INTERFERENCE
1
INTERFERENCE
Topics
 Two source interference
 Double-slit interference
 Coherence
 Intensity in double slit interference
 Interference from thin film
 Michelson’s Interferometer
Text Book:
PHYSICS VOL 2 by Halliday, Resnick and Krane (5th Edition)
MIT-MANIPAL
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2
What is an Electromagnetic wave (EM)?
Electric field (E)
900
Magnetic field (B)
The electromagnetic waves consist of the electric and
magnetic field oscillations.
In the electromagnetic waves, electric field is perpendicular to
the magnetic field and both are perpendicular to the
direction of propagation of the waves.
Properties of electromagnetic waves (EM)
Electromagnetic waves are non-mechanical waves i.e they do not
require material medium for propagation.
They are transverse waves. ie. They travel in the form of ‘crests’
and ‘troughs’.
Examples:
• Light waves
• Heat waves
• Radio and television Waves
• Ultraviolet waves
• Gamma rays, X- rays
They differ from each other in
wavelength (λ) and frequency (f).
In vacuum, all electromagnetic
waves (EM) move at the same
speed and differ from one another
in their frequency (f).
Speed=c=Frequency x wavelength
i.e c= f x λ
c= 3 x 108 m/s
Electromagnetic spectrum
Name
Frequency range (Hz)
Wavelength range
Gamma rays(γ-rays)
5 x 1020- 3 x 1019
0.00006 -0.1nm
X-rays
3 x 1019- 1 x 1016
0.1 nm-30nm
Ultraviolet light
1 x 1016- 8 x 1014
30nm-400nm
Visible light
8 x 1014- 4 x 1014
400nm-800nm
Infra-red
4 x 1014- 1 x 1013
800nm-30000nm
Radio frequencies
3 x 107- 3 x 104
1010-1013 nm
More frequency (f) more energy (E), and lesser wavelength(λ).
Dual Nature of Light
Albert Einstein proposed that light not only behaves as a
wave, but as a particle too.
Light is a particle in addition to a wave-Dual nature of light.
Dual nature of light treated as
1) a wave or
2) as a particle
Light as a stream of particles
Dual nature of light successfully explains all the phenomena
connected with light.
When light behaves as a Wave?
The wave nature of light
dominates when light
interacts with light.
f
The wave nature of light explains the following properties of
light:
• Refraction of light
• Reflection of light
• Interference of light
• Diffraction of light
• Polarization light
When light behaves as a stream of particles?
The particle nature of light
Quantum (bundles/packets of energy) dominates when the light
interacts with matter (like
solids, liquids and gases).
Light as a stream of particles
Particle nature - Photoelectric,
(Photon/quantum)
Compton Effect, Black body
radiation..
The light is propagated in bundles
of small energy, each bundle being
called a quantum.
Each quantum is composed of
many small particles called quanta
or photon.
Photon energy
E = hf
h = Planck’s constant = 6.626x10-34Js
f = frequency of radiation
Light as a wave:
c= f 
Light as a particle:
E = hf
photon
Energy of a photon or light wave:
E  hf 
hc

( f 
c

)
Where h = Planck’s constant = 6.626x10-34Js
f = frequency of a light wave c = velocity of light
λ= wavelength of a light wave -distance between successive crests
Visible Light
400–435 nm
435 nm-440nm
440–480 nm
480–530 nm
530–590 nm
590–630 nm
630–700 nm
Violet
Indigo
Blue
Green
Yellow
Orange
Red
• The color of visible light is determined
by its wavelength.
• White light is a mixture of all colors.
• We can separate out individual colors
with a prism.
Wave Function of Sinusoidal Waves
y(x,t) = ym sin(kx-wt)
ym: amplitude
kx-wt : phase
k: wave number
2
k

w: angular frequency
2
w   2f
T
PRINCIPLE OF SUPERPOSITION
When two waves traveling almost in the same direction
superpose, the resulting displacement at a given point is the
algebraic sum of the individual displacements.
i.e. when waves, y1=A sin ωt & y2=A sin (ωt + ) superpose,
the resultant displacement is
y= y1+y2= a sin (ωt) + a sin (ωt+)
13
INTERFERENCE OF LIGHT
14
TWO-SSOURCE INTERFERENCE
When identical waves from two
sources overlap at a point in
space, the combined wave
intensity at that point can be
greater or less than the intensity
of either of the two waves. This
effect is called interference.
OR
When two waves of same frequency (or wavelength) with
zero initial phase difference or constant phase difference
superimpose over each other, then the resultant amplitude
(or intensity) in the region of superposition is different than
the amplitude (or intensity) of individual waves.
At certain points either two
crests or two troughs interact
giving
rise
to
maximum
amplitude resulting in maximum
intensity.
(Constructive interference).
At certain points a crest and a
trough interact giving rise to
minimum or zero amplitude
resulting in minimum or zero
intensity.
(Destructive interference).
16
TWO-SOURCE INTERFERENCE
Constructive Interference
Destructive interference
Two waves (of the same
wavelength) are said to be in phase
if the crests (and troughs) of one
wave coincide with the crests (and
troughs) of the other. The net
intensity of the resultant wave is
greater than the individual waves.
(Constructive interference).
If the crest of one wave coincides
with the trough of the second, they
are said to be completely out of
phase. The net intensity of the
resultant wave is less than the
individual
waves.
(Destructive
interference).
TWO-SOURCE INTERFERENCE
INTERFERENCE PATTERN PRODUCED BY WATER
WAVES IN A RIPPLE TANK
Maxima: where the shadows show the crests and valleys
(or troughs).
Minima: where the shadows are less clearly visible
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PHASE AND PATH DIFFERENCE
Phase: Phase of a vibrating particle at any instant indicates its state
of vibration.
B
π/2
O
t=0
A
λ/4
F
λ
π
C
λ/2
3π/2
2π
3λ/4
E
λ
D
G
λ
Phase may be expressed in terms of angle as a fraction of 2π.
Path difference  corresponds to phase difference of 2.
Phase difference
Path difference

