10/2/2022
B.TECH FIRST YEAR
ACADEMIC YEAR: 2022-2023
SESSION OUTCOME
COURSE NAME: ENGINEERING PHYSICS
COURSE CODE
“UNDERSTAND THE BASIC
PRINCIPLES OF WAVE OPTICS”
: PY1001
LECTURE SERIES NO : 01 (ONE)
CREDITS
:
4
MODE OF DELIVERY : ONLINE (POWER POINT PRESENTATION)
FACULTY
:
DR. NILANJAN HALDER
EMAIL-ID
:
nilanjan.halder@jaipur.manipal.edu
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INTERFERENCE
Topics
 Two source interference
 Double-slit interference
ASSIGNMENT
QUIZ
MID TERM EXAMINATION –I
END TERM EXAMINATION
ASSESSMENT CRITERIA’S
 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)
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TWO-SOURCE INTERFERENCE
TWO-SOURCE 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.
Maximal constructive
interference of two
waves occurs when their
phase difference is 0, 2,
4 , … (the waves are inphase)
The interference is constructive when the net intensity
is greater than the individual intensities.
The interference is destructive when the net intensity is
less than individual intensities.
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Complete destructive interference of two waves occur when
their phase difference is , 3 , 5 , … (the waves are 180o out
of phase)
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TWO-SOURCE INTERFERENCE
DOUBLE-SLIT INTERFERENCE
INTERFERENCE PATTERN PRODUCED BY WATER
WAVES IN A RIPPLE TANK
A train of plane light waves is incident on two narrow parallel
slits separated by distance d (<<). The interference pattern
on the screen consists of bright and dark fringes.
Maxima: where the shadows show the crests and valleys
Minima: where the shadows are less clearly visible
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DOUBLE-SLIT INTERFERENCE
 Consider two coherent sources S1 and
S2 separated by a distance ‘d’ and kept
at a distance ‘D’ from the screen.
 For D>>d, we can approximate rays r1
and r2 as being parallel.
 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 S1b =
d sin .
Phase difference (Φ) between the interfering waves:
depends on the location of the point P on the screen
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DOUBLE-SLIT INTERFERENCE
DOUBLE-SLIT INTERFERENCE
For maximum at point P
S1b = m
m = 0, 1, 2, . . .
Which can be written as,
d sin  = m
m = 0, 1, 2, . . .
m = 0 is the central maximum.
For minimum at point P
S1b  (m  21 )  m = 0, 1, 2, . . .
Which can be written as,
d sin   (m  21 )  m = 0, 1, 2, . . .
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• For small value of , we can make
following approximation.
sin   
sin   tan  
y
D
• Path difference:
d sin
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  S 1b 
y d
D
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DOUBLE-SLIT INTERFERENCE
DOUBLE-SLIT INTERFERENCE
Separation between adjacent maxima
(for small ) is independent of m
mth
maximum is located at
ym given by
y  ym1  ym
m
y
 m
d
D
or
D
ym  m
d
 (m1)
y 
where m = 0, 1, 2, . . .
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D
d
The spacing between the adjacent
minima is same the spacing between
adjacent maxima.
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DOUBLE-SLIT INTERFERENCE
DOUBLE-SLIT INTERFERENCE
Problem: SP 41-1
The double slit arrangement is illuminated by light of
YOUNG’S DOUBLE SLIT EXPERIMENT
wavelength 546nm. The slits are 12mm apart and the
• 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 his experiment, he allowed sun light
to pass through narrow opening (S0)
and then through two openings (S1
and S2).
screen on which interference pattern appears is 55cm
away.
a) What is the angular position of (i) first minima and (ii)
tenth maxima?
b) What is the separation between two adjacent
maxima?
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D
D
m
d
d
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DOUBLE-SLIT INTERFERENCE
DOUBLE-SLIT INTERFERENCE
Problem: E 41-2
Problem: E 41-5
Monochromatic light illuminates two parallel slits a
A double-slit arrangement produces interference
distance d apart The first maximum is observed at
fringes for sodium light (wavelength = 589 nm) that
an angular position of 15°. By what percentage
are 0.23° apart. For what wavelength would the
should d be increased or decreased so that the
angular separation be 10% greater ? Assume that the
second maximum will instead be observed at 15° ?
angle  is small.
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DOUBLE-SLIT INTERFERENCE
DOUBLE-SLIT INTERFERENCE
Problem: E 41-8
Problem: E 41-11
In an interference experiment in a large ripple tank
(see Fig 41-2) the coherent vibrating sources are
Sketch the interference pattern expected from using two
placed 120 mm
pin-holes rather than narrow slits.
apart.
The distance between
maxima 2.0 m away is 180 mm. If the speed of the
ripples is 25 cm/s, calculate the frequency of the
vibrating sources.
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COHERENCE
COHERENCE
A SECTION OF INFINITE WAVE
For interference
pattern to occur, the
phase difference at
point on the screen
must not change with
time.
A SECTION OF INFINITE WAVE
A WAVE TRAIN
OF FINITE LENGTH L
This is possible only when the two sources are completely
coherent.
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.
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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
INTENSITY IN DOUBLE SLIT INTERFERENCE
A SECTION OF INFINITE WAVE
 Electric field components at P due to S1 and S2 are,
Two waves are said to
E1= E0 sin ωt & E2= E0 sin (ωt + ) respectively.
be coherent when
 Resultant field E = E1 + E2
they are of :
• same amplitude
A WAVE TRAIN
OF FINITE LENGTH L
• same frequency
• same phase or are of
a constant phase
difference
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Laser light is highly coherent whereas
a laboratory monochromatic light
source (sodium vapor lamp) may be
partially coherent.
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INTENSITY IN DOUBLE SLIT INTERFERENCE
INTENSITY IN DOUBLE SLIT INTERFERENCE
Resultant of E1= E0 sin ωt & E2= E0 sin (ωt + )
Resultant of E1= E0 sin ωt & E2= E0 sin (ωt + )
From phasor diagram,
Phasor  Rotating vector.
ADDITION OF TWO VECTORS USING PHASORS
E2
E = E1 + E2
= E sin(t + )
Let two vectors be, E1= E0 sin ωt &
E0
E0
E1
E2= E0 sin (ωt + )
Resultant field E = E1 + E2
ωt + 
ωt
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E0

