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Physics 30 Workbook
Unit4:EMR
UNIT 4:
ELECTROMAGNETIC
WAVES (EMR)
"The speed of Hght it is 186,000 miles
per second. The speed of Ute i's
only 100j000 miles per second."
Einstein discovers that
the speed of light is 65 MPH.
p.67
Unit4:EMR
Physics 30 Workbook
A. TRANSVERSE WAVES
Types of Waves
A wave is a transfer of energy, not matter, from a source to a receiver. It is a travelling
"disturbance" in a medium, where the medium is the material that the wave travels through.
There are two types of waves:
1. Mechanical waves
-waves that MUST travel through matter
-the matter is disturbed (vibrated), but it does not move with the wave
e.g. spring, water, dominoes, sound, earthquakes
2. Electromagnetic Waves
-waves that can travel through a vacuum (no medium is required)
e.g. light, ultraviolet, X-rays
Source Vibrations
The source of all waves is a vibration.
e.g. Sound - vibrating guitar string or vocal chord
Water - dropped rock or ocean floor
The source vibration determines two qualities of a wave:
• it provides the initial energy of a wave
- this energy is then transferred through the medium
• it determines the frequency of the wave (fsource = fwave)
-once the wave is created, the frequency of the wave will never change
(assuming that the source and the receiver are stationary- no Doppler effect)
Frequency and Period
1. Frequency of vibration (f):
-the number of cycles (oscillations, repetitions) in one second
f
= total # of cycles
total time
Units: Hertz (Hz)
2. Period of vibration (T)
-the time it takes for one cycle (repetition)
T=
total time
total # of cycles
Units: Seconds (s)
3. Frequency and Period:
- both are reciprocals of each other
p.68
where 1Hz
1
cycle
1
s
s
Physics 30 Workbook
Unit4:EMR
Anatomy of a Transverse Wave (Ideal)
- in a transverse wave, particles move in a direction perpendicular to the wave motion
Crest
Equilibrium
Source
Vibration
Trough
wave motion
where
A is the amplitude (energy) of a wave
A is the wavelength (length of one cycle)
-particles that are one wavelength apart will move identically (they are in phase)
-due to friction, transverse waves will lose their energy (amplitude), but the frequency will never
change
Speed of a Wave I Universal Wave Equation
If the medium doesn't change, the speed of a wave is constant. Thus, you can change the speed of
a wave only by changing the medium (e.g. stretching the spring, different depths of water, etc.). The
frequency and amplitude of source vibration have no effect on the speed of the wave.
However, the frequency of the source vibration does affect the wavelength, as described by the
universal wave equation :
where
f is the frequency of the source vibration (in Hz)
A is the wavelength of the wave created (in m)
Note: Wavelength and frequency have an inverse relationship
Based on the equation A
Thus,
=
we see that
j'
1
A oc -
f
(assuming v remains constant)
if the source frequency is high, the wavelength is short
if the source frequency is low, the wavelength is long
If it is a light wave, then it travels at the speed of light. In air /vacuum, light travels at
8
v = c = 3.00 x 10 m/s
p.69
Unit4:EMR
Physics 30 Workbook
B. ELECTROMAGNETIC RADIATION (EMR)
B 1. Maxwell's Laws and EMR
Faraday believed that there was a relationship between electricity, magnetism , and light. This was
established by James Maxwell in the 1860's.
Maxwell summarized electromagnetism in 4 laws, described as follows :
1. The distribution of electric charge is related to the net electric field it produces
2. Magnetic field lines are continuous (closed loops), whereas electric field lines begin
and end on electric charges.
3. Electric current can produce a magnetic field (discovered by Oersted).
Also, a changing electric field can produce a magnetic field. ( E---+ B)
4. A changing magnetic field can produce an electric field. (B---+ E)
And thus, it can induce a current in a conductor (based on Faraday's law)
Note:
An electric field can accelerate a charge from rest, and thus induce current,
but a magnetic field can 't. It follows, then, that by changing the magnetic field,
an electric field is created, and this induces a current.
Based on Laws 3 and 4, Maxwell predicted the existence of a brand new kind of wave, which he
called an electromagnetic wave (or electromagnetic radiation, EMR).
• The source of all electromagnetic waves is an accelerating charge
• This creates a changing magnetic field ( B),based on Law #3
• This B creates a changing electric field ( E),based on Law #4
• This E creates a B (Law #3), and the cycle continues indefinitely
Illustrated as:
e-
I
Source
B
E
L1E
B
E
velocity
----------- ---11>-
Electromagnetic Wave
The magnetic field, electric field, and the velocity are always perpendicular to each other.
j
p.70
Physics 30 Workbook
Unit4:EMR
Verified by Heinrich Hertz (1886):
inductJon coil
t
.. ••
'
.
switch
e-l
B
E
B
E
receiver
primary coil
Hertz used a step-up transfonner to induce a high, alternating voltage in the secondary
coils. This high voltage created an accelerating charge across the spark gap, which according
to Maxwell, would create electromagnetic waves. The EMR then travelled across a distance
and induced a current in the receiver wire. The current was detected by a galvanometer.