2

Constructive interference
path difference p= 0
or phase difference  = 0
path difference p = 1
or phase difference  =2
General condition:
Path difference p = m
2λ
path difference p =2
or phase difference  = 4
or phase difference  = 2m
where m = 0, 1, 2, 3,…………
order of interference.
Constructive interference
Maximal constructive interference of two waves occurs
when their:
path difference between the two waves is a whole number
multiple of wavelength.
OR
Phase difference is 0, 2, 4 , … (the waves are in-phase).
DISTRUCTIVE INTERFERENCE
path difference p= /2
phase difference  = 1
path difference p = 3/2
or phase difference  =3
General condition:
Path difference p=(m+1/2)
or
path difference p = 5/2
phase difference  = 5
phase difference =(2m+1)
where m = 0, 1, 2, 3, ………
22
Complete destructive interference of two waves occur
when
the path difference between the two waves is an odd
number multiple of half wavelength.
Or
the phase difference is , 3, 5, … (the waves are 180o
out of phase).
23
COHERENCE
Coherence – necessary
condition for
interference to occur.
Two waves are called coherent when they are of :
• same amplitude
• same frequency/wavelength
• same phase or are at a constant phase difference
24
COHERENCE
For interference pattern
to occur, the phase
difference at point on the
screen must not change
with time. This is
possible only when the
two
sources
are
completely coherent.
A SECTION OF INFINITE WAVE
A WAVE TRAIN OF FINITE LENGTH L
No two independent sources can act as coherent sources, because
the emission of light by the atoms of one source is independent of
that the other.
If the two sources are completely independent light sources, no
fringes appear on the screen (uniform illumination). This is
because the two sources are completely incoherent.
25
COHERENCE
A SECTION OF INFINITE WAVE
A WAVE TRAIN
OF FINITE LENGTH L
Common sources of visible light emit light wave trains of
finite length rather than an infinite wave.
The degree of coherence decreases as the length of wave
train decreases.
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COHERENCE
Laser light is highly coherent whereas a laboratory
monochromatic light source (sodium vapor lamp) may be
partially coherent.
Common sources of visible light emit light wave trains of
finite length rather than an infinite wave.
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COHERENCE
Methods of producing coherent sources:
(1).
Division of wave front: In this method, the wave front is
divided into two or more parts with the help of mirrors, lenses and
prisms.
The common methods are:
a. Young’s double slit arrangement, b. Lloyd's single mirror
method.
(2) Division of Amplitude: In this method, the amplitude of the
incoming beam is divided into two or more parts by partial
reflection with the help of mirrors, lenses and prisms.
These divided parts travel different paths and finally brought
together to produce interference.
The common methods are
a. Newton’s rings,
b. Michelson’s interferometer.
28
DOUBLE SLIT INTERFERENCE
If light waves did not spread out after
passing through the slits, no
interference would occur..
Two narrow slits (can be considered
as two sources of coherent light
waves).
d
Screen
If the widths of the slits are small
compared with the wavelength
distance a (<<) - the light waves
from the two slits spread out (diffract)
– overlap- produce interference
fringes on a screen placed at a
distance ‘D’ from the slits.
29
DOUBLE SLIT INTERFERENCE
A monochromatic light source
B
C
produces two coherent light
sources by illuminating a barrier
containing two small openings
(slits) S1 and S2 separated by a
distance ‘d’ and kept at a distance
‘D’ from the screen.
Waves originating from two
coherent light sources S1 and S2
because maintain a constant
phase relationship.
In interference phenomenon, we
have assumed that slits are point
sources of light.
30
DOUBLE SLIT INTERFERENCE
When the light from S1 and S2
both arrive at a point on the
screen such that constructive
interference occurs at that
location, a bright fringe
appears.
B
C
When the light from the two
slits combine destructively at
any location on the screen, a
dark fringe results.
DOUBLE SLIT INTERFERENCE
O
At center of the screen O, we always get a bright
fringe, because at this point the two waves from slit
S1 and S2, interfere constructively without any
path/phase difference.
32
DOUBLE SLIT INTERFERENCE
- Analysis of Interference pattern
a-is the mid
point of the
slit
We consider waves from each slit that combine at an
arbitrary point P on the screen C.
The point P is at distances of r1 and r2 from the narrow slits
S1 and S2, respectively.
DOUBLE SLIT INTERFERENCE
For
D>>d,
we
can
approximate rays r1 and r2
as being parallel.
The line S2b is drawn so
that the lines PS2 and Pb
have equal lengths.
Path length (S1b) between
the rays r1 and r2 reaching
the point P decides the
intensity at P.
(i.e. maximum/minimum).
34
DOUBLE SLIT INTERFERENCE
Path difference between two waves
from S1 & S2 (separated by a
distance ‘d’) on reaching a point P
on a screen at a distance ‘D’ from
the sources is d sin .
(d sinθ)
The path difference (S1b=d sin)
determines whether the two waves
are in phase or out of phase when
they arrive at point P.
If path length (S1b= d sinθ) is
either zero or some integer
multiple of the wavelength,
constructive interference results at
P.
Constructive Interference: Maximum at P:
P
O
The condition for constructive
interference or Maxima at point
P is
d sin  =m ……….(maxima)
Where m= 0, ±1, ±2…….
m- order number.
Central maximum at O has order m=0.
Central maximum m=0
Each
maximum
above
has
a
symmetrically located maximum below
O; these correspond to m= -1, -2, -3……
Destructive Interference: Minimum at P:
P
O
Central maximum m=0
m=3
m=2
m=1
m=0
m=0
m=1
m=2
m=3
When path length (S1b=d sin ) is an odd
multiple of λ/2, the two waves arriving at
point P are 1800 (=π) out of phase and
give rise to destructive interference.
The condition for
dark fringes, or
destructive interference, at point P is
1
d sin   (m  )..........
..(minima)
2
m  0,1,2..........
..........
....
Negative values of m locate the
minima on the lower half of the
screen.
DOUBLE SLIT INTERFERENCE
Fringe width (Δy)
The distance between two consecutive bright or dark fringes
(for small θ)is known as fringe width Δy.
Δy
Δy
The spacing between the adjacent minima is same the
spacing between adjacent maxima.
DOUBLE SLIT INTERFERENCE
For small value of , we can make
following approximation.
sin   
Q
ym+1
m
O
ym
sin   tan    
D
If d sin=m , mth order
constructive interference will take
place at P.
ym
mD
d(
)  m  y m 
D
d
If (m  1) th order constructive interferen ce takes place at Q then
(m  1)D
y m 1 
d
(m  1)D
D
D
y  y m 1  y m 
m
 Fringewidth  y 
d
d
d
39
DOUBLE SLIT INTERFERENCE
Sample 41-1:Problem 1:
The double -slit arrangement is illuminated with light from a
mercury vapor lamp filtered so that only the strong green line
(λ=546 nm) is visible. The slits are 0.12 mm apart, and the screen on
which the interference pattern appears is 55 cm away.
(a) What is the angular position of the first minimum?
(b)What is the distance on the screen between the adjacent maxima?
λ=546 nm
D=0.55 m
a) θ=? for first minimum.
b) Fringe width Δy=?
d=0.12 mm
DOUBLE SLIT INTERFERENCE
Solution : (a ) At the first min imum (m  0)
1
1
9
(
m

)

(
)
x
(
546
x
10
m)
1
2  2
d sin   (m  )  sin  
 0.0023 radian
3
2
d
(0.12 x10 m)
sin  is small that sin     0.0023 radian  0.130
D (546 x10 9 m)(55x10 2 m)
(b) y 

