atom is about –3.4 eV. What is
the kinetic energy of electron in this state.
19.
What is the energy possessed by an electron for n = ∞?
20.
What is the order of velocity of electron in a H-atom is ground state?
21.
What is the ionisation potential of Hydrogen atom?
22.
Name the series of hydrogen spectrum which lies in the visible region of electromagnetic spectrum.
23.
The electron in the H-atom passes from n = 4 to n = 1 level. What is the maximum no. of
photons tharcan be emitted?
24.
What woud happen if the electrons in an atom were stationary?
25.
What is the ground state energy of electron in 3li7?
26.
Can a Hydrogen atom absorb a photon whose energy exceeds its prinding energy?
Short Questions
Q1. How is a photon different from a neutron?
Q2. Explain the action of a moderator in nuclear reactor.
Q3. Explain the role of control sods in a reactor.
Q4. Explain the concept of nuclear energy with reference to binding energy curve.
Q5. What are delayed neutrons? Discuss their role.
Q6. Why is the density of nucleus more-than that of atom?
Q7. Calculate the disintegration energy Q fro qusion of 98
M0 into two equal fragments 4921Sc by bombarding
42
with a neutron. Given that
m
(
98
42
)
M 0 = 97.90541 u , m
(
49
21
)
Sc = 48.95002 u
mn = 1.00867 u.
Q8. Explain clearly why the presence of a continuous distribution of energy is a pointer to theexistence
of other unobserved products in the decay?
Q9. If a neutron is unstable with a half-life of about 917 second, why do not all the neutrons of
a nucleus decay eventually into protons? How can a nucleus of z protons and (A – z) neutrons
ever remain stable, if the neutrons themsleves are unstable?
Q10. Why every hightemperature is required for fusion reaction to take place.
Q11. Compare the radic of two nuclei with mass numbers 1 and 27 respectivley.
Q12. A radioactive isotope decays in the following sequance
α
0n
A 

→ A1 
→ A2
1
If the mass no. and atomic no. of A2 are 176 and 171 respectively, fidn the mass no. & atomic
no. of A and A1. Which of these are isobars.
142
Q13. Why is the energy distribution of β-rays continuous?
Q14. Explain one similarity and one dissimilarity between nuclear fission & fusion.
Q15. (a) Which property of nuclear forces is responsible for constancy of Ebn.?
(b) What do you mean by charge independent nature of nuclear forces?
Q16. Calculate energy equivalent of 1 a.m. 4 in MeV.
Q17. Show that nuclear density is same for allnuclei.
Q18. Define half-life & average life of a radioactive substance & find the relation between the two.
Q19. A given coin has mass of 3.0g calculate the nuclear energy that would be required to separate
all the neutrons and protons from each other. Coin is made of copper 63
Cu of mass 62.929604.
29
Q20. A radioactive isotope has half-life of Tyears How long will it take the activity to reduce to
(a) 3.125%
Q21. Define distance of closest approach and write its relation also.
Q22. What are the drawbacks of Rutherford ...... of atom?
Q23. Write down the three postulates introduced by Bohr which laid the foundations of quantum
mechanics.
Q24. Find the expression for energy of H-atom in the ground state.
Q25. What are the limitations of Bohr's model.
Q26. In a head on collision between an α-particle and a gold nucleus (z = 79) the minimum distance
of approach is 4 × 10–14m. Calculate the energy of α-particle.
Q27. At what speed must an electron revolve around the nucleus of H-atom so that it may not be
pulled into the nucleus by electrostatic attraction.
Q28. Calculate the frequency of revolution of electron in the second orbit or radius 2.12A°.
Q29. An H-atom rises from n = 1 to n = 4 state by absorbing energy. If P.E. of atom in n = 1 state
be –13.6 eV than calculation (i) P.E. in n = 4 state (ii) energy absorbed by atom in transition
(iii) wavelength of emitted radiation if atom returus to original state.
Q30. A 12.5 eV electron beam is used to bombard gaseous hydrogen at room temperature. What series
of wavelength will be emitted?
Q31. The energy of an electron in an excited H-atom is –34 eV. Calculate the angular momentum
of electron according to Bohr's theroy.
To be Learnt
Short Questions
Q1. A free proton can not decay into (n + e+ + v) because such decay is not energelically allowed.
Yet we observe in nature beta decay with position emission how do we understand the emission
of positrons from nuclei?
Q2. Calculate and compare the energy released by
143
(a) Fusion of 1 kg of hydrogen deep in thin the sun
(b) Fission of 1 kg of
Q3. The half-life of
235
92
U in a fission reactor.
90
38
Sr is 28 years. What is the disintegration rate of 15 mg of this isotope?
Q4. Two nuclei P & Q have equal no. of atoms at t = 0. Their half-lives are 3 hours and 9 hours
respectively compare their rates of disintegration after 18 hours from the start.
Q5. The B.E. per nucleon for dentron 21H and 42He are 1.1 MeV and 7.0 MeV respectively. What
would be the energy released when two dentrons fuse to form a helium nucleus?
Q6. M1 & M2 represent the masses of
M2 = 2M1 or M2 > 2 M1.
20
10
Ne nucleus and
40
20
Ca nucleus respectively. State whether
Q7. Show that Bohr's second postulate. The electron revolves round the nucleus only in certain fixed
orbits without radiating energy can be explained on the basis of de-Broglie hypothesis of wave
nature of electron.
Q8. Find the expression for frequency of electron in Bohr's stationary orbit and calculate its value
also.
Q9. Determine the speed of electron in n = 3 orbit of He+ in the non-relativistic approximation valid.
Q10. Which level of doubly ionized lithium has the same energy as the ground state energy of Hatom? Compare the orbital radii of two levels.
Q11. In a neon-atom the energies of 35 & 3p states are respectively 16.70 eV and 18.70 eV. What
wavelenth corresponds to 3p – 3s transition in neon atom?
Q12. Suppose you are given a chance to repeat the α-particle scattering experiment using a thin sheet
of solid Hydrogen (Hydrogen is solid at temp. below 14 K) in place of gold foil. What result
do you expect?
nh
) is a basic law of nature, it should be
Q13. In Bohr's quantisation postulate (ang. momentum =
2π
equally valid for the case of planetary motion also. Why then do we never speak of quantisation
of orbits of planets around the sum?
Long Questions
Q14. Explain the concept of nuclear binding energy. Draw a curve between mass no. and average.
B.E. pernucleon. Explain the energy release in the process of nuclear fission from this plot.
Q15. Draw a graph showing the variation of potential energy betweenn a pair of nucleons as a function
of their separation. Indicate the regions in which nuclear force is (i) attractive (ii) repulsive.
Q16. Explain the concept of nuclear forces. Discuss their characteristic properties. Which properties
distinguish them from electronstatic forces?
