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PHYSICS 126
Lecture Notes Book 3
Prepared by Kai Wong
Table of Contents
13. Motional EMF
………………………………………………………… 2
14. Electromagnetic Induction
…………………………………………………. 10
………………………………. 18
15. Reflection and Refraction
16 Thin Lenses
…………………………. 27
17. Physical Optics I
18. Physical Optics II
………………… 39
……………………………………………… 45
1
13. Motional EMF
Motional emf
Take a length of metallic rod and move it across a magnetic field as shown:
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
F=evB
×
×
v
×
E
×
×
The free electrons in the rod also move across the field, and will experience a magnetic
force. With the magnetic field pointing into the paper, and the velocity of the rod to the
right, the force on the electrons is downward along the wire. As a result, the electrons
will flow, causing an accumulation of electrons at the lower end and depletion at the
upper end. (The positive ions in the metal also move across the magnetic field and
experience a magnetic force, but they cannot flow along the wire.) The upper end is now
positively charged, and the lower end negatively charged. As a result, an electric field is
created that points downward inside the rod. The electric potential of the upper end
becomes higher than the lower end. This electric field opposes the downward flow of the
electrons. Eventually, when the force due to the electric field becomes equal to the
magnetic force, the electrons in the rod will no longer flow, and we have reached
equilibrium. The magnitude of this electric field, as required by force balance, is equal
to
eE  evB or E  vB
If L is the length of the rod, the potential difference between the two ends is given by
V  EL  vBL
If now the two ends of the rod are joined by a metallic wire, electrons will flow in the
wire from the lower end to the upper end, thereby completing a circuit with current
flowing in the direction indicated:
R
×
I
×
×
×
×
×
×
×
×
×
×
×
×
×
v
×
2
While the positive charge flows in the wire under the influence of the electric field, its
flow along the rod against the electric field is made possible by the magnetic force. The
work done on unit positive charge (+1C) by the magnetic force in the rod as the charge
moves from the lower to the upper ends can be regarded as an emf Є, and is known as the
motional emf. We therefore have
Є=vBL
If the wire has resistance R, from the energy consideration that
Power delivered by the magnetic force = power dissipated in the resistor,
we have
ЄI=I2R,
Or, Є=IR
which allows us to calculate the current.
If the rod is free to glide along frictionless metallic railings connected to a light bulb as
show:
I
F
v
does this mean that it will lit up forever without a source of energy? Obviously this is too
good to be true, and would violate the conservation of energy. To see why this is not true,
notice that because of the upward current in the rod, the rod itself will experience a
magnetic force directed to the left, as shown by the red arrow. The magnitude of this
force is
F  ILB
This force will slow down the rod, eventually shutting down the motional emf, unless an
external force is pulling the rod to the right. To maintain a constant velocity of the rod,
this external force is equal to F in magnitude. The power delivered by this force is
3
Pext  Fv  ILB v  vBLI  Є I
Using a previous relation, w arrive at
Pext  I 2 R
Therefore the external power is responsible for the power generated in the light bulb, and
energy conservation is restored.
Example: If the resistance of the aforementioned light bulb is 5 , the magnetic field is
0.1T, and the length of the rod is 0.3m, how fast should the rod be moving in order that
the power delivered to the light bulb is 2W? How large is the current flowing in the
circuit? How is the external force required to pull at the rod?
Solution: First we find the required emf from
P  I 2R 
I 2R2
 Є2/R
R
so that Є= PR  2  5  3.16V
From the relation Є=vBL, we find
v  Є/ BL  =3.16/ 0.1 0.3 =105m/s
The current is
I  Є/R=3.16/5=0.63A
The external force is
Fext  ILB  0.63  0.1 0.3  0.019 N
As a check, we can find the external force from energy conservation:
Fext v  P
 Fext 
P
2

 0.019 N
v 105
4
Gravity-driven motional emf
If the external force is the weight of the rod in the arrangement as shown:
v
mg
then as the rod falls and picks up speed, the motional emf increases, driving a current that
in turns lead to a magnetic force on the rod that opposes gravity. Eventually, the rod will
reach a steady velocity v , when the magnetic force balances the weight. To find v , we
start with the force balance
ILB  mg ,
mg
BL
This requires an emf given by
and so I 
Є=IR=
mgR
BL
To generate this emf, the speed of the rod must be
v  Є/BL=
mgR
B 2 L2
Example: If the same light bulb, same magnetic field, and the same length of rod as the
previous example are used in the aforementioned circuit in which the mass of the rod is
1g, how fast is the rod falling when its velocity becomes steady? What is the emf
generated and the power delivered to the light bulb?
5
Solution:
mgR 10 3  9.8  5
v 2 2 
 54.4m / s
B L
0.12  0.32
Є=vBL= 54.4  0.1 0.3  1.63V
P=Є2/R=1.632/5=0.53W
In order for the rod to reach the velocity indicated, starting from rest, it would have fallen
a long distance. Neglecting the magnetic force, so that the fall is under gravity alone, the
distance is
v2
x
 151m
2g
AC Generator
The following features are desirable if we rely on the movement of a metallic wire in a
magnetic field to generate emf.
1.
2.
3.
4.
a strong magnetic field
a long length of wire
high speed for the wire movement
a compact volume.
The last item is desirable not least because it is costly to generate strong magnetic field in
a large volume.
There is technological limit to how strong a magnetic field can be created in a large
volume. But the requirements 2,3 and 4 are answered in the design of an AC generator,
which uses rotating coils in a magnetic field.
The sketch below shows a coil placed in an external magnetic field that can rotate about
an axis perpendicular to the field. Leads from the coil connects it to an external circuit
that carries load. As the coil rotates, emf is generated that can deliver current to the load.
6
To calculate the emf, consider the case of a rectangular coil ABCD with width w and
length  . We also label by C’ and D’ the leads out of the coil.
B
A

w
N
C
S
D
D’
C’
When the coil is in the orientation shown below, and is rotating counter-clockwise, the
velocity vectors of the wire BC and DA are upward and downward respectively. If the
coil rotates with the angular velocity  , the magnitude of the velocity is
v  w 2
A
B
A
B
D
C
D’ C’
7
Then, denoting by A  w the area of the coil,
emf generated in BC is ЄBC= vB 
emf generated in DA is ЄDA= vB 
w
1
B  AB into the paper
2
2
w
1
B  AB out of the paper
2
2
emf generated in AB is ЄAB=0 because the positive charge on one half of AB moves
upward and experiences a force into the paper while the charge on other half moves
downward and experience an equal force out of the paper
emf generated in CD is ЄCD=0 for the same reason
Therefore, the net emf is
Є=ЄAB +ЄBC+ЄCD+ЄDA= AB
As shown above in the view from the bottom, this emf is in the direction indicated by the
red arrow, and causes the lead C’ to act as a positive terminal and D’ a negative terminal.
After a quarter period of rotation, the orientation of the coil is as shown below:
A
B
D
C
B
A
D’
C’
The emf is zero because, with velocities of BC and DA in either the same or the opposite
directions as the magnetic field, there is no force on the charge in these segments of the
coil
After another quarter period, the orientation of the coil is as shown:
B
A
B
A
C
D
C’
8
D’
The emf is once again equal to AB in magnitude, but the direction has reversed: the
polarity of the terminals C’ and D’ are reversed compared with half a period earlier.
The general orientation is shown below:
vcosθ
v
A
B
D
C
B
θ
A
D’
C’
In this case, since the force on +1C of charge on the segment BC is (vcosθ)B, the emf of
the loop is Є= AB cos . If time is measured from the position when θ=0, we have
  t . If also the wire is wound so that N coils are stacked up, the emf is
Є= NBA cos t
The emf therefore oscillates with a period of T  2  and a frequency of f   2  .
As a result, the current delivered to the load will oscillate with the same frequency, and is
known as an alternating current. (AC).
Similar to the moving rod, a spinning coil carrying current experiences a torque in a
magnetic field. An external torque is required to counteract this torque to keep the coil
turning. The work done by the external torque comes from the energy supplied by nonelectrical means such as burning fossil fuels or utilizing nuclear power. By heating up
gases or making steam, and allowing the gas to drive a turbine placed in a magnetic field,
electricity is produced. The frequency of AC currents delivered to households is 60Hz,
which is also the frequency of rotation of the turbine.
You can find more information about the generator from the website:
http://www.sunblock99.org.uk/sb99/people/DMackay/ac.html
9
14. Electromagnetic Induction
When a coil moves toward a magnet as shown, it crosses the magnetic field, and a
motional emf is created in the coil that drives a current. Observation shows that the same
current can be created by moving the magnet toward the coil. This means that an emf is
induced in the coil if the magnetic field in its environment is changing in time. This
phenomenon is known as electromagnetic induction. Whether the coil is moving or the
magnet is moving depends on the observer. But the phenomenon of observing a current
in the coil is the same, independently of the observer. Einstein was much impressed by
this fact when he was in high school. This independence of observer is a demonstration of
the relativity principle, and shows that the laws of electromagnetism obeys this
principle.
v
S
v
N
I
S
I
N
Magnetic Flux The statement of the law that governs the emf induced by a changing
magnetic field requires the concept of magnetic flux. Consider a surface, taken to be
planar for simplicity. There are two directions which are both perpendicular to the
surface. Pick one of them as reference and call it the normal direction, and denote it by