 2.5mm
3
d
0.12 x10 m
CHECK YOURSELF
Solve for First maximum : use d sinθ=mλ
put m=1
DOUBLE SLIT INTERFERENCE
Problem 5:
A double-slit arrangement produces interference fringes for sodium light
(λ=589 nm) that are 0.230 apart. For what wavelength would the angular
separation be 10% greater?
(Assume that θ is small).
d sin θ = mλ
sinθ≈ θ
d θ=λ
Use:
Given:
(θ is very small)
(for first maxima , m=1)
λ1/ λ2= θ 1/ θ 2, λ2= λ1 x θ 2/ θ 1,
θ 1 = 0.230 (for 100%)
θ 2 = 0.230 x 1.1 = 0.2530 (10% more: for 110%)
λ2= λ1 x θ 2/ θ 1 = (589 nm)(0.253/0.23)= 647.9 nm
DOUBLE SLIT INTERFERENCE
Problem 11:
Sketch the interference pattern expected from using two pinholes
rather than narrow slits.
In case two pinholes are used instead
of slits, as in Young's original
experiment, hyperbolic fringes are
observed.
If the two sources are placed on a line
perpendicular to the screen, the
shape of the interference fringes is
circular as the individual paths
travelled by light from the two
sources are always equal for a given
fringe.
43
DOUBLE SLIT INTERFERENCE
Tutorial: Problem:2
Monochromatic light illuminates two parallel slits a distance
‘d’ apart. The first maximum is observed at an angular
position of 150. By what percentage should ‘d’ be increased
or decreased so that the second maximum will instead be
observed at 150?
Solution: Use d sinθ=mλ
d1 sin 150=λ
gives the first maximum (m=1)
d2 sin 150 = 2λ put the second maximum (m=2) at the
location of the first.
Divide the second expression by the first and d2 = 2d1. This
is a 100% increase in d1.
Tutorial: Problem 8
DOUBLE SLIT INTERFERENCE
In an interference experiment in a large ripple tank, the
coherent vibrating sources are placed 120 mm apart. The
distance between maxima 2.0 m away is 180 mm. If the
speed of ripples is 25 cm/s, calculate the frequency of the
vibrating sources.
Given:
d = 120 x 10 -3 m, λ= ?
ym=180 x10-3 m
D = 2 m,
v= 25 x 10-2 m. f=?
Use: v=f x λ
To find λ: Use ym= λD/d gives λ=0.0108m
Use: v= f x λ , f= 23 Hz.
45
HRK:958 page: Exercise: 41-2 DOUBLE SLIT INTERFERENCE
Problem 1:
Monochromatic green light of wavelength 554 nm, illuminates two
parallel narrow slits 7.7 μm apart. Calculate the angular position of the
third-order (m=3) bright fringe in radians and (b) in degrees.
Given:
d = 7.7 x 10 -6 m, m=3, λ= 554 nm = 554 x10-9 m
θ = ? in radians and in degrees.
Use: for the bright fringe: d sinθ = mλ (put m=3):
Answers:
(a) θ=0.216 radians (b) θ =12.5 degrees
CHECK YOURSELF
(Solve for third order dark fringe put m=2)
46
DOUBLE SLIT INTERFERENCE
Problem 3:
A double-slit experiment is performed with blue-green light of
wavelength 512 nm. The slits are 1.2mm apart and the screen is 5.4
m from the slits. How far apart are the bright fringes as seen on the
screen?
Given: λ =512 nm
d = 1.2 mm
D= 5.4 m
Fringe width Δy=?
Solution:
Fringe width= Δy = λD/d =
(512x 10-9m)(5.4 m)=(1.2 x 10-3m)
=2.3 x 10-3 m
CHECK YOURSELF: How far apart are the bright fringes as seen on
the screen? Fringe width is the same 2.3 x 10-3 m.
DOUBLE SLIT INTERFERENCE
Problem 4:
Find the slit separation of a double-slit arrangement that will
produce bright interference fringes 1.000 apart in angular
separation. Assume a wavelength of 592 nm.
Given: d=?
θ=1.000
λ= 592 nm
Use:
d =λ/sin θ = (592 x 10-9m)/sin(1.000) =3.39x 10-5m
CHECK YOURSELF
Solve this problem for dark
angular separation.
interference fringes 1.000 apart in
DOUBLE SLIT INTERFERENCE
Problem 6
A double-slit arrangement produces interference fringes for
sodium light (λ = 589 nm) that are 0.200 apart. What is the
angular fringe separation if the entire arrangement is
immersed in water (n=1.33)?
Solution:
Immersing the apparatus in water will shorten the
wavelengths to λ/n. Start with
d sin θ0 =λ; and then find θ from d sin θ= λ/n.
Combining the two expressions,
sin θ/sin θ0 =1/n
gives θ=0.150
Problem 7
DOUBLE SLIT INTERFERENCE
In a double-slit experiment, the distance between slits is 5.22 mm and
the slits are 1.36 m from the screen. Two interference patterns can be
seen on the screen, one due to light with wavelength 480 nm and the
other due to light with wavelength 612 nm. Find the separation on the
screen between the third-order interference fringes of the two different
patterns.
Solution:
The third-order fringe for a wavelength will be located at
ym = mλD/d= 3 λD/d
where ym is measured from the central maximum.
Then Δy is:
y1 - y2 = 3(λ1 -λ2)D/d = 3(612x 10-9m – 480x10-9m)(1.36m)/(5.22x10-3m)
= 1.03x10-4m:
DOUBLE SLIT INTERFERENCE
Problem 9:
If the distance between the first and tenth minima of a double slit
pattern is 18 mm and the slits are separated by 0.15 mm with the
screen 50 cm from the slits. What is the wavelength of the light
Q
used?
So with small angle approximation:
ym+1
m
d sin  =(m+1/2 ) ,
sin   tan  = =ym/D
d sin  =(m+1/2 ) ,
mth order minima will take place at P.
d (ym/D)= =(m+1/2 ) ,
mth order minima will take place at P.
ym = =(m+1/2 ) D/d
O
We are given the distance (n the screen) between the first minima
(m=0) and the tenth minima (m=9). Then
y9-y0 = (9+1/2)  D/d- (0+1/2)  D/d = 9 ( D/d)
Given:
y9-y0 = 18 mm,
d=50 cm=0.5 m,
d= 0.15 mm= 0.15 x10-3 m
Solving for λ = 18 x 10-3 x 0.15 x 10-3/ 9 x 0.5 = 600 nm
Solve this problem with first and 10th maxima.
(CHECK YOURSELF:
Put first maxima m=1, and tenth maximum m=10 )
Coherence
Problem 14:
The coherence length of a wavetrain is the distance over which the
phase constant is the same.
(a) If an individual atom emits coherent light for 1 x 10-8 s, what is the
coherence length of the wavetrain?
(b) Suppose a partially reflecting mirror separates this wave train into
two parts that are later reunited after one beam travels 5 m and
other 10 m. do the waves produce interference fringes observable by
a human eye?
(a) velocity = distance/ time
x = c/t = (3.0 x108m/s)/1 x 108s) = 3 m.
(b) No.
53
YOUNG’S DOUBLE SLIT INTERFERENCE
Thomas Young (1773–1829)
• Double slit experiment was first performed
by Thomas Young in 1801.
• So double slit experiment is known as
Young’s Experiment.
• He used sun light as source for the
experiment.
• In young’s interference experiment, sunlight
diffracted from pinhole S0 falls on pinholes S1
and S2 in screen B.
• Light diffracted from these two pinholes
overlaps on
screen C,
interference pattern.
producing the
54
INTENSITY IN DOUBLE SLIT INTERFERENCE
In this section we derive an expression for the intensity I at any point
P located by the angle ‘θ’.
We know, the intensity of light
wave is proportional to square of
its E-the electric vector.
Intensity α Amplitude2
I α E2
P
O
E0- amplitude of each wave.
Let us consider the electric
components of two sinusoidal waves
r1 and r2 from the two slits S1 and S2
have the same angular frequency ‘ω
(=2πf)’ and a constant phase
difference .
E1= E0 sin ωt
&
E2= E0 sin (ωt + )
55
INTENSITY IN DOUBLE SLIT INTERFERENCE
Then the total magnitude of the
electric field at point P on the screen
is the superposition of the two
waves.
(Assumption: The slit separation
d<<D, the electric vectors from the
two interfering waves are nearly
parallel, and we can replace the
vector sum of the E- fields with the
sum of their components).
The resultant electric field :
E = E1+E2 = Eo sin ωt + Eo sin
(ωt+)
INTENSITY IN DOUBLE SLIT INTERFERENCE
Adding Wave disturbances: Phasors
The combined electric field can be done
algebraically, using a graphical method,
which proves to be convenient in more
complicated situations.
Phasor diagram
The sinusoidal wave can be represented
graphically by a Phasor of magnitude Eo
rotating about the origin counterclockwise
with an angular frequency w.
w
w
E1(=E0 sinwt)
57
Phasor diagrams:
(a) Phasor representation of first wave
disturbance
E1(=E0sinwt)
is represented by the projection of the
phasor on the vertical axis.
The second sinusoidal wave is
E2 = Eo sin (wt + )
It has the same amplitude and frequency
as E1.
Its phase is  with respect to E1.
w
(b)
58
(c) The sum E of wave
disturbances E1 &E2 is the
sum of the projections of
the two phasors on the
vertical axis, placing the tail
of one arrow at the head of
the other, maintaining the
E
proper phase difference.
E is the projection on the
vertical axis of a phasor of
length Eθ, which is the
vector sum of the two
phasors of magnitude E0.