Q17. State and explain the laws of radioactive disintegration hence define disintegration constant and
half-life period establish relation between them.
Q18. State the basic portulates of Bohr's atomic spectra. Hence obtain an expression for the radius
of orbit and energy of orbital electron in H-atom.
Q19. Discuss Geiger Mardsen experiment on scattering of α-particles. How is the size of the nucleus
estimated in this experiment.
144
Pedagogical Remark
1.
A free proton cannot decay into a neutron as Q value of reaction becomes negative. The bound
nucleons behave differently than free nucleons. When we consider the energies of parent and
daughter nuclei involved in beta plus decay (positron decay) the decay is always energetically
allowed.
2.
In fusion 4 Hydrogen nuclei fuse to form a Helium nuclem with a release of 26 MeV of energy
∴ Energy released by fusion of 1 kg of Hydrogen =
E1 = 39 × 1026 MeV
6 × 1023 × 1000
× 200MeV
U
=
Energy released in fission of 1 kg of 235
92
235
E1
∴ E = 7.65.
E2 = 5.1 × 1026 MeV.
2
3.
T = 28 years = 28 × 3.15 × 107s
N, No. of atoms in 15 mg of
Rate of disintegration
90
38
Sr =
dN
.693
= λN =
N = 7.8 × 10Bq
dt
T
23
3
6 23
2
18
18
61×=10
6×10
× 26
15× 10No.
N 0 of P in 18 hours =
1× 1020hives
=6
of half
0 4.
2= ×N
=2 1.0038
,
N
=
N
=
3
0
9290 64
41000
2
4
No. of half lives of Q in 18 hours =
∴ no. of nuclei left undecayed
N1 = N0
∴
5.
R1 λ1 N1 T2 N1
9 N /64 3
=
=
= × 0
=
R 2 λ 2 N 2 T1 N 2
3 N 0 /4 16
For deutron B.E. = 4 × 1.1 = 4.4 MeV
For Helium B.E. = 4 × 7 = 28 MeV
∴ Energy released = 23.6 MeV.
6.
M1 < 16 (mp + mn) and M2 < 20 (mp + mn)
As B.E. of
∴
40
20
Ca > B.E. of
M 2 − 20(m p + mn )
40
⇒ M2 > 2M1
>
20
10
Nc
M1 − 10(m p + mn )
20
145
7.
de-Broglie wavelength for an electron moving on a line of length l with velocity v is
l =
h
h
=
mv p
p =
or
When electron revolves in a circular orbit or radius, r
then
2l = 2πr
∴
p =
i.e., angular momentum (p × r) of electron is integral multiple of
condition of angular momentum.
8.
This is Bohr's quantisation
Freq. is no. of revolutions per second
As
V = rw = r × 2πv
⇒
v =
=
∴
v =
9.
V =
∴
For He Z = 2, n = 3
v = 1.46 × 106 m/s
V
=
C
∴
∴
10.
9 × 109 × 1(1.6 × 10 −19 ) 2
= 6.57 × 1015 rps
1.6.6 × 10−34 × 0.53 × 10 −10
∴ Non relativistic approximate is true
9 1
= ⇒ n = 3 As
n2 1
r3 (li ) [n 2 / z ]li 9 / 3
=
=
= 3.
∴
r1 (H) [n 2 / z ]4
1
11. E = E2 – E1 = 2 eV = 2 × 1.6 × 10–19 J
λ =
= 618.8 nm.
12. Solid hydrogen is much lighter target compared to α-particle acting as projectile.
13. Aug momentum associated with planetary motion are
this will correspond to n
1070
1070 fivearth According to Bohr postulate
∴ Difference in successive energies and angular momenta of the quantised livesls will be so small
that levels will be continuous in stead of discrete.
146
9
1.
2.
Devices in which a controlled flow of electrons can be obtained are the building blocks of electronic
circuits. For example Vacuum diode,Triode,Tetrode, Pentode, Transistor, FET, SCR etc., are some
electronic devices.
Classification of metals, conductors and semi conductors: On the basis of electrical conductivity
(σ) or resistivity (ρ = 1/ σ ) the solids are classified as
(i) Metals – have low resistivity
and conductivity
(ii) Semiconductors – have intermediate resistivity
(iii) Insulators – have high resistivity
and conductivity
and conductivity
3.
Semiconductors
Si and Ge (Elements), CdS, GaAs, CdSe (Inorganic), anthracene, doped pthalocyanines (Organic)
and polyprrole, polyanitine (Organic polymers).
11
−
−
2
11
2
19
8
−
8
19
−
1
−6 4. −1 Energy
bands:band
When atoms come together to form a solid then outer orbit electrons come close and
srEmpty
:10
:10
:105−5to10
to10
to10
to10
Sm
ΩSm
Ω
mΩ
Sm
mm −1 Conduction
interact
exchanging energy. In the crystal each electron has a unique position and a different energy
EC
Eg
level. These different levels form a continuous energy variation and form group of levels called
Energy
energy bands. The energy band of valence electrons is called valence band and higher energy group
gap
EV free electrons is conduction band. There exists a group of levels devoid of electrons between
having
Valence band
Filled
the conduction and Valence band and is called as Energy gap or Forbidden gap.
(Infinite no. of states
Metals:
Conduction band and valence band overlap or lie close so that valence electrons move to
each occupied by
2
electrons)
conduction band easily and move freely and cause charges to flow.
Insulator: A energy gap of 6eV between the two bands with bound electrons in the valence band and
no free electrons in the conduction band.
Semiconductor:A energy gap of 1-1.5eV between valence and conduction band so that valence
electrons may gain external energy to cross the gap and move into the conduction band. This creates
vacancy and possibility of conduction in both the bands.
Fig. 1: Energy band positions in a semiconductors at 0 K
147
5.
Intrinsic semiconductor: Each Si or Ge atom shares one of its 4 valence electrons with 4 nearest
neighbour atoms (diamond like structure) in a covalent bond. At low temperatures no bonds are
broken but at high temperature, the electron breaks free and moves to interstitial space, hence creating
a vacancy in the bond or hole which behaves as an apparent free particle. In intrisic semiconductors,
ne = nh = ni, where ne = no. of electron, nh = no. of holes, ni = intrinsic carrier concentration. An
electron from another bond can occupy this hole site creating hole from the vacated site.
Si atom
covalent bond
Si
Si
Si
electron
pair
Si
Si
Si
Si
Si
Si
Fig 2:Si crystal (electron leaves bond to create hole)
6.
Extrinsic semiconductor: Impurity like pentavalent (As, Sb, P) or trivalent (In, B, Al) are added to
get better conductivity. Depending on the doping we have two type of semiconductors namely n –
type and p – type semiconductor.