This verified the existence of electromagnetic waves. In fact, these waves were later
called radio waves.
B2. The Electromagnetic Spectrum
-using his device, Hertz was able to show that E-M waves had the same characteristics as light waves
e.g. speed= c in vacuum, reflection, refraction, diffraction, interference, and polarization
- this was convincing proof that light and radio waves are both electromagnetic waves
- we have now discovered that there are far more types of electromagnetic waves
----+ they differ only in frequency (determined by the source vibration) and wavelength ( A, = c I f)
lo-s
.5x 10,.0
10"8
10-10
Pinpoint
Protozoans
Molecules
Atoms
Wavelength (m)
10·12
About the size oL
Buildings
Humans
Honey Bee
1o1s
p.71
1Ql6
1Q18
Atomic Nuclei
1Q20
Frequency (Hz)
Unit4:EMR
Physics 30 Workbook
HOMEWORK (EMR)
For the questions below, allforms of EMR travel at the speed of light.
Both A and B
1.
Arrange the following forms of EMR in order of increasing wavelength :
ultraviolet
A. 2.
3.
4.
B. 5.
6.
7.
radio
x-rays
visible
infrared
microwaves
A source vibration completes 2.53 x 1010 cycles in 3.00 minutes. What is the wavelength
of the EMR that it produces?
For the EMR wave shown,
determine the frequency of its
source electrical vibration.
d(km)
A DC transformer is set up to create radio waves. Describe the nature of the EMR when
the switch is closed just once. Why?
Consider the carrier waves of two radio stations: 820kHz (AM) and 96.3 MHz (FM).
Compare these waves in terms of wavelength, speed, and time to travel in 45.0 km (in JlS).
determine the resulting wavelength.
For the source vibration shown,
p
C\
Q
8.40 t(JlS)
v v v
If a radio tower's transmitter is oriented vertically , how should the antenna on your
home radio be oriented for the best reception? Why?
SOLUTIONS
1.
x-rays , ultraviolet, visible, infrared, microwaves, radio
2. f= 1.4056 x 108 Hz ; A,= 2.13 m
4.
3.
4
A-=4.5km; f=6.7 x 10 Hz
When the switch is closed, there is a brief increase in current, which creates a brief change in
magnetic field. This will result in a short burst of EMR . But if the switch remains closed, the
current will become constant, which will result in a constant magnetic field (constant flux).
Constant magnetic fields (currents) do not create EMR, so no EMR will be created.
5. f= 820kHz:
f= 96.3 MHz:
v = 3.00 x 108 m/s ; A= 366m ; t4s Ian = 150 JlS
Same speed ; A= 3.12 m (shorter wavelength, since the frequency is higher)
Same time to travel 45 km (150 JlS)
6.
T= 2.8 JlS ;
f= 3.571 x 10 Hz ;
7.
The antenna should also be oriented vertically (same as transmitter) . The changing electric field
is parallel to the accelerating electrons in the antenna, thus it is vertical as well. If it is to generate
current in a receiver , the conductor must also be vertical.
5
A-= 840 m
p.72
Physics 30 Workbook
Unit4:EMR
C. THE SPEED OF LIGHT (EMR)
C 1. Early Methods
Galileo's Lanterns
The first serious attempt to determine the speed oflight was done by Galileo (1564- 1642). He
and his assistant stood on two distant hilltops with shuttered lanterns . Galileo opened his lantern first,
and when his assistant saw the flash of light, he was to immediately open his lantern . Galileo then used
his pulse to determine the time from the initial opening of the shutter to when he saw his assistant's
flash of light. Finally, knowing the distance between the hills, he would calculate the speed of light.
However, it turned out that light travelled so fast that their reaction times were way too slow to
accurately measure the time . (In fact, light would have taken about 70 !J.S to travel the distance, which
would have been impossible even with digital timers!)
Roemer and Huygens
Olaus Roemer (1644- 1710), a Danish astronomer, observed the motions of the moons around
Jupiter. Although other scientists had already determined their periods of revolution , Roemer wanted
to determine the exact moment that one of the moons would experience an eclipse - when Jupiter
would be in front of the moon. He noticed that as Earth got closer to Jupiter, the times of the eclipse
came earlier and earlier than expected; as Earth moved further away from Jupiter, the times came later
and later.
Roemer concluded that as the Earth got further away, light had to travel further to reach the Earth
and thus the eclipse was seen later. This is illustrated in the diagram below :
diameter of Earth's orbit
------,
,,
,
I
I
I
I
I
,.... T ... ,
,
B
I
I
I
I
I
I
sun
\
'
I
'... _._ .... '
I
I
I
I
I
'
\
Jupiter
(and moon)
,
I
''
'
' ''
...
,,
I
I
I
I
I
He was able to calculate that when Earth was furthest from Jupiter (B), it took 22 min (or 1320 s)
longer than when Earth was closest (A). He concluded that this was (approximately) the time light
took to travel across the diameter of Earth's orbit around the sun.