the unit vector n .

n
θ

B

Suppose the surface is immersed in a uniform magnetic field B , which in general points
in a direction making an angle θ with the normal. The component of the magnetic field
along the normal, or, simply, the normal component of the magnetic field, is given by
Bn  B cos
Let A denote the area of the surface. The magnetic flux through the surface is then
defined to be
  Bn A
The unit of  is T  m 2 , which is also known as the weber (Wb).
10
The orientation of the surface relative to the magnetic field that yields the maximum
magnetic flux is when the normal is parallel to the magnetic field. In this case,   0 and


the flux is   BA . The flux is zero when n and B are perpendicular to each other. In
this case, the magnetic field is parallel to the planar area, so that the area does not


intercept the field. The flux is negative when the angle between n and B exceeds 90º,
so that the component Bn is negative. A useful way to picture the magnetic flux is to
liken it to the amount of water in a jet that is intercepted by an area.
Suppose a length of wire is wound to produce N planar loops of arbitrary shape stacked
up tightly to all lie on a plane. The loops are immersed in a time-varying magnetic field,
so that the magnetic flux through the area inside the loops changes by the amount  in
a time t . Faraday’s law of electromagnetic induction states that the emf induced in the
coil is given by
Є=  N

t
The negative sign refers to the direction of the emf in the coil. The positive direction is
the one conforming to the second right hand rule together with the direction of the chosen
normal for the flux definition. That is, with the right hand forming a fist with stuck-out
thumb, the fingers indicate the positive sense of rotation when the thumb is aligned with
the normal.

n
=positiv
e sense
of
rotation

n out of paper
Lenz’s law gives another method to determine the direction of the induced emf. It states
that the direction of the emf is such that the magnetic field of the induced current
opposes the magnetic flux change that creates the emf.
We can use Faraday’s law and Lenz’s law to give a unified description of both the
situations of a magnet moving toward a coil and the coil moving toward the magnet.
In both cases, the magnetic flux through the coil is changing, because the closer the
11
separation of the coil and the magnet, the stronger is the magnetic field experienced

by the coil. Referring to the sketches in the above, and choosing the normal n to the
plane of the coil to point away from the magnet, then when the distance between the

magnet and the coil is decreasing, the magnetic flux in the direction of n increases.
According to Lenz’s law, to counteract this trend, the magnetic field of the induced
current in the coil must be pointing toward the magnet. From the second right hand
rule, the induced current therefore runs counterclockwise in the coil when viewed
from the opposite direction to the magnet. According to Faraday’s law, since  t
is positive, the induced emf Є is negative. In combination with the direction of the

vector n , this negative sign means that the emf is in the counterclockwise direction
when viewed in the direction opposite to the magnet, which is the same conclusion as
before. However, Faraday’s law is capable of providing quantitative information.
Iinduce
Binduc
S
N

n
d
ed
v
N
Iinduce
d

n out of paper

When the magnet and coil are moving apart, the flux in the direction of n is reducing, the
induced magnetic field should point away from the magnet to replenish the flux, and the
associated induced current should run clockwise.
On the other hand, if the north and south poles of the magnet are reversed as shown, and
the distance between the coil and magnet is decreasing, the magnetic flux in the direction

of n is negative and decreasing (becoming more negative). From Lenz’ law, the
magnetic field of the induced current should now point away from the magnet. The
induced emf is therefore clockwise.
Binduc
ed
N
S

n
v
Iinduce
S
Iinduce
d

n out of paper
d
Example: A 20-turn coil with circular cross-section of radius 0.5m and resistance 25Ω is
placed in a uniform magnetic field perpendicular to the plane of the coil. The field
12
changes from 0.1T to 0.3T in a time interval of 5s, find the averaged induced emf and
average current in the coil during this interval.
Solution: Choosing the normal to the plane of the coil to be in the same direction as the
magnetic field, the initial and final magnetic fluxes in this direction are
2
2
 i   0.5  0.1  0.079Wb
 f   0.5  0.3  0.236Wb
    f   i  0.157Wb
The magnitude of the induced emf is
Є N

t
 20 
0.157
 0.628V
5
The average current is I  Є/R=0.628/25=0.025A
Example: Referring to the previous example and with the magnetic field held constant at
0.3T , the coil is rotated about a diameter through 90˚ in a time of 2s. Find the averaged
induced emf. Repeat the problem when the angle of rotation is 180˚.
Solution: The sketches below shows the configuration before and after rotation through
90˚ and 180˚.

n

n

n out of paper
Initial position
90˚ rotation
180˚ rotation
With the normal direction for the coil chosen as shown, the initial magnetic flux is
 i   0.5  0.3  0.236Wb
2
After 90˚ rotation, the final flux is  f   0.5  0.3  cos 90   0
Induced emf is

0  0.236
 20 
 2.36V
Є=  N
t
2
The positive sign indicates a counterclockwise direction.
2
After 180˚ rotation, the normal component of B is Bn  0.3T , and the final flux is
13
 f   0.5  (0.3)  0.236Wb
2
Therefore, Є=  N

 0.236  0.236
 20 
 0.472V
t
2
The above example shows that we can obtained the emf of an AC generator using
Faraday’s law.
We can also obtain the motional emf of a moving rod using Faraday’s law. As shown in
the sketch below, as the rod on a pair of railings moves right, the area of the loop
included by the railings, the rod, and the resistor increases. Therefore the magnetic flux
through this loop also increases, even though the magnetic field is constant.
L
v
Δ
x
Choosing the normal in the same direction as the magnetic field, which is downward, the
change in flux after the rod has moved through a distance x , is given by
  BA  BLx
The induced emf is therefore
Є= 