(a) n – type semiconductor: Si or Ge with pentavalent doping. Four valence electrons form 4 covalent
bonds but 5th electron is free and weakly bound to parent atom. The ionisation energy (~ 0.01V
for Ge and 0.05V for Si) is small and even at room temperature the electron jumps to conduction
band. The dopant is called donor impurity.
Fig.3: (a) n-type
(b) p - type semiconductor: Si or Ge with trivalent doping means one less electron in the 4 covalent
bonds, so the 4th neighbour has a vacancy or hole that can be occupied by an electron from
another site. Thus a hole is available for conduction. The trivalent atom is called acceptor atom
or impurity.
Fig.3: (b) p-type
148
7.
p – n junction:
(a) Formation of p – n junction: Fusion of a p-type and n– type forms the junction. There is
concentration gradient between p and n sides, majority carrier-holes diffuse from p side to n side
(p ® n) and minority carrier - electrons move from (n ® p) creating a layer of positive and
negative charges on n and p side respectively. This formation deplets the flow of majority carriers
across the junction and is called depletion layer. The potential difference created here acts as a
barrier for the flow of majority carriers and so it is called as Potential Barrier (Vo). External
bias needs to be applied to cause charge flow.
Fig 4: p-n junction
(b) p – n junction under forward bias: When p – side is connected to positive terminal and n – side
to negative terminal of external voltage, it is said to be forward biased. The applied voltage V is
opposite to built in potential V0, hence depletion layer width decreases and barrier height is
reduced to (V0 – V). There is flow of majority carrier flow with a current of the order of mA.
V
W
–
–
–
–
W
p
metallic
contact
–
–
V0–
–
– –
–p –
– –
– –
+
+
+
+
– – ++
+ –+ – + +
+ –+ – + + n
+ –+ – + +
+ + (c) p
+
+
p
n
+
1 (no battery)
+
2 (low battery)
(a)battery)
Forward biased p-n junction,
3 (high
(b) Barrier potential in F.B.
– n junction under reverse bias: The direction of applied voltage is same as direction of
metallic
W depletion barrier potential, so barrier height increases to (V + V). This suppresses flow of majority carriers
contact
0
layer
and so diffusion current decreases but drift of electrons and holes under the electric field remains.
This drift current is of the order of mA. The current under reverse bias is independent of applied
voltage upto a critical value known as breakdown voltage (Vb) when V = Vb diode reverse
current increases sharply. If the reverse current is not limited below this, the diode gets destroyed
due to overheating.
n
2
1
V0
(a) Diode under reverse bias
(b) Barrier potential under R.B.
(d) Study of V – I characteristics of a diode: The circuit to study the variation of current as a
function of applied voltage is shown. In forward bias we use milliammeter and reverse bias we
use microammeter.
149
Voltmeter (V)
p
n
Voltmeter (V)
p
(mA) milliammeter
n
switch
+
microammeter (µA)
switch
–
(a)
– +
(b)
(a) p-n junction in FB
(b) p-n junction in RB
I(mA)
100 80 60 40 20
V
Vbr
10
20
30
0.2 0.4 0.6 0.8 1.0 V (V)
I(µA)
(c)
(c) V - I. characteristics graph
The resistance offered by the device is not a constant and is called as Dynamic resistance given by
Small change in voltage
⇒ rd =
Small change in current
V o l ta g e a t A
Application of junction as a rectifier:
(a) Half wave rectifier: Junction diode allows current to pass through only if it is forward biased,
hence a pulsating voltage will appear across the load only during positive half cycles when diode
is F.B. Here, reverse breakdown voltage of diode must be higher than peak a.c. voltage at the
secondary of the transformer to prevent breakdown of diode.
V o l ta g e a c r o s s
L o ad R L
8.
Input a. c.
time
output voltage
(b) Input ac and output voltage waveforms
time
150
(b) Full wave rectification: The circuit uses two diodes connected to the ends of a centre tapped
transformer. The voltage rectified by the two diodes is half of the secondary voltage i.e., each
diode conducts during one half cycle of input but the net output across the load remains half
sinusoidal with positive values only (as shown ). For positive cycle diode D1 conducts (FB) but
D2 is being out of phase is reverse biased and does not conduct. Thus output across RL is due to
D1 only. In negative cycle of input D1 is R.B. but D2 is F.B. and conducts as with respect to centre
tap point A is negative but B is positive. Hence output across RL is due to D2.
Centretap
transformer A
Input
D1
X
Centretap
RL Output
B
D2
Y
Modification of variable positive sinusoidal signal can be done using capacitor filter. The capacitor is
charged when voltage rises till the peak voltage. When a load is connected the capacitor discharges
t
through
it
and
get
recharged
in the next half cycle. The rate of voltage fall depends on (RLC) and is
Due to Due to Due to Due to
D
D
D
Dcalled
1
2
1
2 We use large C to make time constant large. Here output voltage is near peak
time
constant.
voltage. This filter is used
in power supplies.
t
dc component
X
Input
a. c.
+
–
C
RL
Y
9.
output ac input
Output
waveform
across RL
t
Rectifier
waveform waveform
at B
at A
Fig.9: Full wave rectification circuit
t
t
(a) Full wave rectifier with capacitor filter
(b) Input/output voltage
Special purpose p – n junction diode:
(a) Zener diode: It is designed with heavy doping of p and n sides of p – n junction. Depletion
region is very thin of the order of 10–6 m. Electric field of junction is very high ~ 5 × 106 V/m and
is operated in reverse bias ~ 5V.
151
I – V characteristic: The current increases after breakdown voltage VZ, current varies but voltage
remains constant, so it is used in voltage regulators.
Symbol
for
Zener
diode
Fig.11: I-V character of Zener diode
(b) Zener diode as voltage regulator:
Unregulated
Supply
RS
VL
IL
Regulated
Lood voltage
RL
Zener
VZ
10.
Fig.12: (a) Zener diode as d.c. voltage regulator
The unregulated d.c. voltage is connected to Zener through series resistance such that Zener is
reverse biased. Any increase in input voltage causes current through RS and Zener to increase.
Voltage drop across RS increases without any change in Zener voltage (In breakdown region
voltage across Zener is constant). If input voltage decreases, voltage across RS decreases (as I
decreases) without any change in Zener voltage. Hence in RB Zener voltage remains constant (I
may increase or decrease) so it gives constant voltage across output resistance RL (voltage across
RS changes according to input). The resistance RS is chosen depending upon output requirement.
Optoelectronic junction devices:
(a) Photodiode: It is a p – n junction fabricated with a transparent window to allow light photons to
fall on it. These photons generate electron hole pairs upon absorption. If the juction is reverse
biased using an electrical circuit, these electron hole pair move in opposite directions so as to
produce current in the circuit. This current is very small and is detected by the microammeter
placed in the circuit.