E2
E1
E

E0

E0
ωt
59
C

E2
E
E0

E
E1

ωt
E0
B
A
From the right angled triangle ABC
BC
E
sin( wt  ) 

AC
E
 E  E  sin( wt  )
60
 E  E sin(wt  )
Amplitude
Phase
shift
Wave part
Where E- The resultant electric field
Eθ- Amplitude of the resultant wave
β – the phase difference between amplitude of the resultant wave
and first wave.
From the geometry of the right angled triangle
Eθ
The three phasors, E0,E0,Eθ form an
isosceles triangle
(An any triangle, an exterior angle (φ in this
case) is equal to the sum of the opposite
interior angles( β and β)). So,
or
2 β= φ
β =φ/2
β
β
cos  
E0
E0
φ
( E  / 2)
E0
 E   2 E 0 cos 
61
The amplitude ‘Eθ’ of the resultant wave
disturbance is given by
Eθ = 2 E0 cos β
which determines the intensity of the
interference fringes, depends on ‘β’, which in
turn depends of the value of ‘θ’, that is , on the
location of point P.
θ
Then the intensity of the resultant wave Iθ at P is given by
Iθ α Eθ2
(for resultant wave) and
I0 α E02
(Intensity of each single wave)
Iθ = 4 E02 cos2 β = 4 E02 cos2 ф/2
I θ = 4 I0 cos2 β
Note: The intensity of the resultant wave at any point varies from zero (dark or
minima) to four times (bright or maxima)the intensity I0 (i.e. 4I0=Im) of individual
62
wave.
Intensity distribution in double slit interference
Iθ = 4 I0 cos2 ф/2
Intensity Maxima occur where
cos2ф/2 =1, or
Ф =0 , 2π , 4π,…………..2m π
Iθ = 4 I0 cos2 ф/2
Intensity Minima occur where
cos2ф/2 =0, or
Ф =π , 3π, 5 π …………..(2m+1) π
ф= 2mπ
ф= (2m+1)π
(maxima)
(minima)
Or the path difference:
Or the path difference:
d sinθ= m λ
d sinθ= (m+1/2) λ
m= 0, ±1, ±2……………(maxima)
m= 0, ±1, ±2……… (minima)63
1. The horizontal solid line is I0: this describes the (uniform)
intensity pattern on the screen if one slit is covered up.
2. If the two sources were incoherent, the intensity would be
uniform over the screen and would be 2 I0 indicated by the
horizontal dashed line.
3. For two coherent sources it would be 4I0.
INTENSITY IN DOUBLE SLIT INTERFERENCE
PHASE AND PATH DIFFERENCE
Phase difference Path difference

2

 d sin 
2

   (d sin )
2


 d
or    sin 
2 
The int ensity at any  can be written as
I   4I 0 cos2 
d sin 
I   4I 0 cos (
)

2
I E 
d sin 
Re lative int ensity :  2  cos2 (
)
I0 E 0

2
The phase difference (φ)
between r1 and r2
associated with the path
difference S1b (= d sinθ)
65
Using Trigonometric Identity:
E = E1+E2 = Eo (sin ωt + sin (ωt+))
1

1

sin   sin   2 cos     sin     
2

2

With  = (wt + ), = wt, get:
E  E 0  2 cos(wt    wt ) / 2sin(wt    wt ) / 2
E  E 0  2 cos  / 2sin(wt   / 2)
It can be written as,
Where Eθ

E= Eθ sin(wt+β)
=2 E0 cos β: Amplitude of the resultant
wave.
‘β’ – the phase difference between resultant wave and first wave.
β = ф/2

INTENSITY IN DOUBLE SLIT INTERFERENCE
Problem: SP 41-2
Find graphically the resultant E(t) of the following wave
disturbances.
The phase angle φ between
successive phasors is 150.
E1 = E0 sin wt
E2 = E0 sin (wt + 15o)
E=E1+E2+E3+E4
E3 = E0 sin (wt + 30o)
E4 = E0 sin (wt +
MIT-MANIPAL
E=Eθ sin(wt+β)
45o)
BE-PHYSICS- INTERFERENCE-2010-11
67
To find β:
In any closed n-sided
polygon, the sum of the
interior angles is (n-2)π.
So, in the five sided polygon,
abcdea:
2  b  c  d
 ( n  2) 
 (5  2)180  540 0
b  c  d  165 0
So, 2β=540 - 3(1650)
So, β = 22.50
68
To find Eθ:
β
x3
β-15
x1x2= bc (cosβ-15)
x2x3= cd (cosβ-15)
x3e= de (cosβ)
x2
β-15
ax1= ab cosβ
c
x1
b
69
Eθ = ax1+x1x2+x2x3+x3e
= ab cosβ+ bc (cosβ-150)+ cd (cosβ-150)+de (cosβ)
Substitute
ab=bc=cd=de= E0 and β=22.50
Eθ=3.83 E0
The resultant E(t) is the projection of E on the vertical axis:
E= 3.83 E0 sin (wt+22.50)
70
E  Eh  E v
2
We can use the phasors to
find sum and free to evaluate
the phasors at any time t. We
choose t=0.
sin wt
Where
2
Ey- sum of
vericalcomponents
Alternate method: Verification:
Standard equation:
E(t)= Eθ sin (wt+β)
β
Eθ
cos wt
Eh- sum of
horizontal
components
Horizontal Component :  E h  E 0 cos 0  E 0 cos15  E 0 cos 30  E 0 cos 45  3.53E 0
Vertical comonent : E v  E 0 sin 0  E 0 sin 15  E 0 sin 30  E 0 sin 45  1.46 E 0
E   3.83E 0
Phase angle :   tan 1 (
Ev
1.46
)  tan 1 (
)  22 .50
Eh
3.53
71
Sample Problem: Intensity in double-slit Interference
Problem: E 41-15
Source A of long-range radio waves leads source B by 90 degrees.
The distance rA to a detector is greater than the distance rB by
100m. What is the phase difference at the detector?
Both sources have a wavelength of 400m.
Initially, source A leads source B by
90◦, = 1/4 wavelength (100 m)
100 m=λ/4 corresponds
to π/2 phase
rA
However, source A also lags behind
source B since rA is longer than rB by
100 m, which is 100m/400m
= 1/4 wavelength.
So, the net phase difference between
A and B at the detector is zero.
They are in phase and the reach the detector.
D
E
T
E
C
T
O
R
rB
72
Problem: E 41-16
Find the phase difference between the waves from the two
slits arriving at the mth dark fringe in a double-slit
experiment.
Answer: (2m + 1)π radians.
73
Tutorial Problem: E 41-18
Find the sum of the following quantities (a) graphically,
using phasors; and (b) using trigonometry.
y2=8.0 sin (wt+300)
y1=10 sin wt,
Similar to the equation:
E1= E01 sinwt &
E2= E02 sin (wt+φ)
Resultant electric field E= Eθ sin (wt+β)
Where Eθ- Amplitude of the resultant wave
β= phase difference between first electric field and the resultant
wave.
74
(a) Phasors
To find Eθ:
From the right angled triangle ACB :
AC 2  AD 2  DC 2
y
E   (AB  BD ) 2  DC 2
2
C
 (E 01  E 02 cos ) 2  (E 02 sin ) 2
Eθ
E02
2
φ
β
A
E   (E 01  2E 01E 02 cos   E 02
E01
B
D
x
2
DC E 02 sin 
& the phase angle sin  