A few years later, the Dutch mathematician and scientist Christiaan Huygens calculated the
11
diameter of Earth 's orbit (3.0 x 10 m), and using Roemer 's data, calculated the speed of light as
follows:
d
3 O x l0 11
v= = ·
m = 2.3 x 108 m/s
11!
1320 s
Interestingly , the speed they determined was so fast that most scientists of their day rejected it. It was
not accepted until after their death.
p.73
Unit4:EMR
Physics 30 Workbook
C2. Michelson
In 1905, the American scientist Albert Michelson used a rotating mirror to make a very accurate
measurement of the speed of light. His device is as shown below:
Light source
Rotating
8-sided
Mirror
Fixed
mirror
telescopeD
very large
distance (D)
<
When the mirror is rotated at the minimum frequency (about 32,000 rpm) to see continuous light:
Time for 1/8 revolution = Time for light to travel to the fixed mirror and back
Equation:
Speed of light ::::;
where
d
2D
t
Ysr
D is the distance between the mirrors (in m)
T is the period of rotation of rotating mirror (in s)
Note:
Speed of light ::::;
d
2D
•
For ann-sided mirror, usae the formula
•
Today, we use lasers to calculate the speed of light through a vacuum. The currently
accepted value is c = 2.997 924 562 ± 0.000 000 011 x 108 m/s.
In Physics 30, we round this value to 3.00 x 108 m/s
Ynr
p.74
Physics 30 Workbook
Unit4:EMR
Homework (Speed of light)
Both A and B
1. In a Michelson experiment, the 8-sided mirror is rotating at 166,000 rpm.
If the distance between the mirrors is 5.8 km, then what is the measured speed oflight?
A. 2.
It takes a total time of83.0 ms (milliseconds) for a microwave signal to travel to a satellite
and reflect back to the Earth, then how far is the satellite from the Earth (transmitter)?
3.
During a lightning storm, the temperature of the air is 28° C (speed of sound is 347.8 m/s).
If lightning strikes 16 km away, then compare the times for sound and light to reach you.
4.
In a Michelson-like experiment, a 6-sided mirror is rotating with a period of355 J.lS. If the
measured speed oflight was 3.2 x 108 rn/s, then fmd the distance between the mirrors (in km).
B. 5. A radar tower sends out a signal at an angle
of 41 .0° above the horizontal. The signal
reflects off of a plane and returns to the
transmitter in a total time of 7.50 J.lS .
Determine the height of the plane.
6.
In a Michelson experiment, the two mirrors are located 4.7 km apart. If the speed oflight was
measured at 2.9 x 108 m/s, then at what frequency is the 8-sided mirror rotating?
7.
In a Michelson-like experiment, ann-sided mirror rotates at a minimum frequency
of 435.5 Hz for the light to be seen in the telescope . If the distance between the two mirrors
is 31.0 km and the measured speed of light is 2.7 x 108 rn/s, then determine the
number of sides n.
8. A light year (ly) is the distance that light travels in one year (365 days). If the width of
our galaxy is 5.4 x 1020 m, then how many light years is this?
SOLUTION
1.
2.
3.
4.
5.
6.
7.
8.
T= 3.6145 X 10-4 s ; v = 2.6 X 108 rn/s
Time one-way: 41.5 ms ; d= 1.25 x 107 m = 1.25 x 104 km
Sound: t = 46 s ; Light: t = 5.3 X 10-5 s
n=6;D=9.5x10 3 m=9.5km
t1-way = 3.75 J.lS ; 1-way distance : 1125 m ;
h = 1125 sin 41.0° = 738 m
4
3
T= 2.5931 X 10- s ; f= 3.9 X 10 Hz
T= 2.296 X 10-3 s ; n = 10 (i.e. 10-sided mirror)
15
lly = 9.4608 x 10 m ; The galaxy is 5.7 x 104 ly wide
p.75
Unit4:EMR
Physics 30 Workbook
Light as a Wave or Particle?
Throughout history , the nature oflight has been debated. Many scientists (such as Newton)
thought light was composed of particles, while other scientists (such as Huygens) were convinced that
light was a wave .
We will now investigate the many properties of light that were well known in the times of
"classical physics", which was before the 20 1 h century. Many of these properties gave convincing
evidence that light behaved like a wave.
D. REFLECTION OF LIGHT (EMR) IN PLANE MIRRORS
Dl. Law of Reflection
(Particles and waves)
normal
Angle of
Incidence
Angle of
Reflection
inciden
ray
:
!Y
/'eflected
ray
mrrror
**Angles are always measured with respect to the normal**
- this law is true for both particles and waves
D2. Images in a Plane Mirror
Consider how we see the image of an object:
Object
hnage
: . ..........................IN· ···········.................... . ............................t···········....··-··· ·:;:;,
• The image is located where the
reflected rays appear to converge
ft
- '""... "
...---------:::- irtual Image
• The eye cannot tell that the rays
have been reflected
- assumes they travel in straight lines
_.·
Characteristics of the hnage
• Same size as object
• Vertically upright (but laterally inverted)
• Virtual (not real)
• The image is located the same distance from the mirror as the object
p.76
-formed by
dotted lines
Physics 30 Workbook
Unit4:EMR
Homework (Plane Mirrors)
A.