I4>I3>I2>I1
p
I
(m A)
R. B.
n
µA
Fig.13: (a) Photodiode (R.B.),
I1
I2
I3
I4
V
I (µ A)
(b) I-V character of photodiode for
different illumination intensities
152
(b) Light emitting diode: It consists of heavily doped p – n junction in forward bias. Electrons
move from n ® p and holes from p ® n (minority carriers). Thus, near junction, minority carrier
concentration increases (under no bias it is less) and they combine with majority carriers near the
junction to release energy in form of photons with energy equal to or less than band gap energy.
As forward bias increases, current increases till light intensity reaches maximum. V – I character
is same as junction diode except threshold voltage is high and reverse breakdown voltage is low
(~ 5V). Band gap ~ 1.8 eV, light emitted is 0.4 mm to 0.7 mm (3 eV to 1.8 eV is Eg). Compound
semiconductor – Gallium arsenide – phosphide.
(c) Solar cell:
hν
Top
contact
n
p
Top
contact
(metallised
finger electrode)
bottom
contact
IL
I
Voc (open circuit
voltage)
V
n
p
p
Isc (Short circuit
current)
n
Back
Contact
depletion larges
(i)
11.
(ii)
Fig.14: Solar cell (emf generation), (ii) I-V characteristic
Solar cell generates emf when light falls on its junction (large).
A p – Si wafer (300 mm) over which thin n – Si layer is grown by diffusion process is used. p –
side is coated with metal (back contact) and n – side has a metallic grid electrode (top contact)
which occupies < 15% of area.
Photo voltage generation occurs by 3 steps generation, separation and collection of e – h pairs.
Photocurrent IL flows through load (I – V character shown). Band gap of materials used for solar
cells is 1.5 eV e.g.: Si (Eg = 1.1 eV), Ga As (Eg = 1.43 eV) etc.
Junction Transistor: (a) Types: (i) n-p-n type, (ii) p-n-p type.
(b) Structure terminals: (i) Emitter (E), (ii) Base (B), (iii) Collector (C)
Emitter is heavily doped
Collector is lightly doped
Base is lightly doped and very thin.
(c) Representation:
E
E
B
C
E
E
n p n
B
C
p n p
C
C
B
npn type
B
pnp type
153
(d) Symbol:
E
E
C
C
B
npn type
B
pnp type
(e) Bias of a transistor
Emitter base junction – forward biased
Base collector junction – reverse biased
(f) Common emitter transistor characteristics:
C
IB
R2
+
–
µA
IC
VBE
VBB
mA
R1
+
B
E
IE
–
VCE
VCC
Fig.17: Circuit to study characteristic of a transistor
For input characteristics, base current IB vs base-emitter voltage VBE is plotted while collector
base voltage VCE is kept constant. VCE is kept large 3V to 20V. Input characteristics for various
values of VCE gives almost same curves.
Input characteristics
IB 100
(µA) 80
60
40
20
VCE = 10V
0
0.2 0.4 0.6 0.8
VBE(V)
(b)
(c)
VBE ≅ 0.7 V,
VCE = VBE + VCB
Output character is obtained by varying IC with VCE keeping IB constant. Different curves are obtained
for different values of IB.
AC parameters:
(i) Input resistance =
⇒ ri =
→ dynamic resistance
154
(ii) Output resistance, r0 =
∆VCE
∆IC
IB
(iii) Current amplification factor (b)
;
12.
⇒
Transistor devices:
(a) Transistor as a switch – Operates from the saturation to cut-off state.
V0
cut off
active Saturation
Av
Vi
Fig.18: (a) Transistor as switch circuit
(b) Transfer characteristic
Output voltage, VCE = VCC – IC RC = V0 (VCE = output d.c. voltage)
(i) Vi is low, transistor is not forward biased, V0 = VCC (high) Þ transistor switched off and output
is saturated.
(ii) When Vi is high, transistor is forward biased, V0 = low ∼ zero Þ transistor switched on and
C
∆
R
V
∆
I
∆
I
I
β acC;0βRdcCCC∆CIC RB the output
isI low. C
β
=
C
dc =
ac
βR
∆acV
I B∆
I B∆
I BI
Itwo
So
a
stateRCoperation
B∆
i R
C
+ is possible.
B B VCE I
B
C
R
B Transistor Cas an amplifier
RB
(b)
∼ V–0V0 (C E configuration): Transistor works in active region.
Vi
VBB
I
E
Output,
V0 = VVCC – IC RC ...............(1)
IB
CC
E
Vi ∼
VCC increases, output voltage decreases. Vi and V0 are out of phase.
If input
voltage
VBB
I
E
AV = small signal voltage gain =
...........(2)
Fig.19: Amplifier Circuit
AV = –
=–
where
= a.c. current gain
155
Voltage gain, AV =
13.
14.
; Power gain, AP = bac × AV
Since bac and AV are greater than 1.
Transistor as an oscillator: External output is obtained without input for an a.c. signal by means of
feedback. This is a positive feedback amplifier. The output and nput will be in phase as there will be
p radians of phase change both by the CE amplifier and the coupl;ing ( may be inductive coupling).
Logic Gates: Devices with one or more input but a single output are called Logic Devices. They
work based on logical relationship and not on the magnitude of the input. If in the logical way, “0” is
considered the low/absence/off level and “1” is considered the high/presence/on level, then we say it
is Positive Logic.
A
(i) OR gate: Output is high when one or more input (atleast one) is high.
Y
B
Symbol
Truth Table
Operation
A B
Y
Y=A+B
0 0
0
0 1
1
1 0
1
1 1
1
(ii) AND gate: Output is high only when all inputs are high.
Symbol
A
B
Y
Truth Table
A B
Y
0 0
0
0 1
0
1 0
0
1 1
1
(iii) NOT gate: Output is opposite of input. It is also called as inverter.
156
Operation
Y=A.B
Truth Table Operation
A
0
1
(iv) NAND
Y
Y=
1
0
gate: AND gate followed by a NOT gate. When both inputs are high, output is
Truth Table
low.
A B
Y
0 0
1
0 1
1
1 0
1
1 1
0
(v) NOR gate: OR gate followed by a NOT gate. Output is high only when both inputs low.
Symbol
≤
A
A
A
B
B
15.
Truth Table
A B
Y
0 0
1
0 1
0
Y
1Y 0
0
1 1
0
Both NOR and NAND are called as universal gates as all other gates and logical devices can be
obtained by using these gates only.
Integrated circuits: If an entire circuit is fabricated on a single chip, it is called technologically as
Monolithic IC.
Size 1 mm × 1 mm.
Categories on basis of nature of input signal.