BL x

  BLv
t
t
The negative sign indicates the direction to be counterclockwise. The result is the same as
obtained before from consideration of magnetic forces on moving charges.
Mutual Induction A changing current produces a changing magnetic field. If a wire
loop is placed in the vicinity of the changing current, an emf will be induced on the loop
because of the changing magnetic flux through it. A current will then flow in the coil.
This effect is known as mutual induction. The circuit carrying the first current is called
the primary circuit, and the one on which a current is induced is called the secondary
circuit.
Example: A tiny wire loop of radius 1cm exists concentric with a big loop of radius 20cm
that carries a current in the counterclockwise direction. The current increases from 5A to
14
15A during a time interval of 4ms. Find the magnitude and direction of the average emf
induced in the small loop.
Solution:

n out of paper

Choose the normal n to point out of the paper. The component of the magnetic field
created by the bigger coil at its center is positive, and is equal to
Bn 
0 I
2R
if the radius of the big loop is R and its current I .
Because, the inner loop has a small radius, the magnetic field created by the bigger coil
can be considered to be uniform in its vicinity, and equal to the value above.
The magnetic flux through the loop is
  Bnr 2
where r is the radius of the small loop. The induced emf is therefore
Є=-
7
Bn
 I

15  5
2 4  10
 r 2
 r 2 0
  0.01

 2.5  10 6 V
t
t
2 R t
2  0.2
4  10 3
The negative value means the direction of the emf is clockwise.
Example: Suppose in the previous example, the 15A current in the big loop is reversed in
another 2ms. What is the average emf induced in the small loop during this time interval?
Solution: The normal component of the field Bn changes from
Bn 
0 I
2R
to
Bn  
0 I
2R
15
 I
  I  I
   r 2 Bn  r 2   0  0   r 2 0
R
 2R 2R 
7
 I

 15
2 4  10
 Є 
 r 2 0   0.01
 1.5  10 5 V
3
t
Rt
0.2  2  10
The positive sign indicates that the direction is counter-clockwise.
Thus, in the previous example, when the current in the bigger loop is increasing from 5A
to 15A, the magnetic flux in the direction out of the paper is increasing. To oppose this,
the magnetic field of the induced current in the small loop must be pointing into the
paper. The current in the small loop, and also the induced emf, must therefore be
clockwise to produce such a field. When the current in the big loop changes from 15A
counter-clockwise to 15A clockwise, the magnetic flux in the direction out of the paper
changes from a positive to negative. To counteract this, the magnetic field of the induced
current should be out of the paper. Therefore the induced current and the induced emf are
counter-clockwise.
Self Induction When the current in a single circuit loop is changing in time, the
magnetic field it creates is also changing, and so does the magnetic flux through the loop.
Applying Faraday’s law, an emf will be induced on the loop itself, known sometimes as
the back emf. The phenomenon is called self-induction. The direction of the induced emf
is such as to oppose whatever is responsible for the change in current. Thus, when a
circuit connected to a battery is switched on, the rising current in the direction of the emf
of the battery induces a back emf in the opposite direction that resists the current rise. As
a result, it takes some times before the full current is established in the circuit.
Conversely, when the circuit is switched off, say by opening a switch, the sudden
reduction of magnetic flux induces a back emf that builds high voltage across the opened
gap in the switch, which can sometimes cause a spark.
Transformer
Mutual and self induction take place in circuits carrying alternating currents. The
phenomena can be exploited in a device called the transformer to change the voltage in
the circuits. A transformer consists of a primary coil and a secondary coil wound around
a soft iron core as shown. The purpose of the soft iron core is to confine the magnetic
field due to the current flowing in the circuit, so that the two coils are linked by the same
magnetic flux. If the primary coil is connected to a source of alternating emf, and the
secondary connected to a load, an induced alternating emf will be produced in the
secondary circuit, causing an alternating current to flow. There is a relation between the
emf in the primary and the secondary circuit that can be worked out using Faraday’s law.
16
Let N p , N s be the numbers of turns in the primary and secondary coils respectively.
Applying Faraday’s law to the secondary circuit, the induced emf has magnitude
Єs= N s