1.
If the angle of the incident ray is 3 go with respect to the mirror surface, then what is the angle
of reflection?
2.
For the mirror shown,
predict the angle of
reflection off of mirror B.
97°
B
Both A and B
3.
The objects below are placed in front of plane mirrors. Roughly sketch the image.
a)
b)
Loving
the
Physics
plane mirror
B. 4.
plane mirror
If light bounces off all
three mirrors, find the
angle of reflection off
of mirror C.
B
5.
For the diagram shown,
fmd the angle of incidence
on mirror A.
B
SOLUTIONS
3.
(Plane Mirrors)
a)
b)
Loving
the
Physics
p.77
ii
Q).
mw J
ffi 2 ffFt
Unit4:EMR
Physics 30 Workbook
E. REFRACTION
Refraction
the change in direction of light (or any other wave) as it passes
at an angle from one medium to another
El. Index of Refraction (n)
- if the medium changes, the speed of light changes
-when light travels slowly through a medium, we say that the medium is optically dense
- to determine the optical density of a material, we use the formula
where
c
is the speed oflight in a vacuum (3.00 x 108 m/s)
v is the speed of light in the medium
Note:
• There is an inverse relationship between n and v
-thus, the higher then (the more optically dense), the slower the speed oflight
• For air I vacuum, n = 1
- this is the smallest value for n possible, and thus the highest possible speed of light
E2. Patterns of Refraction
(Waves only- not particles)
1. Less to More Dense (Low n to High n)
normal
Partial Reflection
ei
=
Low n (fast)
8r
Partial Refraction
- the light ray bends
towards the normal
High n (slow)
2. More to Less Dense (High n to Low n)
Partial Reflection
ei = 8r
High n (slow)
Partial Refraction
- the light ray bends
away from the normal
Low n (fast)
p.78
Physics 30 Workbook
Unit4:EMR
Homework (Refraction)
1.
If the index of refraction for a medium is 1.78, then find the time it takes for light
to travel through 250 krn of this medium. Answer in milliseconds.
2.
If light takes 76.3 !J.S to travel through 9.0 km of a medium, then what is the index of refraction
for the medium?
3.
If the index of refraction for a medium is 2.40, then find the distance that light can travel
through it in 7.35 nanoseconds.
4.
Predict the direction of the refracted ray out of the glass.
b)
a)
5.
Predict the direction of the incident ray, if the refracted ray is given.
b)
a)
SOLUTIONS
(Refraction)
1. v = 1.6854 x 108 m/s ; t = 1.48 x 10-3 s = 1.48 ms
2. v= 1.1796 x 108 m/s; n=2.54
3. v= 1.25 x 108 m/s; d=0.919m
4.
a)
b)
R
5.
a)
b)
p.79
Unit4:EMR
Physics 30 Workbook
E3. Snell's Law
or, in general
sin 81 =
=
= .?::!_
sin 82
n 1 v 2 'A 2
Note: Frequency does not change (i.e. f 1 = 6)
E4. Total Internal Reflection (TIR)
-recall, when light travels from a slow to fast medium (high n to low n), there are two rays:
• a reflected ray (where 8i = 8r)
• a refracted ray, which bends away from the normal (8R > 8i)
-but what happens when the angle of incidence is increased?
Critical Angle (8c)
- if we increase the angle of incidence enough, eventually the angle of refraction is 90 o
If 8R = 90°, then 8i is called a critical angle (8i = Elc)
Total Internal Reflection (TIR)
-if the angle of incidence is greater than the critical angle (i.e. 8i > 8c), then there is
no refracted ray (there is only reflection)
R
Critical angle
Total internal reflection
Note:
TIR cannot occur when the ray goes from a fast medium to a slow medium
Why?
-the ray bends towards the normal (8R < 8i)
- so, even if 8i is 89°, 8R will still exist
p.80
Physics 30 Workbook
Unit4:EMR
Homework (Snell's Law I Critical Angle)
A. 1. Light travels into a medium X with an angle of 25° with respect to the surface.
If the angle of refraction is 17°, then find the index of refraction for medium X.
2.
Light travels from water (n = 1.34) into air, and the angle of refraction is 72.0°.
If the wavelength of the light in water is 510 nm, then find:
a) the angle of incidence
b) the wavelength of light in air
c) the frequency of light in water
3.
For the diagram shown,
the ray of light will
reflect off the plane mirror
and then enter into the
medium (n = 1.77).
Plane
muror
What would be the angle
of refraction when it enters
into the medium?
4.
B. 5.
Light travels into an air-glass interface, where the index of refraction for the glass is 1.51.
