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1 INTERFERENCE
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Page 1: 1 INTERFERENCE 2 Topics Two source interference Double-slit interference Coherence Intensity in double slit interference Interference from thin film.

1

INTERFERENCE

Page 2: 1 INTERFERENCE 2 Topics Two source interference Double-slit interference Coherence Intensity in double slit interference Interference from thin film.

BE-PHYSICS- INTERFERENCE-2010-11 2

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

Page 3: 1 INTERFERENCE 2 Topics Two source interference Double-slit interference Coherence Intensity in double slit interference Interference from thin film.

What is an Electromagnetic wave (EM)?

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.

900

Electric field (E)

Magnetic field (B)

Page 4: 1 INTERFERENCE 2 Topics Two source interference Double-slit interference Coherence Intensity in double slit interference Interference from thin film.

Examples: • Light waves• Heat waves

• Radio and television Waves• Ultraviolet waves

• Gamma rays, X- rays

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’.

Properties of electromagnetic waves (EM)

Page 5: 1 INTERFERENCE 2 Topics Two source interference Double-slit interference Coherence Intensity in double slit interference Interference from thin film.

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

Page 6: 1 INTERFERENCE 2 Topics Two source interference Double-slit interference Coherence Intensity in double slit interference Interference from thin film.

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).

Page 7: 1 INTERFERENCE 2 Topics Two source interference Double-slit interference Coherence Intensity in double slit interference Interference from thin film.

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

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.

Page 8: 1 INTERFERENCE 2 Topics Two source interference Double-slit interference Coherence Intensity in double slit interference Interference from thin film.

The wave nature of light dominates when light interacts with light. f

When light behaves as a Wave?

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

Page 9: 1 INTERFERENCE 2 Topics Two source interference Double-slit interference Coherence Intensity in double slit interference Interference from thin film.

When light behaves as a stream of particles?The particle nature of light dominates when the light interacts with matter (like solids, liquids and gases).

Particle nature - Photoelectric, 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.

Light as a stream of particles

Quantum (bundles/packets of energy)

(Photon/quantum)

Photon energy

E = hfh = Planck’s constant = 6.626x10-34Jsf = frequency of radiation

Page 10: 1 INTERFERENCE 2 Topics Two source interference Double-slit interference Coherence Intensity in double slit interference Interference from thin film.

Light as a wave: c= f

Light as a particle: E = hf photon

Energy of a photon or light wave:

)(c

fhc

hfE

Where h = Planck’s constant = 6.626x10-34Jsf = frequency of a light wave -c = velocity of lightλ= wavelength of a light wave -distance between successive crests

Page 11: 1 INTERFERENCE 2 Topics Two source interference Double-slit interference Coherence Intensity in double slit interference Interference from thin film.

Visible Light400–435 nm Violet435 nm-440nm Indigo440–480 nm Blue480–530 nm Green530–590 nm Yellow590–630 nm Orange630–700 nm 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.

Page 12: 1 INTERFERENCE 2 Topics Two source interference Double-slit interference Coherence Intensity in double slit interference Interference from thin film.

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

f2T

2

Page 13: 1 INTERFERENCE 2 Topics Two source interference Double-slit interference Coherence Intensity in double slit interference Interference from thin film.

13

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+)

Page 14: 1 INTERFERENCE 2 Topics Two source interference Double-slit interference Coherence Intensity in double slit interference Interference from thin film.

14

INTERFERENCE OF LIGHT

Page 15: 1 INTERFERENCE 2 Topics Two source interference Double-slit interference Coherence Intensity in double slit interference Interference from thin film.

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. 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.

OR

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16

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).

Page 17: 1 INTERFERENCE 2 Topics Two source interference Double-slit interference Coherence Intensity in double slit interference Interference from thin film.

Constructive InterferenceDestructive 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

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BE-PHYSICS- INTERFERENCE-2010-11

18

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

TWO-SOURCE INTERFERENCE

MIT-MANIPAL

Page 19: 1 INTERFERENCE 2 Topics Two source interference Double-slit interference Coherence Intensity in double slit interference Interference from thin film.