t
In the primary circuit, the back emf is
Єp = N p

t
and is equal to the voltage across the primary coil when the resistance of the coil is
negligible (ideal transformer). From the two relations, it follows that
Єs/ Єp = Ns/Np
Thus, we can use the transformer to step up or step down voltages, depending on the turn
ratio.
If the currents in the coils are denoted by I p and I s respectively, equating the power
delivered to the secondary to the power consumed in the primary, we have
ЄsIs= ЄpIp
so that
Is/Ip=Np/Ns
Therefore, when the voltage is stepped up, the current is stepped down. This is exploited
in long distance power transmission because the I2R resistive loss is much reduced.
Voltages are stepped up to hundreds of thousand volts for this purpose.
17
15. Reflection and Refraction
Light is considered to travel in paths called rays in geometric optics. The rays emanate
from points on a light source, which either itself emits light or reflects light that shines on
it. In a homogeneous medium the rays are straight lines. As evidence for this, consider a
finite-sized opaque disk placed between a point source and a screen. The edge of the
shadow of the disk on the screen can be traced out by straight lines connecting the point
source to points on the perimeter of the disk.
An object is seen through the rays coming from it that enter the pupil of the eye. For each
point of the object that is seen by the eye, there is a cone, or pencil, of rays, that originate
from the point and intercepted by the pupil. An image of the point is formed on the retina.
Two types of reflection occur when light rays are reflected from a surface. In diffuse
reflection, which occurs when the surface is rough, rays are reflected in all directions so
that each point of the surface behaves as a point source of light. In specular reflection,
which occurs when the surface is smooth as in a mirror, the reflected ray for each
incident ray is a single ray that makes an equal angle with the normal to the surface as the
incident ray. Calling the angle between the incident ray and the normal the angle of
incidence and that between the reflected ray and the normal the angle of reflection, the
law of specular reflection states that
Angle of Incidence = angle of reflection
Thus , referring to the diagram below,
i  r
θi
18
θr
Image formation with a plane mirror
Consider a pencil of rays coming from a tiny object (essentially a point) striking a plane
mirror. The bundle of reflected rays, when extended backward, will appear to originate
from a single point behind the mirror. This point is called the image. To an observer
receiving the reflected rays, this is identical to a real object placed at this location. It can
be shown that the image is at equal distance from the mirror as the object. (The distance
between a point and a plane is the shortest distance between the point and any point on
the plane.)
do
di
For an extended object, the image is the collection of points that are the images of
individual points from the object. Since the image cannot be located on a screen, it is
called a virtual image.
Note that the image is not a replica of the object. It is nor possible to bring the image
into coincidence with the object by a rotation. In particular, left and right are
interchanged in the image, as can be seen from the diagram below, showing that the
image of a right hand fist with stuck-out thumb is that of a left hand fist. For both
the object and the image, the thumb is out of the paper, but the fingers wrap in the
counter-clockwise and clockwise directions respectively.
B’
B
A
C
A’
C’
D’
D
19
For a point object to be viewable through a mirror, it is not required that the line joining
the object and the image intersect the mirror. All that is required is for a ray from the
object to enter the eye after reflection. This is illustrated by the following sketch:
Ey
e
It can be shown that if you stand in front of a mirror, you can see your image from head
to toe provided that the length of the mirror is at least half your height and is suitably
place.
Multiple reflections from more than one mirror can create multiple images, because
there could be image of an image. With two parallel mirrors, infinitely many images can
be formed. In the following sketch, the red dot on the left is the image of the blue dot in
the left mirror, while the red dot on the right is the image of the first red dot in the right
mirror, and is therefore the image of an image. A set of rays responsible for the formation
of these images are shown.
For two mutually perpendicular mirrors, there are only three images:
20
Refraction
When a ray in one transparent medium impinges on the interface with another transparent
medium, part of the ray is specularly reflected and another part is transmitted into the
second medium as a refracted ray.
Let  1 = angle between incident ray and the normal
 2 = the angles between the refracted ray and the normal
The Snell’s law of refraction states that
n1 sin 1  n2 sin  2
where n1 and n 2 are the refractive indices of the two media. By definition, the refractive
index of vacuum is unit. The refractive index of air is very close to one. The indices for
some representative materials is as listed:
Medium
Refractive Index
Water
Glass
Diamond
1.33
1.5
2.4
θ1
θ1
n1
n1
n2
n2
θ2
θ2
The incident and refracted rays can be reversed in directions. This is called the principle
of reciprocity. Thus, in applying Snell’s law, it is not necessary to distinguish which is
the incident and which is the refracted ray.
Refraction causes a change in direction, or deflection, of the light ray. In going from a
low to a high index of refraction medium, the ray is bent toward the normal. Conversely,
21
light ray is bent away from the normal going from high to low index of refraction
medium.
Example: Find the angle of refraction of a light ray entering water from air at an angle of
incidence of 40º. Find also the angle by which the incoming ray is deflected.
Solution: n1  1 n2  1.33 1  40 
n
1
 sin  2  1 sin  1 
sin 40   0.483
n2
1.33
 2  28.9 
40   28.9   11.1
The angle of deflection is
40
º
28.
9º
11.
1º
Total Internal Reflection Since light ray is deflected away from the normal in going
from a high to a low refractive index medium, there is a value of the angle of incidence
for which the angle of refraction is 90º. This value is called the critical angle. When the
angle of incidence is equal or greater than the critical angle, there is no refracted ray that
enters the second medium. All the light energy is reflected back to the first medium along
the reflected ray. This phenomenon is called total internal reflection. If the ray goes from
medium 1 to medium 2, the critical angle satisfies
n1 sin  c  n2 sin 90 
 sin  c 
n2
n1
Below we calculate the critical angle of total internal reflection for selected cases:
Case 1: glass to air
n1  1.50 n2  1  c  sin 1
1
 sin 1 0.667  41.8 
1.50
Case 2: diamond to air
22
n1  2.42 n2  1  c  sin 1
1
 sin 1 0.413  24.4 
2.42
Case 3: diamond to water
n1  2.42 n2  1.33  c  sin 1
1.33
 sin 1 0.550  33.3 
2.42
Because the critical angle for diamond in air is small, light rays that have entered a wellcut diamond tend to strike a face at an angle greater than the critical, and be total
internally reflected, eventually transmitted back into air in a direction more or less
opposite to the incoming direction. This explains the brilliance of the diamond. The
larger critical angle for diamond in water means that the diamond will appear less
brilliant.
When driving in a hot day, you might notice the asphalt road surface ahead appears wet.
This is explained by the fact that the hot air immediately in contact with the road surface
has a lower refractive index than the cooler air above. Light rays from the sky traveling
through the cool air to the hot air with a large enough angle of incidence can experience
total internal reflection. Seeing such reflected rays is like seeing an image of the sky with
the road surface acting as a mirror.
A 45º glass prism can be used as a mirror because of total internal reflection:
Another application of total internal reflection is in optical fibres.
Image Formation with refraction When an object under water is viewed from the air
above, the image appears closer. This is demonstrated in the sketch below, which depicts
two refracted rays that enter the eye. These rays, when extended backward, intersect at a
point that forms the image.
23
eye
In general, the location of the image depends on the direction from which it is viewed.
The simplest case is when viewed from directly above. To locate the image, we can
choose one ray to originate from the object and strike the surface vertically up. It will be
transmitted with no change of direction because the angle of incidence is 0º. Another ray
is arbitrary and has an angle of incidence θ1 and the corresponding angle of refraction θ2.
We are considering the ray to travel from medium 1 to medium 2. For this ray to enter the
eye, both angles would be small, so that we can make the approximation
sin 1  tan 1  1
sin  2  tan  2   2
d 0  depth of the object
d i  depth of the image
w  distance between the points of intersection of the two rays with the interface
Rewriting Snell’s law
n1 sin 1  n2 sin  2 in the form
n1 tan 1  n2 tan  2
eye
Let
θ
2
w
θ
di
2
I
θ
1
dO
θ
1
O
and using the relations
tan  1 
n1 n2

do di
w
do
tan  2 
w
.
di
we obtain
Or,
d i n2

d o n1
24
In the case of viewing a submerged object from above, n2 n1  1 1.33  0.74 so that the
apparent depth is only 75% of the real depth. Conversely, for a diver viewing an object
above water, n2 n1  1.33 , so that the image is 33% farther away than the real object.
Transmission of Rays through a finite medium A ray impinging on a slab will emerge
on the other side without deflection, although the transmitted ray is shifted from the
incoming ray:
If the ray is incident on a prism, the emerging ray is deflected from the original direction:
Angle of
deflectio
n
The angle of deflection increases with the apex angle of the prism. It also increases with
the refractive index of the material of the prism. Thus, a prism can be used to change the
direction of a ray.
Dispersion White light is a mixture light of different colors. The refractive index of
materials depends on the color. Selected values for glass are listed below:
Color
Refractive index
Red
Green
Blue
1.520
1.526
1.531
25
As a result, blue light experiences slightly more deflection than red. White light
impinging on a prism will spread out into a spectrum of rainbow colors. This
phenomenon is called dispersion.
screen
The formation of the rainbow is explained by the refraction and total internal reflection
that take place inside a rain drop. Refer to the text book for a description.
26
16. Thin Lenses
Image Formation A pin hole on a screen can be used to form an image on another
screen on the opposite side of the object in a manner as shown:
A
B’
B
A’
The drawback of this method is that the image of a point is a small disk, and so tends to
be fuzzy. It can be made sharper by shrinking the hole. But the brightness is then
reduced.
A far superior way to form images is to use lenses. There are two types of lenses: convex
and concave. They are also called converging and diverging lenses respectively
because incident rays are deflected in the direction of increasing lens thickness. This
tendency can be understood by imagining the lens to be made up of stacked-up
truncated prisms, and the ray is deflected by the prism it happens to impinge on. Since
the apex angle of the prisms increases with distance away from the center of the lens,
rays striking farther away from the center experience more deflection.
It can be shown that if the lens is thin, and the rays make small angles with the optic axis,
which is the straight line through the center, (such rays are called paraxial rays), rays
coming from one point of the object will either converge to a single point or appear to
diverge from a single point on the same side of the object after passing through the lens.
Sharp images can therefore be formed, and the intensity is only limited by the size of the
lens, because the larger the size, the more rays from the object point will be intercepted
by the lens. The image is described as real if it is on the opposite side of the object and
virtual if it is on the same side.
27
Parallel Rays and Focal Plane The farther away the object is, the closer to being
parallel are the rays from each individual point of the object as they arrive at the lens.
They are exactly parallel if the object is infinitely far away. In this case, different sets of
parallel rays are distinguished by the directions they come. But each set, or ray direction,
would form a separate point image by the lens. For a convex lens, these images all lie on
a plane behind the lens called the focal plane. In particular, if the rays are parallel to the
optic axis, the image also lie on the optic axis, and is called the focal point. The distance
between the focal plane and the lens is called the focal length. For a concave lens, the
focal plane is on the same side as the far away object.
28
F
F
Graphical Construction of Images The location of the image formed by a thin lens can
be constructed graphically. We first draw the object as an erected arrow on the optic axis
at the appropriate distance, and mark off on the optic axis the focus and also the mirror
image of the focus on the other side of the lens. It is possible to construct three special
rays originating from the tip of the object arrow.
The first ray is parallel to the optic axis. After passing through the lens, it converges to
the focus F if the lens is convex, and diverges from the focus if the lens is concave:
F
F
The reason for the existence of this ray is directly based on the property of the focus.
The second ray is aimed toward the center of the lens. After passing through the lens, it
simply continues on with no deflection. The existence of this ray is because the center
part of the lens acts as a slab with parallel faces and does not change the direction of the
incident ray. The lateral shift of the ray is negligible because the lens is thin.
29
The third ray is based on the principle of reciprocity, according to which the arrows of
the first ray can be reversed. We obtain the third ray if we also exchange left and right.
Thus, the third ray is aimed toward the mirror image of the focus F’, but upon refraction,
it becomes parallel to the optic axis:
We only
ne
F’
F’
We need only two rays to locate the image, which is the point where the two intersect.
For this purpose, the first two are usually employed. The third ray can be used for
checking. The following diagrams demonstrate the constructions for a convex lens and a
concave lend respectively:
Besides the location of the image, when drawn to scale, the graphical construction also
yields information on the nature of the image (virtual/real or upright/inverted), as well as
the magnification, which is defined as
m
hi
ho
where hi and ho are the sizes of the image and object respectively. The sign convention is
adopted so that while ho is always positive, hi is positive if the object is upright and
negative if inverted.
30
The Lens Equation can be used for calculations. For this equation, the following
conventions apply, to the simple case of a single lens.
Focal length
 converging lens
f 
 diverging lens
Object dis tan ce d o   always
 if on opposite side as object real image 
Im age dis tan ce d i  
 if on same side as object virtual image 
The lens equation is
1
1
1