(a) Linear or analogue IC’s – process analogue signal. Output is directly or linearly proportional to
input e.g.: Operational amplifier.
(b) Digital IC – Signal have 2 levels. IC’s can be subdivided into
(i)
Small scale integration SSI
(ii) Medium scale integration MSI
- Logic gates
- Logic gates
10.
100.
(iii) Large scale integration LSI
- Logic gates 1000.
(iv) Very large scale integration VLSI - Logic gates > 1000.
Exercise
157
Answer Yourself
Very Short Questions
Q1. What are dopants? Give example.
Q2. Draw the schematic representation of n–type material with donor impurity.
Q3. What is diffusion current?
Q4. What is barrier potential?
Q5. What causes the current in F.B.? Why is current high in F. B.more than that in R.B.?
Q6. What is the advantage of using capacitor in full wave rectification?
Q7. Give one application of photodiode.
Q8. What material is used to make a LED?
Q9. What is the difference between the input and output in a CE transistor amplifier?
Q10. Is Oscillator an Amplifier? How?
Q11. In half-wave rectification, what is the output frequency if the input frequency is 50 Hz. What is the
output frequency of a full-wave rectifier for the same input frequency?
Q12. A p-n photodiode is fabricated from a semiconductor with band gap of 2.8 eV. Can it detect a
wavelength of 6000 nm?
Short Questions
Q1. Why do we say that an intrinsic semiconductor is like an insulator at 0 K?
Q2. Compare intrinsic semiconductor with doped semiconductor.
Q3. How is p – n junction formed ? Explain.
Q4. Describe experimental setup to study V – I characteristics of a p–n junction diode.
Q5. What is a Rectifier?How can a junction diode be used as a full wave rectifier.
Q6. What are optoelectronic devices? Name their types.
Q7. Show how intensity of current varies with illumination intensity in a photodiode. Give Reason.
Q8. How LED’s are more advantageous over incandescent lamps?
Q9. What are the important criterion for selection of a solar cell material?
Q10. Draw the symbol and logical equation with truth table of a AND gate.
Q11.
Explain with a neat diagram, the working of a n-p-n transistor.
158
Long Questions
Q1. Draw the characteristic curve of a transistor in CE configuration and thereby identify the regions in
which the transistor be used as an amplifier and switch. Explain the working of transistor as an
amplifier with necessary equations on potential.
Q2. What is the meaning of feedback? Give a method by which feedback is given. Explain how transistor
amplifier when given a positive feedback works as an oscillator.
Q3. What is a rectifier? How is the rectification possible in a semiconductor device? List any two types
of rectification and explain one of them. Show the input and output signals and write the frequency
associated with them.
Q4. Name the device that can be used as a voltage regulator. What is the character possessed by this
device to do this function? With a relevant circuit diagram explain how voltage regulation is done?
To be Learnt
A
B
Q1. Pieces of Cu and Ge are cooled from room temperature to 80 K. What will be the effect on their
resistances?
Q2. Three amplifiers are connected in series. The voltage gain of each is 5. What is the net voltage
amplification?
Q3. The current in the forward bias is known to be more (mA) than the current in the reverse bias (mA).
What is the reason then to operate the photodiodes in reverse bias?
Q4. Why is GaAs preferred over Si for making Solar Cells?
Q5. You are given the A
two circuits as shown in figure. Show that circuit (a) acts as OR gate while the
circuit (b) acts as
AND
gate.
Y
Y
B
(a)
(b)
Q6. Write the truth table for circuit given in figure below consisting of NOR gates and identify the
logic operation (OR, AND, NOT) which this circuit is performing.
A
A
Y
B
B
Q7. What is the type of bias used in a transistor?
Q8. What is the phase difference between the output and input in a CE transistor amplifier?
Pedagogical Remark
1.
Cu is a conductor and Ge is a semiconductor. With increase in temperature in conductors resistance
increases and in semiconductors resistance decreases.
159
2.
3.
4.
5.
6.
7.
8.
Output of one amplifier is the input for the next and so the gain of each amplifier will be multiplied.
On providing energy- illumination, there will be creation of electron-hole pairs. Since in the case of
n type semiconductors electrons are more, the fractional change in the electron number is much less
in comparison to the fractional change in the hole number. This makes the minority carrier dominated
flow is easily detectable than the majority carrier dominated flow. So the photodiodes are operated
in reverse bias.
GaAs is preferred over Si as the energy absorption coefficient is higher in it than Si.
(a) A nad B are given to a NOR gate and the output is further given to a NOT gate. So the final
output will be a OR gate output. The output will be Y= A + B
(b) The inverted signals of A and B are given to a NOR gate. This makes a AND gate. The output
will be Y= A B
The output of the NOR gate is y1= (A+B)’ and the fused NOR gate wil be again an inversion making
the output Y= (A+B). So the combination behaves as a OR gate.
Always for a transistor to work, the emitter should be forward biased and the collector should be
reverse biased.
The output is always 1800 out of phase with the input in CE transistor amplifier. Any increased FB in
the emitter leads to a drop in the output signal.
Practical - An approach
1.
2.
3.
4.
5.
6.
All the apparatus required for doing the experiments are available in boards and study of the characters
can be done easily provided the manual given with the set-up’s are used correctly.
p-n junction chatacteristics, Zener diode characteristics, Transistor characters and Gate models are
available with plug-in and use format.
One need to take care that the potential limit given for each device should be strictly adhered to.
The variation in the potential and current are to be done cautiosly so that we do not overshoot the
prescribed limit.
Teacher may do the experiment once before he/she demonstrates in the class. Getting the ideal
graph for the individual set-up’s is not difficult.
Interpretation of the graph need to be taught as and when the student completes the graph.
Remember the practical work on this chapter makes the student to be clear of the design characters of
the devices used by him.
160
10
●
Communication is the act of transmission of information. A communication system has three
essential elements : transmitter, medium and receiver.
●
Two basic modes of communication: point-to-point and broadcast. In point-to-point, communication
takes place over a link between a single transmitter and a receiver e.g. Telephony. In broadcast
mode, there are a large number of receivers corresponding to a single transmitter e.g. Radio and
television.
●
Transducer: Any device that converts one form of energy into another can be termed as a transducer.
●
Signal: Information converted in electrical form and suitable for transmission is called a signal.
They can be either analog or digital.
●
Noise: Noise refers to the unwanted signals that tend to disturb the transmission and processing
of message signals.
●
Transmitter: A transmitter processes the incoming message signal so as to make it suitable for
transmission.
●
Receiver: A receiver extracts the desired message signals from the received signals at the channel
output.
●
Attenuation: The loss of strength of a signal while propagating through a medium is known as
attenuation.
●
Amplification: It is the process of increasing the amplitude of a signal.
●
Range: It is the largest distance between a source and a destination up to which the signal is
received with sufficient strength.
