Interference-Aug-2023 ( Students Copy).pdf

8/29/2023
COURSE NAME: ENGINEERING PHYSICS
COURSE CODE : PY1001
LECTURE SERIES NO : 01 (ONE)
CREDITS : 3
MODE OF DELIVERY : ONLINE (POWER POINT PRESENTATION)
COURSE COORDINATORS: DR. NILANJAN HALDER & DR. ANUPAM SHARMA
EMAIL-ID : nilanjan.halder@jaipur.manipal.edu
B.TECH FIRST YEAR
ACADEMIC YEAR: 2023-2024
SESSION OUTCOME “UNDERSTAND THE BASIC
PRINCIPLES OF WAVE OPTICS”
ASSESSMENT CRITERIA
ASSIGNMENT
QUIZ
MID TERM EXAMINATION
END TERM EXAMINATION
INTERFERENCE
Topics
 Two source interference
 Double-slit interference
 Coherence
 Intensity in double slit interference
 Interference from thin film
 Newton’s Rings
Text Book:
PHYSICS VOL 2 by Halliday, Resnick and Krane (5th Edition)
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TWO-SOURCE INTERFERENCE
When identical waves from two coherent 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.
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.
A SECTION OF INFINITE WAVE
A WAVE TRAIN
OF FINITE LENGTH L
COHERENCE
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.
For interference
pattern to occur, the
phase difference at
point on the screen
must not change with
time.
A WAVE TRAIN
OF FINITE LENGTH L
COHERENCE
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.
A WAVE TRAIN
OF FINITE LENGTH L
COHERENCE
Two waves are said to
be coherent when
they are of :
• same frequency
• same phase or are of
a constant phase
difference
Laser light is highly coherent whereas
a laboratory monochromatic light
source (sodium vapor lamp) may be
partially coherent.
N.B.: Amplitude may differ
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Conditions for stationary interference pattern
The sources must be coherent; that is, the
waves must maintain a constant phase with
respect to each other.
 This means: the waves have the same
frequency
The sources should be monochromatic; that is,
the waves should be of a single wavelength.
Complete destructive interference of two waves occur when
their phase difference is , 3 , 5 , … (the waves are 180o out
of phase)
Maximal constructive
interference of two
waves occurs when their
phase difference is 0, 2,
4 , … (the waves are in-
phase)
YOUNG’S DOUBLE SLIT EXPERIMENT
• 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).
DOUBLE-SLIT INTERFERENCE
INTERFERENCE PATTERN PRODUCED BY WATER
WAVES IN A RIPPLE TANK
Maxima: where the shadows show the crests and valleys
Minima: where the shadows are less clearly visible
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A train of plane light waves is incident on two narrow parallel
slits separated by distance d (of the order of ). The interference
pattern on the screen consists of bright and dark fringes. Phase difference (Φ) between the interfering waves:
depends on the location of the point P on the screen
 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 .
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
m = 0, 1, 2, . . .
Which can be written as,
m = 0, 1, 2, . . .



 )
m
(
sin
d 2
1


 )
m
(
b
S 2
1
1
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• For small value of , we can make
following approximation.
• Path difference:



sin
D
y
tan
sin 



𝑑 sin 𝜃 = 𝑆 𝑏 =
𝑦 𝑑
𝐷
mth maximum is located at
ym given by
where m = 0, 1, 2, . . .
d
D
m
y
or
D
y
d
m
m
m




Separation between adjacent maxima
(for small ) is independent of m
d
D
y
d
D
m
d
D
)
1
m
(
y
y
y m
1
m










 
The spacing between the adjacent
minima is same as the spacing
between adjacent maxima.
Problem: SP 41-1
The double slit arrangement is illuminated by light of
wavelength 546nm. The slits are 12mm apart and the
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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Problem: E 41-2
Monochromatic light illuminates two parallel slits a
distance d apart The first maximum is observed at
an angular position of 15°. By what percentage
should d be increased or decreased so that the
second maximum will instead be observed at 15° ?
Problem: E 41-5
A double-slit arrangement produces interference
fringes for sodium light (wavelength = 589 nm) that
are 0.23° apart. For what wavelength would the
angular separation be 10% greater ? Assume that the
angle  is small.
Problem: E 41-8
In an interference experiment in a large ripple tank
(see Fig 41-2) the coherent vibrating sources are
placed 120 mm 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.
Problem: E 41-11
Sketch the interference pattern expected from using two
pin-holes rather than narrow slits.
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Properties of Electromagnetic Waves
• An EM wave has both an electric field and a magnetic field
• The electric and magnetic fields are perpendicular to each other
and oscillate in phase
• The direction of propagation of the wave is perpendicular to both
Light : An Electromagnetic Wave
Phasor: Rotating Vector INTENSITY IN DOUBLE SLIT INTERFERENCE
 Electric field components at P due to S1 and S2 are,
E1= E0 sin ωt & E2= E0 sin (ωt + ) respectively.
 Resultant field E = E1 + E2
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Resultant of E1= E0 sin ωt & E2= E0 sin (ωt + )
Phasor  Rotating vector.
ADDITION OF TWO VECTORS USING PHASORS
Let two vectors be, E1= E0 sin ωt &
E2= E0 sin (ωt + )
Resultant field E = E1 + E2
ωt
E0
E0
ωt + 
E1
E2
INTENSITY IN DOUBLE SLIT INTERFERENCE
Resultant of E1= E0 sin ωt & E2= E0 sin (ωt + )
From phasor diagram,
E = E1 + E2
= E sin(t + )
= 2E0 cos  sin(t + )
But  = /2. So above eqn can be
written as,
E = 2 E0 cos(/2) sin(wt+/2) ωt