do do
f
The magnification equation is
m
di
do
Example: The focal length of camera lens is +0.05m. Find the location of the image of a
man 1.70m tall, standing 2.50m in front of a camera and determine the nature of the
image.
d o  2.50m
Solution: Given f  0.05m
1
1 1
1
1
 


 20.00  0.40  19.6
di
f d o 0.05 2.5
1
 di 
 0.051m
19.6
Thus the image is 0.051m behind the lens, and is real.
d
0.051
The magnification is m   i  
 0.02
do
2.5
 hi  mho  0.02  1.7  0.034m
Thus, the size of the image is 34mm. It is inverted.
Image formation with convex lens
Let us investigate how the location and nature of the image changes as an object is moved
from far away toward a convex lens.
31
Case 1: The object is at infinity: d o   .
From the lens equation, we find d i  f and also m   d i d o  0 . The image is on the
focal plane, real, and infinitesimal.
Case 2: d 0  2 f The object is farther away then twice the focal length.
From the lens equation, we observe that as d o goes down, d i goes up. That is, as the
object approaches the lens, the image recedes from the lens, starting at the focal plane
behind the lens. The image is inverted and real. The magnification also increases away
from zero. In this range, the magnification is still less than one. This is the range where
camera operates.
Case 3: 2 f  d o  f . The object is between one and two focal lengths from the lens. The
magnification is now greater than one. This range is appropriate for a slide projector.
When d o  f , the image recedes to infinity.
Case 4: f  d o . The object is within a focal length from the lens. This causes the image
distance d e to become negative. The image then appears on the same side as the object,
and is upright and virtual. The magnification is greater one. This is the range for a
magnifying glass.
32
Image Formation with Concave Lens The image is always on the same side and in
front of the object. It is virtual, and smaller than the object.
Vision and Its Correction
The essential optical components of the human eye are the eye lens and the retina, on
which an inverted image of an object is formed and sensed by optic nerves. The focal
length of the lens is variable, being controlled by muscles. To view far away objects, the
muscle is relaxed, and the lens is the weakest, meaning that its focal length assumes the
longest possible value. The far point is defined to the location of an object that has its
image on the retina with relaxed eye muscle. For a normal eye, since objects infinitely far
away are focused on the retina, the far point is at infinity. If the object is at a finite
distance away, its image would be behind the retina and out of focus. To remedy for this,
the eye muscle is strained, reducing the focal length of the lens, and once again causing
the image to fall on the retina. This action is known as accommodation. However, there
is a limit to accommodation, resulting in a smallest object distance within which an
object cannot be focused onto the retina any more. This location is called the near point.
For a normal eye, the near point is 25cm.
For a near-sighted person, either because the lens is too strong even in its relaxed state,
or because the retina is too far behind the lens, objects infinitely far away are focused in a
location between the lens and the retina, and so cannot be seen clearly. As the object
moves closer, causing the image to recede, there will come a point when the image is on
the retina. This location is the far point of the person, and is a finite distance instead of
infinity as for normal vision. To correct for the vision, a lens can be placed in front of the
eye that can weaken the focusing power when used in combination with the eye lens, so
that the image of object an infinity is pushed back to the retina. This means the use of a
concave lens. Since it is the image of the object by this lens that the eye will see, the
33
design for this concave lens should be to produce for an object at infinity an image on the
far point of the person, as the following example shows:
Example: The far point of a myopic person is 10cm. What kind of lens and of what focal
length should be worn?
Solution: The lens should form an image at 10cm in front of the eye for an object at
infinity. Thus, d o   , d i  10cm
1
1
1
1 1


 
 f  d i  10cm
f do di  di
This is a diverging lens with focal length 10cm.
Optician uses the dioptre D to measure the strength of a lens. The definition is
D
1
f
f in metre
Thus, the strength of the lens in the above example is -10 dioptre.
A far-sighted person has reduced ability of accommodation compared with a normal
person. He or she cannot strain the eye muscle enough to focus object at 25cm away, the
near point of normal vision. Instead, the near point is more than 25cm away. The lens to
correct this vision should strengthen the focusing power, and so is a convex lens. Its
design should be to create for an object at 25cm an image at the near point of person so
he or she can see clearly.
Example: Find the focal length of a lens suitable for a far-sighted person whose near
point is 40cm.
Solution: The design requirement is d o  0.25m d i  0.4m
1
1
1
1
f 
 0.67m


 4.00  2.50  1.50
1.50
f 0.25 0.4
Simple Optical Instruments Optical instruments are often designed to allow objects to
be seen better, which most of the time means larger in appearance. The perceived size of
an object is the angle it subtends at the eye. For a distant object, the angle in radian is
given by

h
d
h
d
θ
eye
where h is the actual size of the object and d its distance away. For the sun, this angle is
about 0.5º.
Example: Find the size of the image of the sun formed by a convex lens of focal length
5cm.
34
Solution: Since the sun is very far away, its image is on the focal plane. The figure shows
two sets of parallel rays from the two diametrically opposite points A and B on the sun
respectively, forming images A, and B’ on the focal plane. For calculation purpose,
another figure shows only the rays passing through the center of the lens. They include
the angle θ=0.5º=0.5×2π/360 rad = 0.0087 rad between them, which is the angle
subtended by the sun. From the triangle between the lens and the focal plane, we
conclude that the distance between A, and B’ is 0.0087×5cm=0.04cm.
B’
A
B’
θ
A’
B
A’
f
Magnifying Glass An obvious way to increase the apparent size of an object is to bring it
close to the eye. The closest distance while maintaining clear vision is the near point
distance N(=25cm for normal vision). Then the angle subtended at the eye is
h
 o
N
The object can be made to appear larger when it placed within a focal length of a convex
lens and viewed from the other side of the lens. If the eye is right up against the lens, as
shown in the diagram below, the angle subtended by the image at the eye is equal to the
angle subtended by the object, and is given by
 