●
Bandwidth: It refers to the frequency range occupied by the signal. E.g. speech signal requires
a bandwidth of 2800 Hz, video signals require 4.2 MHz & TV signals (voice and picture) require
6 MHz of bandwidth.
●
Modulation: At the transmitter, information contained in the low frequency message signal is
superimposed on a high frequency wave, which acts as a carrier of the information. This process
is known as modulation.
●
Demodulation: The process of retrieval of information from the carrier wave at the receiver is
termed demodulation.
●
Repeater: It is a combination of a receiver and a transmitter. It, picks up the signal from the
transmitter, amplifies and retransmits it to the receiver. They are used to extend the range of a
communication system.
●
Detector: It separates the modulating wave (the message) and the carrier wave. It consists of a
rectifier and a R-C CIRCUIT.
161
Types of Communication Systems
●
According to the nature of information source
Speech transmission (as in radio)
Picture transmission (picture including moving and live pictures)
Facsimile transmission (Fax) data transmission
●
According to the mode of transmission
Analog communication
Digital communication
●
According to the transmission channel
Line communication
Two wire transmission line
Coaxial cable
Optical fibre cable
Space communication
●
According to the type of modulation
Amplitude modulation (AM)
Frequency modulation (FM)
Phase modulation (PM)
Propagation of Electromagnetic waves
Fig. 1 Various propagation modes for em waves
162
●
Ground wave—To radiate signals with high efficiency, the antennas should have a size comparable
to the wavelength λ of the signal (at least – λ/4). At longer wavelengths (i.e., at lower frequencies),
the antennas have large physical size and they are located on or very near to the ground, the wave
glides over the surface of the earth. The mode of propagation is called surface wave propagation.
A wave induces current in the ground over which it passes and it is attenuated as a result of absorption
of energy by the earth. The attenuation of surface waves increases very rapidly with increase in
frequency.
●
Sky wave—In the frequency range from a few MHz up to 30 to 40 MHz, long distance communication
can be achieved by ionospheres’ reflection of radio waves back towards the earth. This mode of
propagation is called sky wave propagation and is used by short wave broadcast services.
●
Space wave—Electromagnetic waves of frequencies higher than 30 MHz penetrate the ionosphere
and escape. A space wave travels in a straight line from transmitting antenna to the receiving
antenna. Space waves are used for line-of-sight (LOS) communication & satellite communication.
At frequencies above 40 MHz, communication is essentially limited to line-of-sight paths .Television
broadcast; microwave links and satellite communication are some examples of communication
systems that use space wave mode of propagation.
Fig. 2 Line of sight communication by space waves
●
Modulation and its Necessity
The purpose of a communication system is to transmit information or message signals. These are also
called baseband signals. No signal has single frequency but it spreads over a range of frequencies called
its BANDWIDTH. We can not transmit signal frequency less than 20 kHz over a long distance directly
due to following difficulties.
●
SIZE OF THE ANTENNA/AERIAL—Antenna should have size comparable to the wavelength
of the signal (at least λ/4)
●
EFFECTIVE POWER RADIATED BY THE ANTENNA—The power radiated by the antenna
is inversely proportional to λ2 .This implies that for the same antenna length, the power radiated
by short wavelength or high frequency should be large.
●
MIXING UP OF SIGNAL—Suppose many transmitters are transmitting baseband information
signals simultaneously. All of these signals will get mixed up and there is no way to distinguish
between them.
So there is a need for translating the original low frequency baseband message or information
signal into high frequency wave before transmission such that the translated signal continues to possess
the information contained in the original signal. This is achieved by a process known as modulation.
●
Amplitude Modulation
In amplitude modulation the amplitude of the carrier wave is made to vary proportional to the message
signals.
163
Let
c (t) = AC sin ωCt – represents carrier wave
m (t) = Am sin ωmt – represents message signal
Where
ωm = 2πfm is angular frequency of the message signal
cm(t) = ( AC + Am sin ω mt ) sin ω C t
=
= AC sin ω C t + µAC sin ω m t sin ω C t
where
µ =
called modulation index, m ≤ | to avoid distortion
cm(t) =
The frequncy we is known as carrier frequency and (ωC + ωm) and (ωC – ωm) are known as upper
and lower side band respectively.
●
Block Diagram of a Simple Modulator for Obtaining an am Signal
Fig. 3 Block diagram of a simple modulator for obtaining an AM signal
●
Block Diagram of a Detector for am Signal
Fig. 4 Block diagram of a detector for AM signal. The quantity on y-axis can be current or voltage
164
●
Block Diagram of a Transmitter
Fig. 5 Block diagram of a transmitter
●
Block Diagram of a Receiver
Fig. 6 Block diagram of a receiver
Answer Yourself
Very Short Questions
Q1. What is communication?
Q2. Name the three basic units of any communication system.
Q3. What is a communication channel?
Q4. What is a transducer? Give an example.
Q5. Name the type of communication in which the signal is a discrete and binary coded version of
the message.
Q6. How does the power radiated by an antenna vary with wavelength?
Q7. What do you mean by bandwidth?
Q8. Find the length of the dipole antenna for a carrier wave of wavelength λ?
Q9. What is audio frequency range?
Q10. What is modulation?
Q11. What is demodulation?
Q12. What is a modem?
165
Q13. Which is more efficient mode of transmission FM or AM?
Q14. What is a ground wave?
Q15. What is a sky wave?
Q16. How does the effective power radiated by an antenna vary with the wavelength?
Q17. Why are labers preferred for optical fiber communication?
Q18. What is a geostationary satellite?
Q19. What is a transponder?
Q20. What type of modulation is required for television broadcast?
Short Questions.
Q1. Distinguish analog and digital transmission.
Q2. Write the functions of transponder and repeater in Communication systems.
Q3. Distinguish amplitude modulation and frequency modulation.
Q4. Explain the need for modulation.
Q5. What is the meant by the term attenuation?
Q6. Why ground wave transmission is restricted to 1500 kHz?
Q7. Define critical frequency in sky wave propagation.
Q8. Why AM radio reception is affected by electrical disturbances while FM radio reception is not?
Q9. Draw a block diagram showing the basic communication system.
Q10. What is the range of frequencies used for FM radio transmission?
Q11. Draw a block diagram showing a simple amplitude modulator
Q12. Explain how a message signal is extracted from Amplitude Modulated signal.
Q13. Distinguish Sky wave propagation and Space wave propagation.
Q14. Name the different layers of earth’s atmosphere.
Q15. Mention two ways in which the range of transmission can be increased.
Q16. Explain the term modulation index and describe its significance.
Q17. List four devices used for communication each of which differ from other in the mode of
communication.
Q18. Derive an expression for amplitude modulated wave.