E2
E
= 2E0 cos  sin(t + )
But  = /2. So above eqn can be
written as,

E

E0
E1
E = 2 E0 cos(/2) sin(wt+/2)
ωt
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INTENSITY IN DOUBLE SLIT INTERFERENCE
INTENSITY IN DOUBLE SLIT INTERFERENCE
 E = 2 E0 cos(/2) sin(wt+/2)
PHASE AND PATH DIFFERENCE
 So intensity at an arbitrary point P on the screen due to
interference of two sources having phase difference ;
I

Phase difference Path difference

2


4 E 02 cos 2  
2
Path difference  corresponds

4  0 cos 2  
2
2
where   E is intensity due to single source
0
0

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
to phase difference of 2.
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INTENSITY IN DOUBLE SLIT INTERFERENCE



4  0 cos 2  
2
where   E 2 is intensity due to single source
0
0
Since   2  dsin  /  ,
 
2
4  0 cos 
 d sin  




From above equation,
At maxima :   2 m 
or
At minima :   ( 2 m  1) 
d sin 
or
 m
d sin 
Light intensity (I) versus d sin θ for a double-slit
interference pattern when the screen is far from the
two slits (D>> d).
 (m  1 ) 
2
where m  0,  1,  2, . . .
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INTENSITY IN DOUBLE SLIT INTERFERENCE
INTENSITY IN DOUBLE SLIT INTERFERENCE
Problem: SP 41-2
Find graphically the resultant E(t) of the following wave
disturbances.
E1 = E0 sin t
E2 = E0 sin (t + 15o)
E3 = E0 sin (t + 30o)
E4 = E0 sin (t + 45o)
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INTENSITY IN DOUBLE SLIT INTERFERENCE
INTENSITY IN DOUBLE SLIT INTERFERENCE
Problem: E 41-18
Find the sum of the following quantities (a) graphically,
using phasors; and (b) using trigonometry:
y1 = 10 sin (t)
y2 = 8.0 sin (t + 30°)
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.
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INTERFERENCE FROM THIN FILMS
Thickness and color in a thin film
 A film of thickness of the order of a
micron.
 Thickness of the film is comparable
with the wavelength.
 Greater thickness spoils the coherence
of the light to produce colour.
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A soapy water film on a
vertical loop viewed by
reflected light
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INTERFERENCE FROM THIN FILMS
INTERFERENCE FROM THIN FILMS
Phase change on Reflection
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).
If the medium beyond the interface has a lower index of
refraction, there is no phase change of the reflected wave.
Phase changes on reflection at a
The region ac looks bright or dark for an observer depending
junction between two strings of
on the path difference between the rays r1 and r2.
different linear mass densities.
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INTERFERENCE FROM THIN FILMS
INTERFERENCE FROM THIN FILMS
OPTICAL PATH
Distance traveled by light in a medium in the time interval
Equations for Thin Film Interference:
of ‘t’ is d = vt
Normal incidence (i = 0)
•
Refractive index n = c/v
Path difference = 2 d + (½) n
•
Hence, ct = nd
•
nd  Optical path.
Constructive interference:
•
Optical path is the distance traveled by light in vacuum in
2 d + (½) n = m n
same time ‘t’.
Destructive interference:
If n is wavelength in the film of refractive index n and  is
2 d + (½) n = (m+½) n
•
•