ho
do
ho
θ'
do
ey
e
di
The magnifying power is defined by
M
  ho d o N N N


 
 ho N d o
f di
where the last step follows from the lens equation. Since the eye can view the image as
long as it is further away than the near point, we have the choice
35
N
if d i  
f
N
M   1 if d i   N
f
M 
Thus, the image is magnified so long as the focal length is less than the near point
distance.
Microscope The magnifying power can be increased if the object is pre-magnified to
form a real image behind a convex lens in an arrangement similar to that of slide
projector. This real image is then viewed by another lens acting as a simple magnifying
glass. The first lens is called the objective, and the second the eyepiece. The rays used to
construct the image in this two lens system is as shown. The image of the objective is
shown in green, and the final image is in red.
The image of the objective does not have to show up on a screen. The rays that actually
pass through the two lenses are shown in the following:
Telescopes are designed to enlarge the images of far away objects. A simple telescope
can also be constructed using an objective and an eyepiece that are both convex lenses.
However, for objects far way, the idea of pre-magnification does not work. In this case,
the objective always forms a real image at the focal plane, as the previous example on the
image of the sun on the focal plane of a convex lens shows. The eyepiece is again used
to examine this image. It is also arranged that this image falls on the front focal plane of
the eyepiece, so that the final image is at infinity. The telescope does not cause the
distance of the image closer than the object. It only increases the angle subtended at the
eye. The following shows the rays used to construct the images. The blue rays show the
formation by the objective of a point image from an infinitely far away point object from
a certain direction. The red rays show the formation of an image of this image by the
eyepiece. The last image is virtual and at infinite distance from the eyepiece.
36
The magnifying power of the telescope can be derived from the following diagram,
showing only the rays that pass through the centers of the lenses. In addition, the optic
axis can be considered a ray from another point on the object at which the telescope is
directly aimed.
θ
A
B
θ
θ'
D
C
fo
fe
We see that whereas the object subtends the angle  at the objective, which is essentially
the angle subtended at the unaided eye, the image subtends the angle   at the eye. Let
f o and f e be the focal lengths of the objective and the eyepiece respectively. From
triangle ABC,

h
fo
where h is the length BC. From triangle DBC, we find
 
h
fe
Therefore the magnifying power is
M
  h fe fo


 h fo fe
and can be much greater than one is the objective has much longer focal length than the
eyepiece. That is why astronomical telescopes are long. They are also big because the
brightness of the image depends on the amount of light received by the objective. Note
also that the image is inverted.
37
You can find information and images on some large telescopes from the following
website
http://www.munisingwebsites.com/lookum/history.html
38
17. Physical Optics
Simple Harmonic Motion and Phase
We learn simple harmonic motion in the first part of this course. An example comes from
the spring mass system. Recall that the displacement of the mass from its equilibrium
position at any time can be found from the projection onto the x-axis of a fictitious
particle travelling in a circle with uniform angular velocity and centered at the origin. If
the angular velocity is  rad/s and the radius of the circle is A , the position of the object
at any time t is given by
x  A cost   
where  is the angular position of the fictitious particle at t  0 .
ωt
A
α
x
The period and frequency of this motion are given by
T
2

and f 
1 

T 2
The angle t   is called the phase of the motion at the time t , so that  can be called
the initial phase. The quantity A , representing the maximum displacement from
equilibrium, is called the amplitude.
Wave phenomena, such as a vibrating string, sound waves in air, or ripples on the surface
of a swimming pool, involve oscillations of particles of the medium. In a monochromatic
wave, the oscillations are simple harmonic motions with the same frequency and
amplitude. This common frequency is called the wave frequency f . However, the phases
of these oscillations at different locations are different. The distance between two
locations where the phase difference is 2 is called the wavelength  Particles of the
medium at these locations oscillate together, and are said to be in phase. The same
terminology applies to phase differences of  2 ,4 , . . The pattern of vibration gives
the impression of a sinusoidal structure travelling in one direction. The speed of this
apparent movement is called the wave speed v , and is a property of the medium. The
39
same medium can accommodate monochromatic waves of different frequencies, The
fundamental relation
f  v
exists between the wavelength, the frequency, and the wave speed.
Light waves The velocity of light in vacuum is
c  3  10 8 m / s
The wavelengths of light of different colors are tabulated:
Color
Wavelength (nm)
Red
660
Yellow
580
Green
550
Blue
470
Violet
410
For a three-dimensional monochromatic wave, the locations where the oscillations have
the same phase is said to form a wave front. If the wave originates from a point source,
such as the ripples generated in a swimming pool when one location is continuously
agitated, the wave fronts are spherical surfaces centered at the source. For visualization
purposes, we draw the wave fronts that correspond to the wave crests (maximum
displacement one way, say upward) and the wave troughs (maximum displacement the
other way, say downward) at one moment in time.
λ
The distance between successive wave fronts for crests = the wavelength 
The distance between the front of a crest and the nearest front of a trough =  2
40
At far distances from a point source, the curvature of the wave fronts becomes small, so
that the fronts are approximately planes. Waves with plane wave fronts are called plane
waves
We also define rays to be the family of curves that are normal (perpendicular) to the
wave fronts. We see that for spherical and plane waves, the rays are straight lines. They
are shown in red in the figures. There is a phase difference between two points on a
ray. If the ray is a straight line, and the distance between the two points is L , the phase
difference is given by
  2
L

because for each length increment of a wave length, the phase changes by 2 .
Propagation of wave fronts The pattern of wave fronts moves as the time changes. For
spherical waves, new fronts are continually born from the source at the center, and spread
outward. After a time interval t , the radius of each front grows by the amount vt . For
a plane wave, each front has traveled a distance of the same amount. Thus, after half a
wave period, the locations of the crests are taken up by the neighboring troughs, and vice
versa, and after one full period, a crest front becomes occupied by the crests that were
one wavelength behind.
Huygens Principle provides a method to follow the propagation of a wave front,
obtaining the new front after a given time interval from the old front. In this construction,
each point on the existing front is considered to be a source of spherical wave. The waves
from these points are called wavelets, and are generated at the same phase as the old
front. After a time interval t , these wavelets all have radius vt . The envelope of these
wavelets, which is the surface tangent to all of them, is then the new front. The figures
below show how this works for spherical waves and plane waves.
41
The wave theory of light can explain phenomena of geometrical optics such as the laws
of reflection and refraction.
Specular Reflection Consider a plane wave incident on a plane reflecting surface. The
rays of the plane wave are parallel straight lines making the same angle  i with the
normal to the reflecting surface. As the incident wave travels forward, each front of the
wave will sweep through the surface, with different parts of the front contacting the
surface at different times. At the time of contact, the point on the front acts as a source of
wavelets according to Huygens principle. These wavelets spread back into the medium
from which the plane wave arrives. The figures below show the wave front and rays at
two times at interval t apart. These times correspond respectively to the arrivals of the
points A and B on an incident front at the reflecting surface. On the latter figure are also
shown the location of the new front A’B’ (in red) as well as the old front A’’B’ had there
not been a reflecting surface. The largest semicircle represents the wavelet generated at
the point A. The incident and reflected rays are also shown.
A’
B
B’
A
A’’
The law of specular reflection can be demonstrated from the diagram below, obtained by
merging the two previous figures together. The proof is as follows:
First we observe that
AA  BB  vt ,
42
because by the time when B reaches B’, the wavelet from A has reached the point A’, and
the waves travel at the same speed v in the same medium.
The triangle AA’B’ and B’BA are therefore congruent.
 ABA  B AB
Since  i  ABA
 r  BAB
It follows that  i   r , the law of reflection.
θi
θr
A
A’
B
θr
θi
B’
It turns out that there is a 180º change of phase between the reflected and the incident
wave. The wave front A’B’ of the reflected wave is actually 180º out of phase with the
wave front AB. But this change of phase does not affect the proof.
Refraction Consider a plane wave in a transparent medium 1 incident on another
transparent medium 2. In general the wave speeds are different, and are denoted by v1
and v2 .The figure below shows an incident wave front and the transmitted wave front
after time interval t Also shown are the incident and refracted rays together with the
angles of incidence and refraction are  1 and  2 .
B
θ1
A
θ1
B’
θ2
A’
In this case,
BB  v1t
θ2
AA  v2 t
Also, 1  ABA
 2  BAB
From triangle ABB’, sin  1 
BB v1 t