Q19. What should be the length of a dipole antenna for a carrier wave having frequency 3 × 108
Hz?
Q20. Explain how communication is possible with the help of sky wave propagation?
Q21. Derive the expression for modulation index.
166
To be Learnt
Very Short Questions
Q1. Which is more efficient mode of transmission FM or AM?
Q2. In a carrier frequency of 100 kHz and a modulating frequency of 5 kHz, what is the bandwith of
AM transmission?
Q3. What is data retrieval?
Q4. What is a modem?
Q5. Name the type of modulation technique preferred for digital communication.
Q6. Name the type of radio wave propagation involved when TV signals, broadcast by a tall antenna,
are intercepted by the received antenna directly.
Q7. What is the cnt-off frequency beyond which the ionosphere does not reflect electromagnetic
radiations?
Q8. Why is it necessary to use satellites for long distance TV transmission?
Q9. What is transponder?
Q10. What is the function of cladding in a typical optical fibre?
Short Question
Q11. Why is an FM signal less susceptible to noise than an AM signal?
Q12. Deduce an expression for the distance upto which the TV signals can directly be received from
a TV tower of height hT.
Q13. Give an expression for the maximum LOS distance between two antennas for space wave
propagation.
Q14. Define the term critical frequency.
Q15. A bandwith of 5MHz is available for A.M. transmission. If the maximum audio signal frequency
used for modulating the carrier is not to exceed 5 kHz, how may station can be broadcast within
this band?
Q16. What should be the height of a transmitting antenna if the TV signals is to cover radius of 128
km?
Pedagogical Remark
1.
2.
3.
4.
FM
10 kHz
It is process of recovering the orignal data from the modulated wave.
It is a device which can represent digital data by analog signals and vice-versa.
167
5.
Pulse code modulation.
6.
Line of sight communcation.
7.
40 MHz.
8.
TV signals are not reflected by ionosphere due to high frequency.
9.
A device which receives the signal and retransmits it after amplification.
10.
It helps is multiple total internal reflections.
11.
In FM, amplitude remains constant, and noise is a form of amplitude variations due to atmosphere.
So amplitude limiters can be used to reduce the noise.
12.
d=
13.
d=
14.
fc =
15.
Bandwidth = 2 × 5 = 10 kHz.
2h T R
No. of stations which can be broadcast =
16.
d2
= 1280 m.
hT =
2R
❑ ❑ ❑
168
Life of Science Articles
2.
3.
The science teachers have been facing great difficulty in maintaining the stock of science materials.
Majority of them do not know the life of various science equipments. To help, the teachers, the
“Science Branch News Bulletin” has reproduced the copy of the circular issued by the Director
of Education order No. F-4 (52)/-61 Edn (P) dated 2nd August, 1962 which is as follows :
The Director of Education is pleased to accept the recommendation of the Committee, appointed
by him, to fix the life of furniture articles and other non-consumable articles (science equipments
etc.) used in the Directorate of Education, Delhi/and the Govt./Aided schools in the Union Territory
of Delhi, vide order no. F(52)/61-Edn({) dated 1.9.1961 and accordingly fixes the life of each
item, as shown in the Annexures “A”, B, C, D, E, F, H and I to the extent mentioned therein.
In case there is any other item, the life of whih has not been fixed, the same may kindly be intimated
to this Branch urgently, so that the same may also be considered.
In the case of schools, which are/were in tents, the life of the articles, to be condemned, may
be reduced by 25 p.c. of the life, as fixed in the attached Annexures, provided the condemnation
Board are satisfied for the proper use of the articles in question. A certificate to this effect must
be endorsed on he lists. (This para was subsequently added vide letter no. F.4(52)/61-Edn(P), dated
December, 1962).
Annexure ‘A’ (Years)
1.
2.
3.
4.
5.
6.
7.
8.
9.
10.
11.
Almirah wooden
Almirah Iron
Black Board
Black Board Stand
Benches
Buckets (Tin)
Bicycle
Bicycles stand (iron)
Bicycle stand (wooden)
Chairs wooden seat
Chairs iron/steel seat
20
50
3
3
5
2
8
10
3
5
10
12.
13.
14.
15.
16.
17.
18.
19.
20.
21.
22.
169
Chairs cane seat
Cash Box (wooden)
Cash Box (iron)
Chauki (Takhat)
Desk Single Shift
Desk Double Shift
Stool
Durries
Rack (wooden)
Rack (iron)
Officers Table
5
5
25
10
7
5
3
5
10
25
20
23. Teachers Table
7
35. Notice Board
5
24. Office Table
10
36. Tray (wooden)
5
25. Library Table
10
37. Tray (iron)
5
26. Physics Table
10
38. Paper Stand
7
5
27. Chemistry Table
8
39. Foot Rest
28. Domestic Science Table
8
40. Hat Hanger & Looking glass
10
10
41. Confidential Box
10
2
42. Teapoy wooden
10
10
43. Carpet
10
32. Newspaper Stand
5
44. Trunk
10
33. Waste Paper Basket (Tin)
5
45. Map Stand
5
34. Waste Paper Basket (wooden)
3
46. Cash Safe
50
29. Biology Table
30. Table Cloth
31. Screens
Annexure ‘B’ Physics Apparatus (Non-consumable)
1.
Balance (Spring)
5
21. Screen (Glass)
2
2.
Balance (Physical)
5
22. Lens Stand (wooden)
2
3.
Weight Boxes
5
23. Optical bench (wooden)
2
4.
Boyle’s Law Apparatus
5
24. Spectomdeter
5
5.
Vernier Callipers
5
25. Wire Gauge Stand Iron
5
6.
Fortin’s Barometer
5
26. -do-Wooden
3
7.
Metallic Cylinder
5
27. Travelling Microscope
8.
Metal Sphere
7
28. Copper Calorimeter
5
9.