AC
E
Substitute E01=10, E02=8, & φ=300
Eθ=17.39
β= 13.30
The resultant electric field E= 17.39 sin (wt+13.3)
75
Where
2
We can use the phasors to
find sum and free to evaluate
the phasors at any time t. We
choose t=0.
sin wt
E  Eh  E v
2
Ey- sum of
vericalcomponents
Alternate method: Trigonometry
Standard equation:
E(t)= Eθ sin (wt+β)
β
Eθ
cos wt
Eh- sum of
horizontal
components
 E  10 cos 0  8 cos 30  10  6.92  16.92
h
E  10 sin 0  8 sin 30  0  4  4
v
E  17.39

  tan ( E / E )  13.3
1
0
v
h
E  17.39 sin(wt  13.3)
76
REFLECTION PHASE SHIFT
Reflection causes a /2 phase shift
Incident wave
Reflected wave
It has been observed that if
the medium beyond the
interface has a higher index
of refraction, the reflected
wave undergoes a phase
change of  (180o) ) or a
path length λ/2.
Reflection causes no phase shift
Incident wave
Reflected wave
If the medium beyond the
interface has a lower index
of refraction, there is no
phase/path length change
of the reflected wave.
77
INTERFERENCE FROM THIN FILMS
We see color when sunlight falls
on a bubble, an oil slick, or a soap
bubble - the interference of light
waves reflected from the front and
back surfaces of thin, transparent
films.
The film thickness is typically of
the order of magnitude of the
wavelength of light.
A soapy water
film
on
a
vertical
loop
viewed
by
reflected light
Thin films deposited on optical components- such
as camera lenses-reduce reflection and enhance the
intensity of the transmitted light.
Thin coatings on windows can enhance the
reflectivity for infrared radiation while having less
effect on the visible radiation.
It is possible to reduce the heating effect of sunlight
on a building.
79
INTERFERENCE FROM THIN FILMS
n1
n2 n1
A thin film (a soap film or thin
film of air between two glass
plates)
is viewed by light
reflected from a sourer S.
Waves reflected from the front
d
Source of light
surface (r1) and back surface (r2)
interfere and enter the eye.
80
n1
n2
n1
The incident ray ‘i’ from the
source enters the eye as ray r1
after reflection from the front
surface of the film at ‘a’.
The incident ray ‘i’ also enters
the film at ‘a’ as refracted ray
and is reflected from the back
surface of the film at ‘b’.
Source of light
d
d- thickness of the film.
It then emerges from the
front surface of the film at
‘c’ and also enters the eye,
as ray r2.
n1
Source of light
n2
d
n1
Interference in light reflected
from a thin film is due to a
combination of rays r1 and r2.
As the waves originated from
the same source by division of
amplitude, hence they are
coherent and they are close
together.
The region ac looks bright or
dark for an observer depends
on the phase difference
between waves of rays r1 and r2.
82
As r1 and r2 have travelled over
paths of different lengths, have
n1
n2
n1
traversed different media, and
have suffered different kinds of
reflections at ‘a’ and ‘b’.
The phase difference between
d
Source of light
two reflected rays
determine
interfere
r1 and r2
whether
they
constructively
destructively.
or
83
EQUATIONS FOR THIN –FILM INTERFERENCE
n1 n2 n1
If a wave travelling from a
medium of index of refraction
‘n1’ toward a medium of index
of refraction ‘n2’ undergoes a
d
Source of light
phase change 1800
reflection
when
n2>n1
upon
and
undergoes no phase change if
n2<n1.
84
The wavelength of light ‘λn’
λ

n 
n
in a medium whose index
of refraction is ‘n’
n 

n
λ - wavelength of the light in free space.
85
n1 n2 n1
To obtain Equations for thin
film
Interference,
let
us
simplify by assuming near normal incidence θi=0.
The ‘r2’ travels a longer path
(2d) than ‘r1’’ as ‘r2’ travels
d
Source of light
twice
through
the
before reaching the eye.
film
n1
n2
d
Source of light
n1
The path difference due to the
travel of ray r2 through the film
is approximately ‘2d’.
Other possible contributions to
the total path difference
between r1 and r2 : the phase
difference of π (or path
difference
of
one-half
wavelength) that might occur
on reflection at the front/back
surface of the film.
87
A general equation for Thin film Interference:
The path difference between rays r1 and r2 is:
?
?
Path difference = 2d + λn/2 + λn/2 …………(1)
Front
surface
Back
surface
Note: Depending on the relative index of refraction of the
film in comparison with what is on either side of the film, we
might need to include neither of the extra terms, or perhaps
one of them, or perhaps both of them.
Examples:
(a) Interference in a thin soap film
Air n
Air
For the interference from a
thin soap film of index of
No phase
change
refraction ‘n’ surrounded
by air, we must add the
extra half wavelength for
1800 phase
change
the front surface reflection,
d
but not for the back surface .
Interference in a thin soap film
The
total
path
difference=
2d+λn/2=mλn ...(maxima)
where m=1,2,3………
Where we have dropped the
m=0 solution because it is not
physically meaningful.
The
total
path
difference:
= 2d + λn/2 =(m+1/2) λn
……… (minima)
where m=0,1,2,3…..