AB AB
43
From triangle AA’B’, sin  2 

AA v2 t

AB AB
sin  1 v1

sin  2 v2
If we define the refractive index of a medium by
n
c
v
where c is velocity of light in vacuum and v velocity of light in the medium, it follows
that n  1 for vacuum, and

sin  1 n2
which is Snell’s law.

sin  2 n1
Thus, the wave theory of light not only explains Snell’s law, but also provides an
interpretation of the refractive index.
The frequencies of the incident and refracted waves are expected to be the same since the
particles in the interface must vibrate with a definite frequency. Using the fundamental
relation f  v , we have
In vacuum:
In medium:
f  c
medium f  v
v

  medium   
c
n
Thus the wavelength is reduced in a transparent medium.
44
18. Physical Optics II
Principle of Superposition
This states that when two waves meet, the resulting amplitude is the sum of the
amplitudes of the two, provided the amplitudes are small. The two waves are also said to
interfere with each other. If at a certain location in space, the amplitudes of the two waves
with the same frequency and maximum amplitude are
x1  a cost  1 
x2  a cost  2 
then the resulting amplitude is
x  x1  x2  2a cos
1   2
2
  2 

cos t  1

2 

where we have used the trig identity
cos A  cos B  2 cos
A B
A B
cos
2
2
We see that the resulting oscillation is a sinusoidal oscillation with the same frequency.
But the maximum amplitude depends on the phase difference between the two waves.
The intensity of a wave, which corresponds to its energy, is determined by the square of
the amplitude. The square of the maximum amplitude is
2
xmax
 4a 2 cos 2
1   2
2
Two cases are of special interest.
Case 1: The waves are in phase. 1  2  0,2 ,4 ,
2
 4a 2 is as large as possible. This is known as constructive interference.
Then xmax
Case 2: The waves are 180º out-of-phase. 1  2   ,3 ,5 ,
2
 0 . The amplitude remains zero at all time. This is known as destructive
Then xmax
interference.
Interference of waves from two point sources
Consider two point sources S1, S2 of waves with the same wavelength  and maximum
amplitude. The two are said to be coherent if the phase difference between them does not
change in time. Spherical waves are generated from the two sources. The interference of
these waves causes some locations to have large amplitudes and others to have small
45
amplitudes and even zero amplitude. At a point P in space whose distances from the
sources are r1 and r2 respectively, we have
r
Phase of wave 1 at P – phase of wave 1 at S1 = 2 1

Phase of wave 2 at P – phase of wave 2 at S2 = 2
r2

If the two sources are in phase, we have
Phase of wave 2 at P – Phase of wave 1 at P =
2

r2  r1 
As a result, we can rewrite the phase condition for constructive and destructive
interference in terms of the path difference r2  r1 as follows:
m constructive int erference

r2  r1  
1
 m  2  destructive int erference


where m  0,1,2,
Young’s Double Slit Experiment
In optics, two coherent sources can be created by allowing a monochromatic plane wave
of light to impinge on a screen with a pair of narrow slits. The monochromatic wave can
come from a laser, or, in Young’s experiment, from an illuminated slit placed at the focus
of a convex lens. The pair of slits acts as in-phase sources of cylindrical waves. If a
screen is placed at some distance behind the slits, a pattern of alternate bright and dark
strips called fringes will be formed, as a result of interference.
r1
S1
P
O
r2
S2
46
The distance of the screen is much larger than the distance between the slits, and the
point P is usually rather close to the point O, which is equidistant from the slits. The rays
S1P and S2P are nearly parallel to each other in the vicinity of the slits, as shown as
below:
S1
θ
d
From this we see that the path difference can be written as
r2  r1  d sin 
where d is the distance between the slits, and  is the angle that defines the direction of
the rays. The conditions for constructive and destructive interference at P can be
rewritten as
m constructive int erference

d sin   
1
 m  2  destructive int erference


Since in most cases, the wavelength is much smaller than the slit distance, and the order
m for the fringes of interests are not large, the angle  would be small. Then we can
employ the small angle approximation
sin   tan  
provided  is measured in radian. For example, the condition for a bright fringe of order
m can be written
 
m
d
The point O on the screen corresponds to the angle   0 , and is therefore the location of
a bright fringe, called the central bright fringe. The distance PO , denoted by y , can be
related to the angle  by examining the following diagram:
47
P
y
θ
θ
O
L
The result is
y  L tan 
where L is the distance between the slits and the screen. In the small angle
approximation, this becomes
y  L
Upon elimination of the angle  , the location of the bright fringes are given by
ym

d
L
The spacing between successive bright (and also dark) fringes is
y 

d
L
The angular separation between successive bright fringes is  

d
For example, if we use red light of wavelength
  664nm  664  109 m  6.64  107 m  0.664m
and the distance between the slits is
d  1.20  104 m  120m
the angular separation between successive bright fringes is
6.64  107
 
 5.53  103 rad
4
1.20  10
If the screen-slit distance is L  2.50m , the spacing of the fringes on the screen is
48
y  L  5.53  103  2.50  0.013m  1.3cm
The double slit experiment confirms the wave nature of light.
Diffraction Grating
The spacing between the bright fringes can be increased by reducing the spacing between
the slits. This is achieved in a diffraction grating, which is obtained by etching many
parallel lines on a glass plate, creating a multi-slit pattern. The distance between these
lines can be of the order of microns. The number of lines per unit length is called the
grating constant. For example, if the grating constant is 1.0  106 lines / m , the distance
between successive slits is
d
1
1

 1.0  10 6 m  1.0m
6
grating cons tan t 1.0  10
When a diffraction grating is used, the image brightness at a point P on the screen is
determined by the interference between a very large number of rays, one from each slit.
For most locations, the waves will cancel out, resulting in darkness. On the selected
locations determined by the condition
d sin   m
m  0,1,2,
y  L tan 
and
the path difference between a pair of rays from any two slits is an integral multiple of the
wavelength. Therefore, all rays will enhance each other, resulting in a bright line on the
screen. These bright lines are far apart.
For example, using the grating with d  1m , the first order lines for red light of
wavelength 0.66 m would occur at the angle
sin 1

d
 sin 1
0.66
 41.3
1.0
and there is no second order line because with m  2 .
m

d
 1.32
so that the equation d sin   m has no solution for  .
49
If violet light (   410nm  0.41m is used, the angle for the first order line would be