If the wavelength of light is the glass is 720 nm, fmd:
a) the critical angle
b) the frequency of the light in air
Light with a wavelength of 660 nm travels into glass with an angle of 30.0° with respect
to the surface. If the index of refraction for glass is 1.57, then fmd:
a) the angle of refraction inside the glass
b) the wavelength of light inside the glass
c) the frequency of the light inside the glass
6.
Light in medium A (n = 1.81) has a frequency of2.50 x 1014 Hz. When it enters into
medium B (n = 1.27), the angle of refraction is 56.0°. Find:
a) the wavelength of the light in medium B
b) the angle of incidence
7.
For the diagram shown, the ray oflight
will reflect off the plane mirror
and then enter into the medium (n = 2.06).
If the angle of refraction
is 12.0°, then fmd e.
p.81
mirror
Unit4:EMR
Physics 30 Workbook
8.
If the critical angle for a liquid-air interface is 37°, then fmd the time it takes for
light to travel through 2.1 km of the liquid.
9.
The index of refraction for medium A is 1.92, while the speed oflight in medium B
is 2.50 x 108 m/s. Find the critical angle for the A-B interface.
10. A horizontal ray oflight enters a piece of glass
(n = 1.52) shaped as an equilateral triangle.
What is the angle of refraction when it leaves
the glass on the other side?
SOLUTIONS
(Snell's Law)
1.
B1=65o ; n2=3 .1
2.
a)
b)
C)
3.
45.2°
683 nm
Vj = 2.2388
X
108 rn/s ; ji = 4.39
X
14
10 Hz
The angle of incidence as it enters the medium is B1 = 33o
Then, B2 = 17.9°
;
14
b)
A- 2 = 1087.2 nm ; f2 = 2.76 x 10 Hz
4.
a)
41.5°
5.
a)
b)
c)
33.5°
420 nm
14
14
Method 1: Ji = 4.55 x 10 Hz ; f2 = Ji = 4.55 x 10 Hz
14
Method 2: v2 = 1.91 x 108 rn/s ; f2 = 4.55 x 10 Hz
6.
a) fs = 2.5 X 10 Hz ;
b) 35.6°
7.
The angle of incidence as it enters the medium is B1 = 25.36°
8.
n1 = 1.6616 ;
9.
ns = 1.20 ;
14
VJ
VB= 2.3622 X
= 1.8054 X 108 rn/s ;
108 m/s ; As
t= 1.2 X
=
10-S S
Be= 38.7°
10. The angle of refraction as it enters the glass is 19.205°
The angle of incidence as it leaves the glass is 40.7951 o
The angle of refraction leaving the glass is 83.3°
p.82
9.45 X 10-7 m
Then, B= 17.4°
Physics 30 Workbook
Unit4:EMR
F. CURVED MIRRORS I LENSES
Fl. Types of Curved Mirrors I Lenses
There are two types of curved mirrors I lenses:
Converging Lenses (convex)
1. Converging Mirrors (concave)
<
PA
>
f+
Diverging Lenses (Concave)
2. Diverging Mirrors (Convex)
'
......................... . ........ .....................................................
PA
·-----··
V _,.--_,/p
<
!-
....
C
)
<
Note : Spherical Aberration
if the mirror is spherical, the parallel
incident rays will not converge on
the same focus
the further the rays are away from the
principal axis, the further they go
away from the principal focus
a better focus is obtained using
parabolic mirrors
- similar with lenses
p.83
f-
Unit4:EMR
Physics 30 Workbook
F2. Images in Curved Mirrors I Lenses
Step 1: Sketch the following reflected I refracted rays:
• Ray parallel to the PA reflects I refracts through F (or diverges from F)
-
c
-----
F
:
;<
F
• Ray through centre reflects back I refracts on same path
..............----- ---
--- ---
.........................
··················· ·········•···························· ·· . ... . ...................
C
F
...... ........................................................
F
• Ray through F reflects I refracts parallel to PA
..........
C
----···
.. . .............................. ...............................................
..
F
Step 2: Locate the image
• image is always located where the reflected I refracted rays converge (intersect)
• sketch the image from the PA to the intersection
Step 3: Describe the image
• Size - larger I smaller I same size as the object
• Attitude -inverted (flipped vertically) or upright
• Type- real (formed by real rays) or virtual
p.84
Physics 30 Workbook
Unit4:EMR
SUMMARY FOR CURVED MIRRORS I LENSES
1. Converging Mirrors (for lenses, C = 2F)
Beyond C
OnC
Between C and F
OnF
Between F and V
Image Characteristics
Location of Image
Smaller, Inverted, Real
Same size, Inverted, Real
Larger, Inverted, Real
No image
Larger, Upright , Virtual
Btw F and C
OnC
Beyond C
---------Other side of mirror
2. Diverging Mirrors I Lenses
All locations
In general:
•
•
Other side of mirror
Smaller, Upright, Virtual
Real images are always inverted.
Virtual images are always upright.