n 
n
Note: These equations apply when the index of refraction of
the film is greater than the index of refraction of the material
on either side.
90
THIN FILM INTERFERENCE -WEDGE SHAPED FILM
A thin wedge of air film can be
formed by two glasses slides on
each other at one edge and
separated by a thin spacer (a
thin wire or a thin sheet) at the
opposite edge.
A thin film having zero thickness
at one end and progressively
increasing to a particular
thickness at the other end is
called a wedge.
91
The arrangement for observing
interference of light in a wedge
shaped film. The wedge angle is
usually very small and of the
order of a degree.
d
wire
When a parallel beam of
monochromatic
light
falls
normally on a wedge shaped film
part of it is reflected from upper
surface and some part from lower
surface (division of amplitude).
92
Ray BC reflected from the top –
NO phase change.
i
r
Back
surface
Ray DE (the back surface
reflection), undergoes a π phase
change and λ/2 (half wave length)
at the air to glass boundary due to
reflection.
These two coherent waves
superpose-producing constructive
and destructive interferences, the
positions of which depend on the
thickness of the film.
Constructive Interference
The total path difference:
Path difference = 2d + λn/2 = mλn …..(maxima)
where m=1,2,3……
m=0 dropped, physically not meaningful.
Destructive Interference
Path difference = 2d + λn/2 = (m+1/2) λn …(minima)
where m=0,1,2,3………………………………
Sample problem:41-3
A soap film (n=1.33) in air is 320 nm thick. If it is illuminated
with white light at normal incidence, what color will it
appear to be in reflected light?
Solution:
The wavelengths which are
maximally reflected are
constructively interfered.
No phase
change
No phase
change
Air
n=1.33
1800 phase
change
d
n
2d 
 m n (max ima )
2
Air
2nd
851 nm


1
1
(m  ) (m  )
2
2
Constructive interference maxima occur for the following
wavelengths:
1702 nm (=1),
567 nm(m=2),
340 nm (m=3) and so on.
Only the maximum corresponding to m=2 lies in the
visible region (between about 400 nm and 700 nm); light of
wavelength 567 nm appears yellow-green.
96
Sample Problem: 41-4. Page 951
Lenses are often coated with thin films of transparent
substances such as MgF2 (n=1.38) to reduce the reflection
from the glass surface. How thick a coating is needed to
produce a minimum reflection at the center of the visible
spectrum (λ=550 nm)?
Given: λ=550 nm
n=1.38
Minimum reflection:
(Destructive interference)
Thickness of coating: d=?
Solution:
Light strikes the lens
incidence (θ).
at near-normal
For both the front and back surfaces of the
MgF2 film the reflection have additional
path difference (λ/2).
The path difference for destructive
interference is therefore Path difference:
2nd+λn/2+λn/2=(m+1/2)λn…(minima)
98
Where m=1,2,3……………………………,
Dropped m=0 solution : physically not meaningful.
We seek the minimum thickness for destructive
interference. For m=1, we obtain
1
( m  )

550nm
2
d


 100nm
2n
4n 4 x 1.38
INTERFERENCE FROM THIN FILMS
Problem: E 41-23
A disabled tanker leaks kerosene (n=1.20) into the Persian
Gulf, creating a large slick on top of water (n = 1.33).
(a)If you look straight down from aeroplane on to the
region of slick where thickness is 460nm, for which
wavelengths of visible light is the reflection is greatest?
(b) If you are scuba diving directly under this region of
slick, for which wavelengths of visible light is the
transmitted intensity is strongest?
MIT-MANIPAL
BE-PHYSICS- INTERFERENCE-2010-11
100
White light (400 nm700nm)
Normal incidence
Reflected
light (λ=?)
R.I=1.0
n=1.2
d=460nm
R.I=1.33
Thin film of oil
water
Solution:
The reflected light from the
film is brightest at the
wavelength (λ) for which the
reflected rays are in phase with
one
another.(constructive
interference).
(Both Front & back surface
reflections have phase change)
.
101
The total path difference for maxima:
Path difference = 2d + λn/2+ λn/2 = mλn ,
2d=(m-1) λ/n
(λn=λ/n)
λ=2nd/(m-1)
Where (m=1,2,3….)
Find λ for d=460 nm, n=1.2 & m=1,2,3…
λ - for m=1 (not possible)
For
m=2: λ=1104 nm
(IR region)
For
m=3: λ=552 nm
(Green light-visible)
For
m=4: λ=368 nm
(UV light)
So, Green light appears in the reflected light
102
Minimally
reflected
light
R.I=1.0
n=1.2
R.I=1.33
The wavelengths which are minimally
reflected are maximally transmitted, and
vice versa. Maximally transmitted
wavelengths is the same as finding the
minimally reflected wavelengths.
Thin film of oil
water
The total path difference for minima:
Path difference
= 2d + λn/2+ λn/2 = (m+1/2)λn
Maximum
Transmitted
light =?
where (m=1,2,3….)
substitute : (λn=λ/n)
λ=2nd/(m-1/2)
103
Find λ for d=460 nm, n=1.2, & m=1,2,3…
λ - for m=1: λ=2208 nm (not visible region)
For m=2:
λ=736 nm (IR region)
For m=3:
λ=442 nm
(Blue light-visible)(maximum transmitted)
104
Problem: E 41-25
If the wavelength of the incident light is λ=572 nm, rays A
and B are out of phase by 1.50 λ. Find the thickness d of
the film.
The total path difference for minima:
Path difference =
2d +λn/2= (m+1/2)λn
2d= mλn
m=0 not possible
Take m=1
d=λ/2n=215nm
105
OR
In the given total phase of 1.50 λn.
The top surface contributes a phase difference π=0.5λn
So, phase difference because of the thickness= 2d=λn=2π
2d=λn
d=λ/2n=572 nm/2 x 1.33=215 nm
106
INTERFERENCE FROM THIN FILMS
Problem: E 41-29
A broad source of light (wavelength = 680nm) illuminates
normally two glass plates 120 mm long that touch at one
end and are separated by a wire 0.048mm in diameter at
the other end. How many bright fringes appear over 120
mm distance?
λ=680 x10- 9m
d=0.048 x 10-3 m)
θ
MIT-MANIPAL
x=120 mm
BE-PHYSICS- INTERFERENCE-2010-11
107
Constructive int erference occurs when :
n
2d 
 m n
2
m?
2nd 1 2 x 1 x (0.048 x10 3 ) 1
m
 
  141 .67  141
9

2
680 x10
2
.
108
INTERFERENCE FROM THIN FILMS
Problem: E 41-27
A thin film of acetone (n = 1.25) is coating a thick glass plate
(n = 1.50). Plane light waves of variable wavelengths are
incident normal to the film. When one views the reflected
wave, it is noted that complete destructive interference
occurs at 600nm and constructive interference at 700nm.
Calculate the thickness of the acetone film?
Put: λ =λ/n
n
Thin film
n=1.25
MIT-MANIPAL
Glass plate
1.50
BE-PHYSICS- INTERFERENCE-2010-11
109
Constructive interference
2d + λn/2 + λn/2=m λn
Gives : 2nd=(m-1) λ
2nd=(m-1)(700nm ) ……………(1)
Destructive interference
2d + λn/2 +λn/2 =(m+1/2 )λn
2nd=(m-1/2) λ
2nd=(m-1/2)(600nm ) ……………(2)
Divide (1)/(2): m=4, d=840 nm
110
Problem 21:We wish to coat a flat slab of glass (n = 1.5) with a
transparent material (n=1.25) so that light of wavelength
620nm (in vacuum) incident normally is not reflected. What
should be the minimum thickness of the coating?
Air=1
Film=
1.25
Glass=1.5
111
Both the reflected rays r1(front surface
reflection) and r2(back surface reflection)
have additional path difference (λ/2).
n n
1
2d 

 (m  ) n (min ima )
2
2
2
m=1
1
( m  )