sin 1 (0.41)  24.2 . In addition, there will now be a second order line at
sin 1 0.82  55.1 .
The location of the first order violet line on a screen 2.5m away would be
y  L tan   2.5 tan 24.2  1.12m .
Normally, we do not form images of the lines on a screen, but instead view them from the
appropriate directions. If light with mixed colors is used, the grating can resolve it into
the component colors and spread them out as lines in various directions, starting with the
violet end of the spectrum as the view angle increases away from zero. These lines are
sharply defined if the mix of colors have well-defined wavelengths. In this capacity, the
diffraction can be used to determine the wavelengths in the spectrum of light emitted
from an unknown source. It is then known as a grating spectrometer.
The regular arrangement of atoms in a crystal acts like a grating as the atoms scatter
electromagnetic waves incident on the crystal. Diffraction patterns are formed as a result,
with enhanced intensity observed in selected direction. The spacing between the atoms in
a crystal is of the order of the size of the atoms, typically 1.0  1010 m . The wavelength of
the electromagnetic radiation that exhibits the effect of interference must be comparable.
Such short wavelength puts the radiation in the X-ray region. If the wavelength of the Xray is known, we can infer the structure of the crystal by studying the X-ray diffraction
pattern. This branch of science is called crystallography.
Thin Film Interference
Interference also occurs when light is reflected from a thin film of a transparent material,
sandwiched between two extended transparent media. Some examples of the film and the
media are:
Soap film in air, the media are air- soap solution-air
Gasoline film on water, the media are air-gasoline-water
Air gap between glass plates, the media are glass-air-glass.
The interference is between light reflected from the first and the second interfaces. The
latter has been transmitted into the film before reflection. The two rays thst interfere are
shown in blue and red:
50
The analysis is particularly simple if the film is viewed directly from above. In this case,
Path difference between the red and blue rays = 2t
where t is the film thickness.
The phase difference depends on whether or not there is 180º phase change for two
reflections. Such a phase change occurs if the reflection is from a medium of low to one
of high refractive index. As a result, we have
Phase difference between red and blue rays = 2
2t
f

if there is only one 180º reflection phase change. Otherwise, the phase difference is just
2
2t
f
.
In these expressions,  f is the wavelength in the film, and is related to the wavelength in
vacuum by
f 

nf
where n f is the refractive index of the film.
Using the phase condition for interference, we obtain the conditions
m f constructive

2t 
 
1
2  m   f destructive
2

m  0,1,2,
f
in case a single 180º reflection takes place. Otherwise, the term  f 2 is missing.
Example: Determine the minimum thickness of a film of gasoline (refractive index=1.40)
floating in water (refractive index=1.33) from which no reflection is observed for an
incident red light (wavelength 660nm).
Solution: The reflected ray from the gasoline film back to air experiences 180º phase
change but not the ray reflected back into the film from water. The condition for
destructive interference that corresponds to minimum thickness is given by
51
2t 
f
t 
2
f
2

3
f
2


2n f

660
 236nm
2  1.4
If white light shines on the film in the above example, the red color component is
removed from the reflected light, which then will appear with the complementary color of
red. If the thickness of the film varies from place to place, different colors are removed
from the light reflected from various parts of the film. As a result, the film appears multicolored.
When light is reflected from the air wedge between two glass plates making a small angle
with each other, the thickness of the film of air varies from zero to a maximum value.
The reflected light then shows a large number of dark fringes, at locations where the
thickness of the film satisfies the condition for destructive interference of the reflected
light.
If a plano-convex lens is placed on a flat glass plate, with its apex touching the plate, the
air trapped in between again has varying thickness. In this case, locations with equal
thickness form rings around the point of contact. The interference pattern of reflected
monochromatic light consists of a series of dark rings known as Newton’s rings. The
center appears dark because of 180º reflection phase change for the ray that undergoes
reflection from air to glass.
52
Diffraction
Because of its wave nature, light will spread out after passing through a hole on a screen.
On encountering an obstacle, it will also bend so that it infiltrates the shadow that was
predicted to be a sharply-defined region had it traveled only in straight lines. These
phenomena are known as diffraction.
The simplest diffraction phenomenon occurs when a monochromatic light of wavelength
 is incident perpendicularly on a screen with a slit of width w . The plane wave front of
the light arriving at the screen is blocked except for the region where the slit is located.
To predict the intensity of light at any location behind the screen, we can use Huyghen’s
principle, and consider each point in the exposed wave front at the slit to emit spherical
wavelets, which interfere among themselves to yield the resultant oscillation at the
location being considered. If these locations are all on another screen, we should see a
pattern of maximum and minimum intensity. Such an arrangement is known as Fresnel
diffraction.
P
Alternatively, we can place a convex lens far behind the slit screen, focusing the
transmitted light onto another screen placed at the focal plane of the lens, producing an
intensity pattern there. This arrangement is called Fraunhofer diffraction.
P
In Frauenhofer diffraction, the intensity at any point P on the screen depends on the angle
of the parallel rays coming out of the slit that are eventually focused onto the point by the
lens. Let  be the angle between such rays and the normal to the slit screen. At   0 ,
the path difference between any pair of rays out of the slit is zero, and so the rays will
interfere constructively, yielding maximum intensity. As  increases, the rays are not all
in phase, resulting in reduction in intensity. As seen from the diagram, the quantity
53
w sin  is the path difference between the ray from the top and from the bottom of the
slit. When the condition
w sin   
is satisfied, if we divide the slit into the upper and the lower halves, then for each ray
from the upper half, there is a corresponding ray from the lower half that has a path
difference of  2 with the former. Example of such a pair is colored blue in the diagram.
The two rays therefore destructively interfere and cancel each other out. This cancellation
occurring for each pair means complete cancellation for rays from the whole slit. We then
expect dark fringes at either edge of the central maximum.

w
wsinθ
In most cases for light waves, w   , this angle that gives the direction of the first dark
fringe can be approximated by
1 

w
As  increases beyond 1 , the intensity rises again, reaching a smaller maximum than the
central one before dropping down to zero again at 21 . In fact, zero intensities are
obtained when   n for any integer n . The intensity patterns as a function of  and
on the screen are as shown:
54
Similar analysis holds when the slit is replaced by a circular hole. In this case, the pattern
on the screen consists of concentric dark rings with a central bright circular region. The
first dark ring is formed from a cone of rays coming out of the hole with half angle
 1  1.22

D
where D is the diameter of the hole.
Previously, when we analyze the double slit experiment, the width of the slits is
considered to be zero. When we include the slit width w in the analysis, the resulting
angular distribution of intensity is obtained by superposing the single-slit and the double
slit patterns. The single slit pattern from both slits are the same. It is shown in the
following figure as a red curve. The double slit pattern is shown as the blue area, which is
seen to have its intensity modified by that of the single slit. In this figure, b is the slit
width and a is the slit separation. If b is much less than a , we should see many dark and
bright fringes within the central diffraction peak.
55
Diffraction also occurs when monochromatic light is incident on a blockage in the form
of a circular disk, resulting in some intensity even in the shadow on a screen predicted by
geometrical optics. A curious fact is that a tiny bright spot exists at the center of the
shadow. This results from constructive interference from the unblocked wave front,
which can be divided into concentric rings. Being equidistant from the center of the
shadow, points from each ring emit rays which interfere constructively.
Resolving Power
In geometrical optics, a light ray can be traced through an optical instrument according to
the rule of reflection and refraction. Because of diffraction, in going through an opening
of finite size, a light ray does not maintain its well defined direction. If the opening is a
circle of diameter D , the incoming light ray spreads out into a cone of half angle as
given earlier around the incoming direction. If another ray comes through the hole from a
direction too close to the first ray, the resulting spreads would blur the distinction
between the transmitted rays. The images from the two incoming rays then overlap each
other. The figure below shows the overlapping images of two stars close together in the
sky. An estimate of the smallest angular separation between the incoming directions that
can be resolved can be taken to be
 min  1.22

D
so that the central maximum of the diffraction pattern from one ray coincides with the
dark fringe from the other. This is known as Rayleigh’s criterion, and the angle  min is
taken to be a measure of the resolving power. Rayleigh’s criterion poses a fundamental
limit on resolving power, that can be overcome only with increasing aperture (size of
opening). Thus a good astronomical telescope has a large objective lens to increase both
the resolving power and the brightness of images. Because the wavelengths of radio
waves are large, radio telescopes often rely on arrays of antenna receivers spaced far
apart.
56