<------->·
------------->
do
:
d;
F3. Equations for Curved Mirrors I Lenses
1
1. 1
-+-=do
J
di
h;
R
Also,
=
2f
Sign Convention
d
+
f
+
h , Mag +
Virtual
Real
}
Measured from V I optical centre
Converging
Diverging
Upright
Inverted
}
Measured from PA
Note:
If Mag < 1, then the image is smaller than the object
If Mag= 1, then the image is the same size as the object
If Mag > 1, then the image is larger than the object
p.85
Unit4:EMR
Physics 30 Workbook
Homework (Curved Mirrors I Lenses)
Both A and B
1.
For each, sketch the image, describe its characteristics , and show where the eye is located.
a)
Converging mirror / lens: Object between C and F (between F and 2F)
I
-- -
----- ---------------------- - --- ------------------------------- .
I
Image characteristics:
b)
Diverging mirror / lens
----------------- .l ------------------------------ - --------- -I
Image characteristics:
c)
------
Converging mirror / lens : Object beyond C (2F)
I
--- ------------------------------ ----------------------------- ------
--- ------- --------------- - ---------------- ,---- -------I
I
I
------
. -----------------------------------------
1-------- ----
-------- ---------- --- ----- -- ----- ---- --- ------- -- ,--- - -------I
I
I
Image characteristics :
d)
Converging mirror / lens: Object inside focal length
----- -------- - ----------- - -· - ·--- + ---------- --- - --------- -1
I
--------- - ----------------- ---- "I --------- --- - ------------- ---- ---1
I
-------------- - ------ -- ------- -- T ------------------ ----------- -- - ·· ··
I
---------------------------------- ·t · ---- - ----------- ----- - --
Image characteristics :
I
_
p.86
Physics 30 Workbook
2.
Unit4:EMR
Based on the descriptions below, identify the type of mirror I lens and the
approximate location of the object, if possible.
a)
c)
e)
b)
image is virtual and smaller
d)
image has the same size
image has a magnification of -0.25
image is bigger and inverted
image is virtual and the magnification is 3
f) there is no image
For the rest of the questions, be certain to interpret all signs.
A. 3.
A 19 em high object is located 24 em in front of a concave (converging) mirror.
If the mirror's radius of curvature is 36 em, then fmd:
a) the distance to the image
b) the height ofthe image
4.
An 8.0 em high object is placed in front of a convex (diverging) mirror. If the virtual image
is 3.0 em high and is located 5.0 em from the mirror, then find:
a) the distance to the object
b) the mirror's radius of curvature
5.
A convex (diverging) mirror has a focal length of 17 em. If an object is placed 30 em in front
of the mirror, then what is the magnification of the image?
6.
An 11 em high object is placed 7.0 em in front of a concave (converging) mirror. If the virtual
image has a height of 55 em, then what is the mirror's radius of curvature?
7.
A convex (diverging) mirror has a radius of curvature of 42 em. If a 72 em object
is placed in front ofthe mirror and the resulting image is located 16 em from the mirror,
then what is the height of the image?
B. 8.
9.
An object is placed 21 em in front of a concave (converging) mirror. If the image is real and
is 4.0 times larger than the object, then find:
a) the distance to the image
b) the mirror's focal length
A converging (convex) lens has a radius of curvature of 12 em. If an object is placed 15 em in
front of the lens, then what is the magnification of the image?
10. An object is placed 17 em in front of a concave (diverging) lens. If the image is 20% of
the size of the object, then what is the lens's radius of curvature?
11. A converging (convex) lens has a radius of curvature of 40 em. If the virtual image produced
is 18 em tall and 50 em from the lens, than fmd the height ofthe object.
12. A 30 em high object is located in front of a concave (diverging) lens. Ifthe height ofthe
image is 11 em and it is located 5.0 em from the lens, then what is the lens's focal length?
*13. A very large compound microscope is composed of two converging lenses that are 1.65 m
apart. The focal length of the first lens is 20 em, while the focal length for the second lens is
15 em. If the 10 em object is placed 23 em in front ofthe first lens, then determine the total
magnification. Hint: The image for the first lens becomes the object for the second lens
p.87
Unit4:EMR
Physics 30 Workbook
SOLUTIONS
(Curved Mirrors)
1.
Image characteristics: Larger, real, inverted
Image characteristics: Smaller, virtual, upright
c)
Image characteristics : Smaller, Real, Inverted
p.88
Physics 30 Workbook
d)
Unit4:EMR
: - ;;;:-;:_:_: =-
'
... -=- ....
-- ").(_:_::
----- '
.
.
.
.