Phase: Phase of a vibrating particle at any instant indicates its state of vibration.

Phase may be expressed in terms of angle as a fraction of 2π.

PHASE AND PATH DIFFERENCE

F

CA G

λ

λB

π/2 π 3π/2

D

E

2π O

Path difference corresponds to phase difference of 2.

differencePath

2

differencePhase

λ/4λ/2

3λ/4λ

t=0

Page 20: 1 INTERFERENCE 2 Topics Two source interference Double-slit interference Coherence Intensity in double slit interference Interference from thin film.

Constructive interference

path difference p= 0 or phase difference = 0

path difference p = 1or phase difference =2

path difference p =2 or phase difference = 4

General condition:

Path difference p = m

or phase difference = 2m

where m = 0, 1, 2, 3,…………order of interference.

Page 21: 1 INTERFERENCE 2 Topics Two source interference Double-slit interference Coherence Intensity in double slit interference Interference from thin film.

Maximal constructive interference of two waves occurs when their: path difference between the two waves is a whole number multiple of wavelength.

ORPhase difference is 0, 2, 4 , … (the waves are in-phase).

Constructive interference

Page 22: 1 INTERFERENCE 2 Topics Two source interference Double-slit interference Coherence Intensity in double slit interference Interference from thin film.

22

path difference p= /2 phase difference = 1

path difference p = 3/2 or phase difference =3

DISTRUCTIVE INTERFERENCE

path difference p = 5/2phase difference = 5

General condition:

Path difference p=(m+1/2) or

phase difference =(2m+1)

where m = 0, 1, 2, 3, ………

Page 23: 1 INTERFERENCE 2 Topics Two source interference Double-slit interference Coherence Intensity in double slit interference Interference from thin film.

23

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).

Page 24: 1 INTERFERENCE 2 Topics Two source interference Double-slit interference Coherence Intensity in double slit interference Interference from thin film.

24

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

COHERENCE

Page 25: 1 INTERFERENCE 2 Topics Two source interference Double-slit interference Coherence Intensity in double slit interference Interference from thin film.

25

A SECTION OF INFINITE WAVE

A WAVE TRAIN OF FINITE LENGTH L

COHERENCE

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.

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.

Page 26: 1 INTERFERENCE 2 Topics Two source interference Double-slit interference Coherence Intensity in double slit interference Interference from thin film.

BE-PHYSICS- INTERFERENCE-2010-11 26

A SECTION OF INFINITE WAVE

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.

MIT-MANIPAL

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BE-PHYSICS- INTERFERENCE-2010-11 27

COHERENCE

Common sources of visible light emit light wave trains of finite length rather than an infinite wave.MIT-MANIPAL

Laser light is highly coherent whereas a laboratory monochromatic light source (sodium vapor lamp) may be partially coherent.

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28

(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 area. Newton’s rings, b. Michelson’s

interferometer.

COHERENCE

(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.

Methods of producing coherent sources:

Page 29: 1 INTERFERENCE 2 Topics Two source interference Double-slit interference Coherence Intensity in double slit interference Interference from thin film.

29

DOUBLE SLIT INTERFERENCE

Screen

Two narrow slits (can be considered as two sources of coherent light waves).

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.

d

If light waves did not spread out after passing through the slits, no interference would occur..

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A monochromatic light source 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.

DOUBLE SLIT INTERFERENCE

B C

Page 31: 1 INTERFERENCE 2 Topics Two source interference Double-slit interference Coherence Intensity in double slit interference Interference from thin film.

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.

When the light from the two slits combine destructively at any location on the screen, a dark fringe results.

B C

DOUBLE SLIT INTERFERENCE

Page 32: 1 INTERFERENCE 2 Topics Two source interference Double-slit interference Coherence Intensity in double slit interference Interference from thin film.

3232

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.