620nm
2
d


 124nm
2n
4n
4 x 1.25
112
Problem 22:
A thin film in air is 410nm thick and is
illuminated by white light normal to its surface. Its index of
refraction is 1.50. What wavelength in the visible spectrum
will be intensified in the reflected beam?
Air
n=1.50
Air
No phase
change
1800 phase
change
d=410 nm
113
n
2d 
 m n (max ima )
2
2nd
( 2)(1.5)(410nm)
1230nm



1
1
1
(m  )
(m 
(m  )
2
2)
2
The result is only in the visible range when
m = 3, so λ= 492 nm.
114
Problem: 24:
In costume jewelry, rhinestones (made of glass with n =1.50)
are often coated with silicon monoxide (n = 2.0) to make
them more reflective. How thick should the coating be to
achieve strong reflection for 560 nm light incident normally?
Air=n=1
Film=n= 2.0
Glass=n=1.5
115
n
2d   m n (max ima )
2
1
( m  )
 560nm
2
d


 70 nm
2n
4n
4 x2
116
Problem 26:
Light of wavelength 585nm is incident normally on a thin,
soapy film (n=1.33) suspended in air. If the film is
0.00121mm thick, determine whether it appears bright or
dark when observed from point near the light source.
Air
n=1.33
Air
No phase
change
1800 phase
change
d=0.00121 mm
117
m should be an integer.
n
2d 
 (m  1 / 2) n (min ima )
2
2nd
m
 2(1.33)(1. 21x10 -6 m)/(585 x 10 -9 m)  5.5

So the interference is NOT dark.
n
2d 
 m n (max ima )
2
2nd 1
-6
-9
m
  2(1.33)(1. 21x10 m)/(585 x 10 m)  6.00

2
So the interference is bright.
118
Problem 28:
White light reflected at perpendicular incidence from a soap
film in air has, in the visible spectrum, an interference
maximum at 600nm and a minimum at 450nm with no
minimum in between. If n = 1.33 for the film, what is the film
thickness?
Air
n=1.33
Air
No phase
change
1800 phase
change
d=?
119
With First source (λ=600nm):
n
2d 
 m n (max ima )
2
600
2nd 
 600 m
2
2nd  300  600 m.......... ...(1)
With second source (λ=450nm): 2d   n  (m  1 ) (min ima )
n
2
2
1
2nd  225  (m  )450 .......... (2)
2
Dividing eqn (1) by (2) :
Use equation (1) or (2) to
600
We get m 
2
solve for d:
2(600  450 )
From equation (2)
d= 2 x 450/2 x1.33=338 nm
120
Problem 31: Light of wavelength 630nm is incident
normally on a thin wedge shaped film with index of
refraction 1.50. There are ten bright and nine dark fringes
over the length of the film. By how much does the film
thickness changes over the length?
d1
x
d2
B1 B2 B3 B4 B5 B6 B7 B8 B9 B
10
m1=1
m2=10
Number of bright bands in ‘x’ mm length. 10 bright & 9
dark bands. Film thickness over the length d2-d1=?
Constructive int erference occurs when :
n
n
1
2d 
 m n  2nd 
 m n  2nd  ( m  )
2
2
2
Let for the first bright band minimum value of m  1
& the last bright band be m  10
1
Then, 2nd 1  (m 1 - ) .....( 1)
2
1
& 2nd 2  ( m 2  ).......... ..( 2)
2
Eqn ( 2)  eqn( 2) gives
(d 2  d 1 )  9/2n  9(630 nm)/2 (1.50)  1.89 m
122
INTERFERENCE FROM THIN FILMS
Instead of wedge shaped films,
interference is possible even in
curved films also.
Circular interference fringes can
be produced by enclosing a very
thin film of air of varying
thickness between a plane glass
plate and a plano -convex lens of
a large radius of curvature.
If monochromatic light is
allowed to fall normally
and viewed, dark and
bright circular fringes
known as Newton’s Rings
are produced.
The fringes are circular
because the air film has
circular symmetry.
124
Newton’s Rings:
When the light is incident on the
plano-convex lens part of the light
incident on the system is reflected
from glass-to-air boundary (say at
point D).
The reminder of the light is
transmitted through the air film, and
it is again reflected from the air-toglass boundary (say from point J).
The two rays are (1 and 2 ) reflected
from the top and bottom of the air
film interfere with each other to
produce darkness and brightness .
The interference effect is due to
the combination of ray 1, reflected
from the flat surface, with ray 2,
reflected from the curved surface
of the lens.
Air-Glass interface
Phase change (π)
Ray 1 undergoes a phase change
of 1800 upon reflection (because it
is reflected from a medium of
higher index of refraction),
whereas ray 2 undergoes no phase
change (because it is reflected
from a medium of lower refractive
index).
126
The condition for constructive
interference remains unchanged.
Path difference:
2 d

 m
2
For normal incidence r  0, cos 0  1
air film n  1
1
2 d  (m  ) ................(1)
2
The radius (r) of the bright ring:
Consider the section AB of
the lens, wherein it forms
an air film of thickness d.
C
Let mth order bright ring of
radius ‘r‘ forms here.
R
A
D
O
B
Let C be the center of
curvature of the planoconvex lens.
R(=CB=OC) be the radius of
curvature of the planoconvex lens.
128
C
We can write: d= OD=OC-CD
d
R

R
R r
2
2

 r  
R  R 1    
R 

2
A
D
O
B

1
2
If r / R  1,
we can exp and the squarebracket by the binomial theorem,
keeping only two terms , or
 1  r 2

r2
d  R  R 1     .....  
 2R 
 2R
129
Substituting ‘d’ value in equation (1) and solving for ‘r’:
1
2 d  ( m  ) .......... ......( 1)
2
r2
1
2(
)  ( m  ) (for air n  1) gives
2R
2
1
r  R(m  )
2
D  2r
…….(For bright ring)
The radii/diameters of the bright rings.
1. D- is the diameter of the mth bright ring.
2. m=0 is physically not possible,
3. So, m=1,2……………………..
4. center ring must be the dark.
5. Radii/diameters of bright rings are
proportional to square root of odd numbers.
131
Problem 33: In Newton’s ring experiment, the radius of
curvature R of the lens is 5m and its diameter is 20mm.
a) How many rings are produced?
b) How many rings would be seen if the arrangement is
immersed in water (n=1.33)?
(Assume that wavelength = 589nm)
Given: λ = 589 nm
Radius of curvature R= 5 m
diameter of the ring=d=20mm
Therefore, radius r= 10 mm=0.01 m
(a) m=? (b) n=1.33, m=?
132
1
(a ) r  R(m  )
2
r2 1
0.012
1
m
 

 34 is the number of rings observed.
9
R 2 5 x 589 x 10
2
(b) Putting the apparatus in water eectively changes the wavelength to