Image characteristics: Larger, Virtual, Upright
2.
a)
Diverging mirror I lens ; Object can be anywhere
b)
Converging mirror I lens ; Object is between C and F (between 2F and F)
c)
Converging mirror I lens ; Object is on C (on 2F)
d)
Converging mirror I lens ; Object is between V and F (between 0 and F)
e)
Converging mirror I lens ; Object is beyond C (beyond 2F)
f)
Converging mirror I lens ; Object is on F
3. f= 18 em (real)
4.
d; = -5.0 em (virtual)
5. f= -17 em (virtual) ;
6.
d; = -35 em (virtual)
a)
d; = 72 em (real)
b)
a)
do= 13 em (real)
b) f= -8.0 em ; R = -16 em (virtual)
d; = -10.85 em (virtual)
; f= 8.75 em
, R = 18 em
h; =-57 em (inverted)
Mag=0.36 (upright, smaller)
(real)
7. f=-2lcm (virtual); d;= -16cm (virtual); do = 67 .2cm (real); h;=l7cm (upright)
8.
Mag= -4.0 (inverted)
a)
d; = 84 em (real)
b) f= 17 em (real)
9. f= 6 em (real) ; d; = 10 em (real); Mag= -0.67 (smaller, inverted)
10. d; = -3.4 em (virtual) ; f= -4.25 em (virtual) ; R = -8.5 em (virtual)
11. f= 20 em (real) ; do= 14.2857 em ; h0 = 5.1 em (upright)
12. d; = -5.0 em (virtual) ; do= 13.636 em (real) ; f= -7.9 em (virtual)
13. Lens 1: d; = 153.33 em ; h; = -66.667 em
Lens 2: d; = -52.5 em ; h; = 300 em
Mag : 30x
p.89
Unit4:EMR
Physics 30 Workbook
G. Diffraction , Interference, and Polarization of Light (EMR)
Gl. Diffraction
(Waves only- not particles)
When waves go through an opening or around a comer, they bend . This bending is called
diffraction. The longer the wavelength of the incident waves , the more bending takes place .
short A
Around comers:
long A
---;1less bending
Through openings:
more bending
The smaller the opening, the greater the diffraction
Wide opening:
Narrow opening:
Narrow opening, shorter A
Significant
Diffraction
Nearly
straight
Less
Diffraction
G2. Interference (Waves only- not particles)
Waves can pass right through each other. However , when they occupy the same position
(superposed), they interfere with each other.
Principle of Superposition
- the resultant displacement of a particle is equal to the sum of its separate displacements
There are two types of interference:
1. Constructive Interference (both+, both -)
2. Destructive Interference (one+, one-)
- add to a higher amplitude (brighter)
light wave
- produce a smaller amplitude
- if equal, they cancel each other
u
p.90
(\ +
no wave
Physics 30 Workbook
Unit4:EMR
G3. Double-Slit Interference (Young's Double-Slit Experiment)
(Waves only- not particles)
When a monochromatic wave (i.e. a wave undergoing SHM; a wave of one frequency) passes
through two adjacent, narrow slits, there is diffraction and interference on the opposite side. When a
screen is placed beyond the barrier, we observe bright dots I lines on the screen.
Constructive
2 crests
Interference
2 troughs
I
"
\
' '
Destructive Interference (dim)
crest and trough
Jr---------------------------·
ent
front
----------------------------
---------------------------·
trough
crest
Maxima Minima -
This is a bright dot I line. The difference in the distances from the two slits is a multiple
of the wavelength, so the waves arrive in phase and undergo constructive interference.
This is a dark region. The difference in the distances from the two slits is a multiple
of0.5 'A, so the waves arrive out of phase and undergo destructive interference.
Mathematics ofYoung's Double-Slit
L
-------·-····
T
d
1
Min 1 (n = 0.5)
Max 1 (n = 1)
X
where tan () = -
Equations:
L
A= dx
nL
if () < 10°
p.91
Min 2 (n = 1.5)
Max 2 (n = 2)
Unit4:EMR
Physics 30 Workbook
G4. Dispersion of Light (EMR) using Prisms and Diffraction Grating
When light is shone through a triangular prism or diffraction grating, the different wavelengths of
light bend by different amounts. This causes the light to separate into its different wavelengths, which
is called dispersion.
Dispersion using a Triangular Prism (Waves only)
When light passes through a triangular prism, the
different wavelengths of light refract by different amounts.
The shorter the wavelength,
the greater the angle of refraction.
R
0
y
When the light disperses into its component
wavelengths, all wavelengths are observed with no gaps in
between them. This is called a continuous spectrum.
G
B
I
v
Dispersion using Diffraction Grating (Waves only)
Diffraction grating is a transparent glass or plastic that has many fme wires within it, all parallel to
each other and a very small distance (in j..Lm) apart from each other. This results in many small slits, all
side by side.
When light passes through the diffraction grating, the light behaves much like a double-slit
experiment. Each wavelength of light will have maxima and minima shown on the screen. The location
of the maxima are given by the equations
A.= d sine
n
and
A.= dx
nL
Based on the first equation, it follows that:
The shorter the wavelength, the smaller the angle of deviation
The dispersion pattern (spectrum) for diffraction grating is as shown below:
llight
diffraction
In reality, these
would overlap
ROYGBIV ROYGBIV ROYGBIV
Note:
VIBGYOR VIBGYOR VIBGYOR
If the light is monochromatic (i.e. one wavelength of light), then you would see only that
wavelength on the screen. This would give a pattern like Young's double-slit experiment.
p.92
Physics 30 Workbook
Unit4:EMR
G5. Polarization of Light (EMR)
(Waves only)
When a wave travels through a filter that allows vibrations only in one plane, then the resulting
wave that is produced will vibrate in that plane only. We call this light polarized.