Metres rod (wooden)
1
29. Hypsometer (Copper)
5
10. S.G. Bottle
5
30. Thermometer
1
11. Spherometer
5
31. Max. Min. Thermometer
5
12. Screw Gauge
5
32. Magnet (bar)
5
33. Compass needle
3
13. StopWatch
10
10
14. Inclined plane
5
34. Compass (for lines of force)
5
15. Gravesand’s apparatus
5
35. Deflection Mangnetometer
5
16. Young’s modulus
5
36. Ammeter
10
17. Concave Mirror
2
37. Voltameter
10
18. Convex Lens
5
38. Galvanometer
7
19. Glass Prism
5
39. Accumulator
2
20. Glass Slab
5
40. Laclanche Cell
2
170
41. Electric Bell
2
64. Tunning Forks
2
42. Electrophorus
5
65. Resonance apparatus
5
43. Gold leaf electroscope
5
66. Stove (Oil)
5
44. Glass rod
1
67. Binoculars
10
45. Ebonite rod
1
68. Soldering Rod (Fire)
5
46. Silk and Cat Skin pieces
1
69. Solder (Electric)
2
47. Proof plane
2
70. Graduate cylinder
2
48. Slide wire Bridge
5
71. Glass Plate Machine
5
49. Potentiometer
5
72. Spirit Level
5
50. One Way and two way keys
5
73. Battery Clamps
3
51. Resistance box
5
74. Siren
7
52. Rheoslat
5
75. Hydrometer
5
53. Resistance Coil
5
76. Lactometer
5
54. Stading Key (two keys)
5
77. Drawing Board
2
55. Tangent Galvanometer
7
78. Barometer Tube
2
56. Induction coil
5
79. Photographic Camera
20
57. Torch lamp holder
2
80. Telescope
20
58. Switches
1
81. Newton’s Disc
10
59. cutout fuses
1
82. Pin hole camera
10
60. Pliers
5
83. Microscope
30
61. Spirit Lamp
4
84. Epidiascope
20
62. Tripod Stand
5
85. Radio set
10
63. Retort stand and clamps
5
86. TV. Set
10
G
University Physic : H.D. Young, M.W.
Zemansky and F.W. Sears, Narosa Pub. House.
G
Physics - Foundations and Frontiers : George
Gamow and J.M. Ciearland, Tata Mcgraw Hill.
171
172
2a
2b
Name
3
Date of
purchase
4a
Normal
in
building
4b
used in
tents/
double
Shifts
Total Life in lived
8
required
Pagein stock
Total
9
required to
be written off
No./quantity
10
article in
Stock
Balance of
serviceable
11
stock/property
Register
Rate as per
Consumable/Non-Consumable Articles To be Written Off
1
No. No.
No. & Name of
the Article
5
life of the
Articles
Fixed
Vide letter
No.
accordance
of 4(52) 61
Edn
Dated
2.8.82
12
Total cost of the
articles required
to be written off
List A (Whose Life Has been Laiddown) (Unserviceable Articles)
Name of the School——————
PROFORMA OF CONDENMATION
13
Remarks if any
affecting the life
of the articles
not covered by
other Columns
6
Whether the
articles
mentioned in
Col. 9
have
completed
their life in
accordance with
the life
mentioned
in Co. 5
7b
Page
14
of the head of the
Insititution
Recommendation
7a
S.No.
S. No & Page in
the
Property Register
List of Non-consumable
173
Certified that in the above statement only those items have been included whose life has been fixed vide letter No. F4 (52)761
Edn 2-8-62 as amended.
Certified that the articles mentioned in Col. 9 has already completed their life as fixed vide letter No. F(52)/61 Edn. 2-8-62
amended.
Item Nos. ----------- have been physically verified to be unserviceable and are recommended for being written off.
Certified that the articles contained in the list have been checked from the stock register and the date of the purchase price etc.
have been certified.
3.
4.
5.
6.
Member of the Condemnation Board
Certified that the no. sanction in respect of the articles in question was received by the school concerned in the part was any
of these articles includes in list pending sanction.
-------- there has not been any negligence leading to breakage of or damage to the articles on the part of any officer concerned
with the charge of the articles,
Certified that the articles mentioned in the Col 9 have become unserviceable due to normal wear and tear certificate further that
2.
1.
Certificate :
Suggseted Readings
●
Factes of Physics : A conceptual Approach, West Publishing Company.
●
Fundamentals of Physics : David Halliday, Robert Resnick, Jearl Walker, Asian Books Pvt. Ltd.,
New Delhi.
●
University Physics : H.D. Young, M.W. Zemansky and F.W. Sears, Narosa Pub. House.
●
Physics - Foundations and Frontiers : George Gamow and J.M. Clearland, Tata Mcgraw Hill.
●
College Physics : R L Weber, K.V Manning, M.W. White & G.A Weygard, Tata Mcgraw Hill.
●
College Physics : Reymond A. Sarvey and Jerry S. Fanghan Harcourt Brace & Co. Principles
of Physics : Raymond A. Serway & John W. Jewett, Jr.
●
The Elements of Physics : I.S Grant & W R. Phillips.
●
Physics can be fun : Y. Perelman, Mir Publishers.
●
Advanced level Physics M. Nelkon & RParker, Arnold - Heinemann.
●
Success in Physics - Tom Duncan, John Murray Publications Ltd.
●
Success in Electronics - Tom Duncan : John Murray Publications Ltd.
●
Concepts of Physics - H.C. Verma : Bharti Bharti Bhawan Publishers.
●
3000 Solved Problems in Physics : Alvin Halpern.
●
Schaum’s solved problem series : Tata Mcgraw Hill.
●
Mcgrwaw Hill’s Dictionary of Physics.
●
Physics - Resnick and Halliday
Net Resources
The following is just a suggestive list:
●
www.physicsclassroom.com
●
www.learn4good.com/kinds/high_school_science_physics
●
www.hsphys.com
●
faraday.physicssuiowa.edu/resource.html
●
www.library.aucktand.ac.nz/subjects/physics/phymeta.com
●
http://en.wikipedia.org/wiki/Optics
●
http://www.ee.umd.edu/~taylor/optics.htm
174
●
http://www.walter-fendt.de/ph14e/mwave.htm
●
http://www.school-for-champions.com/science/ac.htm
●
http://www.aiiaboutcircuits.com/vol_2/cht_1/1.html
●
ed.org/Education Resources/HighSchooi/Electricity/altematingcurrent.htm
●
http://theory.uwinnJnpeg.ca/physics/bohr/node1.html
●
http://en.wikipedia.org/wiki/Atomic_physics
●
http://en.wikipedia.org/wiki/Wage-particle_duaiity
●
L11/L8/de_Broglie_Wages/de_brog!ie-waves.htm!
●
http://physics.nrnt.edu/~raymond/ciasses/ph.13xbook/node189.html
●
http://www.search.com/refernce/Geiger%2DMarsden_experiment?redir=1
●
http://einstein.stanford.edu/content/faqs/maser.html
●
http://en.widipedia.org/wiki/Electromagneticjadiation
●
http://www.eo.ucar.edu/rainbows
●
http://www.atoptics.co.uk/bows.htm
●
http://astro.nineplanets.org/bigeves.html
●
http://users.ece.gatech.edu/~alan/ECE3080/Lectures/ECE3080-L-1Introduction%20to%20Electronic%20Materials%20Pierret%20Chap%201%20and%202.pdf-Georgia
Tech
●
www.pptsearch365.com/diode-properties-ppt.html
●
ebookfreetoday.com/bulk-semiconductor~0~ppt.html
●
downppt.com/ppt/semiconductor.html
and many more which you may get on the internet
175