BACK SURFACE
the wavelength in vacuum then n =  / n
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m = 1, 2, 3, . . .
(maxima)
m = 0, 1, 2, . . .
(minima)
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INTERFERENCE FROM THIN FILMS
What should be the minimum thickness (in terms of wavelength used) and refractive
index of a non- reflective coating on lens made up of glass?
WEDGE SHAPED FILM
Light is reflected from both surfaces of the coating. In both reflections
the light is reflected from a medium of greater index than that in which it
is traveling, so the same phase change occurs in both reflections.
In wedge – shaped thin film,
constructive interference occurs in
certain part of the film [2 d + (½) n =
m n] and destructive interference in
others [2 d + (½) n = (m+½) n].
Then bands of maximum and
minimum intensity appear, called
The thickness of the non-reflective coating can be a quarter-wavelength. (t = λ/4) .
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fringes of constant thickness.
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Air Wedge – the air between two sheets of flat
glass angled to form a wedge
Air Wedge – the air between two sheets of flat
glass angled to form a wedge
Minima (destructive)
2t  m 
Maxima (constructive)
2t = (m - ½) 
𝑥 =
𝜆
1
𝑚 −
2𝛼
2
tan α = t /x
t  x
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x
m


2
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INTERFERENCE FROM THIN FILMS
INTERFERENCE FROM THIN FILMS
Problem: SP 41-4
Problem: SP 41-3
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
required to produce a minimum reflection at the center
of the visible spectrum? ( wavelength = 550nm)
A soap film (n=1.33) in air is 320nm thick. If it is
illuminated with white light at normal incidence, what
color will it appear to be in reflected light?
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INTERFERENCE FROM THIN FILMS
INTERFERENCE FROM THIN FILMS
Problem: E 41-23
Problem: E 41-25
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?
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If the wavelength of the incident light is λ = 572 nm,
rays A and B in Fig 41-24 are out of phase by
1.50 λ. Find the thickness d of the film.
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Newton’s Rings
INTERFERENCE FROM THIN FILMS
• Newton's rings are formed due to interference
between the light waves reflected from the
top and bottom surfaces of the air film formed
between a plane and curved surface of large
radi of curvature.
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?
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Newton’s Ring Experiment
Why different colors? for any given difference in path length, the
condition ΔL = (m-1/2)  n might be satisfied for some wavelength but
not for some other. A given color might or might not be present in the
visible image.
A plano-convex lens of large radius of curvature is placed on a plane glass plate to get
an air film of circular symmetry. This set up is placed below a traveling microscope.
The air film is illuminated normally by reflecting the horizontal beam of sodium
light using an inclined glass plate.
The traveling microscope is focused and the Newton’s rings
(bright and dark circular interference fringes) are observed.
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INTERFERENCE FROM THIN FILMS
INTERFERENCE FROM THIN FILMS
Newton’s rings (sample problem 41-5):
Constructive interference
2d = (m - ½) 
(n = 1 for air film)
d


r R  1
R 
R
2
 r
binomial
2


1  r 
d  R  R 1 

  . . .
2 R 


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Substituting d in
2d = (m - ½) 
we get
2
2

 r  
R  R 1  
 
 R  

using
Newton’s rings
1
2
r 

1
2
m  1, 2 , . . .
expansion

m
r2
2R
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
R
(maxima)
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INTERFERENCE FROM THIN FILMS
Problem: E41-33
In a Newton’s ring experiment, the radius of curvature R of
MICHELSON’S INTERFEROMETER
 Light from an extended monochromatic
source P falls on a half-silvered mirror M.
 The incident beam is divided into reflected
the lens is 5.0m and its diameter is 20mm.
(a) How many ring are produced?
and transmitted beams of equal intensity.
 These
(b) How many rings would be seen if the arrangement is
almost in
reflected normally from movable mirror
(M2) and fixed mirror (M1).
(Assume wavelength = 589nm)
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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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travel
perpendicular directions and will be
immersed in water (n = 1.33)?
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two beams
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
For the center of the fringe pattern to
change from bright dark and to bright
again, the path difference between two
beams must change by one
wavelength, which means that mirror
M2 moves through a distance of /2. 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. Thus microscopic
length measurements can be made by
this interferometer.
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Note, that the light rays going to mirror M2 traverse the beam splitter
three times before reaching the observer, whereas the rays going to mirror
M1 traverses it only once.
In order to achieve exact quality of path difference through glass, the
compensator plate of exactly the same thickness , refractive index and at
same inclination as Beam Splitter is introduced between M1 and beam
splitter.
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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?
Experimental set up
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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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QUESTIONS – INTERFERENCE
Draw a schematic plot of the intensity of light in a double-slit
interference against phase-difference (and path-difference).
Explain the term reflection phase-shift.
Obtain the equations for thin-film interference.
Explain the interference-pattern in the case of wedge-shaped
thin-films.
Obtain an expression for the radius of mth order bright ring in
the case of Newton’s rings.
Explain Michelson’s interferometer. Explain how microscopic
length measurements are made in this.
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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 ?
Obtain an expression for the fringe-width in the case of
interference of light of wavelength λ, from a double-slit of slitseparation d.
Explain the term coherence.
Obtain an expression for the intensity of light in double-slit
interference using phasor-diagram.
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