E1
E2
E
E


E0
E0
 E = 2 E0 cos(/2) sin(wt+/2)
 So intensity at an arbitrary point P on the screen due to
interference of two sources having phase difference ;





 

2
cos
E
4 2
2
0
I
Ι ∝ 4 Ι cos
𝜑
2
where Ι ∝ E is intensity due to single source
.
.
.
2,
1,
0,
m
where
)
2
1
(m
sin
d
)
1
2
(
:
minima
At
m
sin
d
2
:
maxima
At
equation,
above
From
sin
cos
4
,
/
dsin
2
Since
2
0




































or
m
or
m
d
source
single
to
due
intensity
is
E
where
2
cos
4
2
0
0
2
0







 



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Light intensity (I) versus d sin θ for a double-slit
interference pattern when the screen is far from the
two slits (D>> d).
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)
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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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°)
 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.
A soapy water film on a
vertical loop viewed by
reflected light
INTERFERENCE FROM THIN FILMS
Thickness and color in a thin film
The region ac looks bright or dark for an observer depending
on the path difference between the rays r1 and r2.
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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
junction between two strings of
different linear mass densities.
Optical Path & Geometrical Path
• Distance traveled by light in a medium in the time interval
of ‘t’ is d = vt  Geometrical path
• Refractive index n = c/v
• Hence, ct = nd
• nd  Optical path
• Optical path is the distance traveled by light in vacuum in
same time ‘t’.
• If n is wavelength in the film of refractive index n and  is
the wavelength in vacuum then n =  / n
Equations for Thin Film Interference:
Normal incidence (i = 0)
Path difference = 2 d + (½) n

BACK SURFACE
Constructive interference:
2 d + (½) n = m n m = 1, 2, 3, . . . (maxima)
Destructive interference:
2 d + (½) n = (m+½) n m = 0, 1, 2, . . . (minima)
thin film with one & two phase changes
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8/29/2023
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?
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.
The thickness of the non-reflective coating can be a quarter-wavelength. (t = λ/4) .
WEDGE SHAPED FILM
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
fringes of constant thickness.
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8/29/2023
Air Wedge – the air between two sheets of flat
glass angled to form a wedge
m =1 2 3 4 5
Δ𝑥
Air Wedge – the air between two sheets of flat
glass angled to form a wedge
⇒ 𝑡 = 𝑥𝛼
Now, tan α = t /x
For small α; tan α ≈ α
Minima (destructive)
m
x
2

 

Maxima (constructive)

m
t 
2
2t = (m - ½) 
𝑥 =
𝜆
2𝛼
𝑚 −
1
2
𝜶
x
t
Problem: SP 41-3
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?
Problem: SP 41-4
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)
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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?
Problem: E 41-25
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.
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?
INTERFERENCE FROM THIN FILMS Newton’s Rings
• 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.
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8/29/2023
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.
Newton’s Ring Experiment
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.
Newton’s rings (sample problem 41-5):
Constructive interference
2d = (m - ½)  (n = 1 for air film)
2
1
2
2
2
R
r
1
R
R
r
R
R
d




















R
2
r
.
.
.
R
r
2
1
1
R
R
d
1
R
r
2
2
expansion
binomial
using




















Substituting d in
2d = (m - ½) 
we get
 
(maxima)
.
.
.
,
2
,
1
m
R
m
r 2
1




Newton’s rings
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8/29/2023
Problem: E41-33
In a Newton’s ring experiment, the radius of curvature R of
the lens is 5.0m and its diameter is 20mm.
(a) How many ring are produced?
(b) How many rings would be seen if the arrangement is
immersed in water (n = 1.33)?
(Assume wavelength = 589nm)
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 slit-
separation d.
Explain the term coherence.
Obtain an expression for the intensity of light in double-slit
interference using phasor-diagram.
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.
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