For example, imagine a rope that is moved up and down and then it is moved side-to-side. If this
rope passes through a vertical slit, then only the vertical waves will pass through; the side-to-side
waves would be absorbed or reflected. If these vertically polarized waves were then passed through a
second slit that is horizontal, then the waves cannot pass through at all. They would be completely
absorbed or reflected and the transmission would almost be entirely stopped.
Vertical
slit
Horizontal
slit
Light waves radiate in all directions, and this light is referred to as unpolarized. Much like the
rope, when it goes through a horizontal polarizing filter, it is plane-polarized (specifically, in this case,
it is horizontally polarized), which means it oscillates entirely on one plane. If it then goes through a
vertical polarizing filter ("cross-polarized "), then the light energy is almost completely absorbed.
plane polarized
light
unpolarized light
horizontal
polarizer
near zero intensity
vertical
polarizer
Light can also be partially plane-polarized when it reflects off of a surface or refracts into a new
medium.
Applications of Polarization
•
Polarizing filters are used in cameras and sunglasses to remove some of the light that
reflects off of surfaces or refracts (scatters) through the atmosphere, which has been
partially plane-polarized. This reduces the glare.
•
LCD displays (in digital watches and calculators) have polarizing film within them.
Millions of microscopic crystals float between electrodes. When the electrodes are charged,
the electric field created causes the crystals to line up "cross-polarized" with the film.
This results in dark areas (since the light cannot exit), which is seen as numbers and text.
•
Many insects (such as bees) use polarized sky light for navigation, since it is always
perpendicular to the direction of the sun.
•
Radio waves transmitted from towers are typically polarized (defined by the orientation of the
electric field). AM and FM are vertically polarized, while television is horizontally polarized.
p.93
Unit4:EMR
Physics 30 Workbook
Homework (Young's Double Slit Experiment)
A. 1.
Monochromatic EMR is shone through diffraction grating (distance between the slits is
2.50 !liD). If the angle to the 2nd-order maximum is 11.0°, then determine the EMR frequency.
2.
620 nm EMR is shone through diffraction grating (distance between the slits is 9.9 !liD). If the
distance from the grating 80 em, determine the distance between maxima. Assume B < 10°.
3.
470 nm EMR is shone through diffraction grating that is 36 em from the screen and the
distance between the maxima is 11 em. Determine the distance between the slits.
Both A andB
4.
Which type of EMR (AM or FM) would have better reception on the other side of a hill?
5.
Two microwave transmitters are placed side by side, each emitting EMR in phase with a
frequency of 7.5 GHz. If a receiver is placed 12 em from one transmitter and 22 em from the
other transmitter, would there be a strong signal? Explain.
6.
Unpolarized light is sent through a vertical polarizing filter and it then travels to a second
polarizing filter. Describe the light you would see as the second filter is rotated .
B. 7. Monochromatic, 5.90 x 1014 Hz EMR is shone through diffraction grating which is rated
at 200,000 slits/m. Determine the angle to the third order maximum.
8.
For a Young's double-slit experiment, the diffraction grating has 800 slits per mm. When
monochromatic light with a wavelength of 600 nm is shone through the grating, the distance
between the maxima is 7.00 em. What is the distance (in em) from the grating to the screen?
9.
A Physics 30 student manipulates the distance between the slits and measures the resulting
distance between maxima . For this relationship , determine the equation for the straightened
curve, determine the significance of the slope, and provide the units for the slope.
SOLUTIONS
IL = 2.3851 x 10-7 m ; f= 1.26 x 1015 Hz
n = 1 ; x = 5.0 em
6
3. B = 16.99° ; n = 1 ; d= 1.6 x 10- m = 1.6!-lm
4. In order to have good reception, the waves must be able to diffract significantly around the hill.
Since AM has a longer wavelength, it refracts more. So, AM would have better reception .
5. A= 4.0 em ; The difference in distances between the receivers is 10 em. Since this distance is
a multiple of 0.5 A , the waves will arrive out of phase . This results in destructive interference,
which means there will be no signal.
6. The light that leaves the first filter will be vertically polarized . Thus, when the second filter is
oriented vertically , the light will go through and be bright. However, as it is rotated towards the
horizontal , less light will make it and it will go dimmer. When the second filter is horizontal, the
light is cross-polarized and no light will make it through .
7
6
7. d = 5.0 x 10- m ; A= 5.0847 x 10- m ; B= 17.8°
6
x
8. d=l.25 x 10- m; n=1 ; 0=28.7° ; L = 12.8cm
slope
2
9. Equation:
x = -- ; Slope = IL L ; Units for slope: m
d
1.
2.
1
d
p.94