(589 nm) / (1.33)  443 nm (  n  )
n
so the number of rings will now be
(0.01) 2
1
m
  45
(443 nm) x (5m) 2
133
Problem 34: The diameter of the tenth ring in a Newton’s
rings apparatus changes from 1.42 to 1.27 cm as a liquid is
introduced between the lens and the plate. Find the
refraction of the liquid.
Given: Radius of tenth bright ring in air: rair = 1.42 cm
Radius of tenth bright ring in liquid: rliquid = 1.27 cm ring
Refractive index of the liquid: nliquid = ?
134
We know :  n 
 n liquid
r

n
i.e.  liquid 
 air
n liquid
 air

 ? ................(1)
 liquid
R ( m 
1
)  rair 
2
1
) ...........( 2)
2
1
R liquid (10  ) .....(3)
2
R air (10 
rliquid 
Divide eqn ( 2) / eqn (3) :
 air
r
1.42 
 ( air ) 2  
 1.25

 liquid
rliquid
1.27 
2
n
135
Problem 35: A Newton’s ring apparatus is used to
determine the radius of curvature of a lens. The radii of
the nth and (n+20)th bright rings are found to be 0.162cm
and 0.368cm, respectively, in light of wavelength 546nm.
Calculate the radius of curvature of the lower surface of
the lens.
Given: λ = 546 nm
Radius of the nth ring=rn =0.162cm
Radius of the (n+20)th
=rn+20=0.368cm
m=? Radius of curvature R= ?
136
For bright ring : r 
(0.162 cm) 
R ( m 
1
)
2
1
( n  ) R .......... ........( 1)
2
1
(0.368 cm)  (n  20 - ) R .......... ....( 2)
2
Divide one by another, we get
2
 (0.368 cm) 
( n  19 .5)

 5.308

( n  0.5)
 (0.162 cm) 
solve for R using the value of m  5.308.
Then R  r 2 / (n - 1/2) 
 (0.162 cm) 2 (5.308 - 0.5)(546 nm)  1.00m
137
INTERFERENCE FROM THIN FILMS
OPTICAL PATH
•
Distance traveled by light in a medium in the time
interval of ‘t’ is d = vt
•
Refractive index n = c/v
•
Hence, ct = nd
•
nd  Optical path.
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d
138
MICHELSON’S INTERFEROMETER
Purpose
Interferometers are basic
optical tools used to
precisely
measure
wavelength, distance, index
of refraction of optical
beams.
It is a device working on the
principle of interference of
light and is used in precise
measurements of length or
changes in length.
139
MICHELSON’S INTERFEROMETER
 Light
from
an
extended
monochromatic source P falls on a halfsilvered mirror M.
 The incident beam is divided into
reflected and transmitted
beams of
equal intensity (Division of amplitude).
 These two beams travel almost in
B
perpendicular directions and will be
reflected
normally
from
movable
mirror (M2) and fixed mirror (M1).
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G
140
MICHELSON’S INTERFEROMETER
 The two beams finally proceed
towards a telescope (T) through
which interference pattern of
circular fringes will be seen.
 The interference occurs because
the two light beams travel
different paths between M and
M1 or M2.
 Each beam travels its respective
path twice. When the beams
recombine, their path difference
is 2 (d2 – d1).
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MICHELSON’S INTERFEROMETER
The path difference can be
changed by moving mirror M2.
As M2 is moved, the circular
fringes appear to grow or shrink
depending on the direction of
motion of M2.
New rings appear at the center
of the interference pattern and
grow outward or larger rings
collapse disappear at the center
as they shrink.
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MICHELSON’S INTERFEROMETER
Each fringe corresponds to a
movement of the mirror M2
through one-half wavelength.
The number of fringes is thus
the same as the number of
half wavelength.
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If N fringes cross the field of
view when mirror M2 is moved
by d, then
d = N (/2)
d is measured by a micrometer
attached
to
M2.
microscopic
Thus
length
measurements can be made by
this interferometer.
144
Michelson
interferometer
equation

N  d
2
2 (d)  N
The interferometer is used to measure changes in
length by counting the number of interference
fringes that pass the field of view as mirror M2 is
moved.
Length measurements made in this way can be
accurate if large numbers of fringes are counted.
145
Applications
Determination of wavelength
The fact that whenever the movable mirror moves by

,a fringe originates or vanishes at the center is used
2
to determine  from the equation 2d = N , where d is
the distance moved and N, the number of fringes
originated or vanished.
146
Applications
Determination of refractive index (n)
When a thin film (whose refractive
index n is to be determined) of
thickness ‘d’ is introduced on the path
of one of the interfering beams, an
d
additional path difference
(nd–d)2= 2d(n-1) will be introduced.
As a result there will be shift of fringes.
If ‘m’ fringes shift, then,
2d(n -1) = m from which ‘n’ can be
determined.
147
MICHELSON’S INTERFEROMETER
Problem: SP 41-6
Yellow light (wavelength = 589nm) illuminates a
Michelson interferometer. How many bright fringes
will be counted as the mirror is moved through 1.0
cm?
The number of fringes is the same as the
number of half wavelengths in 1.0000 cm.
Nλ= 2d
N= 2 d/λ=2(1.0000 x10-2 m)/(589 x 10-9m)
= 33,956 fringes
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MICHELSON’S INTERFEROMETER
Problem: E41-39
If mirror M2 in Michelson’s interferometer is
moved through 0.233mm, 792 fringes are
counted with a light meter.
What is the
wavelength of the light used?
Nλ= 2d
λ= 2 d/N=2(0.233 mm)/792= 588 nm
= 588 nm
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MICHELSON’S INTERFEROMETER
Problem: E41-40
An airtight chamber 5.0 cm
long with glass windows is
placed in one arm of a
Michelson’s interferometer as
indicated in Fig 41-28 . Light of
wavelength λ = 500 nm is used.
The air is slowly evacuated
from the chamber using
a
vacuum pump. While the air is
being removed, 60 fringes are
observed to pass through the
view. From these data find the
index of refraction of air at
atmospheric pressure.
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Answer:
The change in the optical path length is
2(nd – d)
2d(n-1)=mλ
n=mλ/2d+1=1.00030
151
Problem HRK-41-38:
A thin film with n=1.42 for light of wavelength
589nm is placed in one arm of a Michelson
interferometer. If a shift of 7 fringes occurs, what is
the film thickness?
Solution:
2d(n-1) = m
d = mλ/ 2(n-1)
d = 7 x (589 x10-9 m)/ 2(1.41-1)
= 4.9 x 10-6 m
QUESTIONS – INTERFERENCE
What is the necessary condition on the path length difference
(and phase difference) between two waves that interfere (A)
constructively and (B) destructively ?
[2]
Obtain an expression for the fringe-width in the case of
interference of light of wavelength λ, from a double-slit of slitseparation d. [5]
Explain the term coherence.
[2]
Obtain an expression for the intensity of light in double-slit
interference using phasor-diagram.
[5]
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QUESTIONS – INTERFERENCE
Draw a schematic plot of the intensity of light in a double-slit
interference against phase-difference (and path-difference).
[2]
Explain the term reflection phase-shift.
[1]
Obtain the equations for thin-film interference. [2]
Explain the interference-pattern in the case of wedge-shaped
thin-films.
[2]
Obtain an expression for the radius of mTH order bright ring in
the case of Newton’s rings. [5]
Explain Michelson’s interferometer. Explain how microscopic
length measurements are made in this. [4]
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INTERFERENCE – ANSWERS TO PROBLEMS
SP41-1: 0.12degree, 2.5mm
E41-1: 0.22radian, 12 degrees
E41-5: 650nm
E41-9: 600nm
SP41-2: E(t)= 3.83 Eo sin(wt+22.5o )
E41-15: 0 degree
E41-19: 1.21, 2.22, 8.13m
SP41-3: 567nm (Yellow-green)
SP41-4: 100nm
E41-23: 552nm, 442nm
E41-27: 840nm
E41-29: 141
E41-33: 34, 45
SP41-6: 33956 fringes
E41-39:
588nm
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