O

DOUBLE SLIT INTERFERENCE

Page 33: 1 INTERFERENCE 2 Topics Two source interference Double-slit interference Coherence Intensity in double slit interference Interference from thin film.

- Analysis of Interference pattern

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

a-is the mid point of the slit

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34

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).

DOUBLE SLIT INTERFERENCE

Page 35: 1 INTERFERENCE 2 Topics Two source interference Double-slit interference Coherence Intensity in double slit interference Interference from thin film.

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 .

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.

(d sinθ)

DOUBLE SLIT INTERFERENCE

Page 36: 1 INTERFERENCE 2 Topics Two source interference Double-slit interference Coherence Intensity in double slit interference Interference from thin film.

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.

Each maximum above has a symmetrically located maximum below O; these correspond to m= -1, -2, -3……

Central maximum m=0

Page 37: 1 INTERFERENCE 2 Topics Two source interference Double-slit interference Coherence Intensity in double slit interference Interference from thin film.

Destructive Interference: Minimum at P: 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

........................2,1,0m

)ima..(min..........)2

1m(sind

Central maximum m=0m=0m=1

m=2m=3

m=3m=2m=1m=0

Negative values of m locate the minima on the lower half of the screen.

P

O

Page 38: 1 INTERFERENCE 2 Topics Two source interference Double-slit interference Coherence Intensity in double slit interference Interference from thin film.

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

Page 40: 1 INTERFERENCE 2 Topics Two source interference Double-slit interference Coherence Intensity in double slit interference Interference from thin film.

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

d=0.12 mm

a) θ=? for first minimum.

b) Fringe width Δy=?

DOUBLE SLIT INTERFERENCE

Page 41: 1 INTERFERENCE 2 Topics Two source interference Double-slit interference Coherence Intensity in double slit interference Interference from thin film.

mm5.2m10x12.0

)m10x55)(m10x546(

d

Dy)b(

13.0radian0023.0sinthatsmallissin

radian0023.0)m10x12.0(

)m10x546(x)21

(

d

)21

m(sin)

2

1m(sind

)0m(imumminfirsttheAt)a(:Solution

3

29

0

3

9

CHECK YOURSELFSolve for First maximum : use d sinθ=mλ put m=1

DOUBLE SLIT INTERFERENCE

Page 42: 1 INTERFERENCE 2 Topics Two source interference Double-slit interference Coherence Intensity in double slit interference Interference from thin film.

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 (θ is very small) d θ=λ (for first maxima , m=1)

Use: λ1/ λ2= θ 1/ θ 2, λ2= λ1 x θ 2/ θ 1,

Given: 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

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43

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.

DOUBLE SLIT INTERFERENCE

Page 44: 1 INTERFERENCE 2 Topics Two source interference Double-slit interference Coherence Intensity in double slit interference Interference from thin film.

Solution: Use d sinθ=mλd1 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:2Monochromatic 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?

DOUBLE SLIT INTERFERENCE

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45

Tutorial: Problem 8 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 mD = 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.

DOUBLE SLIT INTERFERENCE

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46

HRK:958 page: Exercise: 41-2Problem 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)

DOUBLE SLIT INTERFERENCE

Page 47: 1 INTERFERENCE 2 Topics Two source interference Double-slit interference Coherence Intensity in double slit interference Interference from thin film.

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.

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 nmd = 1.2 mmD= 5.4 mFringe width Δy=?

DOUBLE SLIT INTERFERENCE

Page 48: 1 INTERFERENCE 2 Topics Two source interference Double-slit interference Coherence Intensity in double slit interference Interference from thin film.

Use:d =λ/sin θ = (592 x 10-9m)/sin(1.000) =3.39x 10-5m

CHECK YOURSELFSolve this problem for dark interference fringes 1.000 apart in angular separation.

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

DOUBLE SLIT INTERFERENCE

Page 49: 1 INTERFERENCE 2 Topics Two source interference Double-slit interference Coherence Intensity in double slit interference Interference from thin film.

Problem 6A 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/ngives θ=0.150

DOUBLE SLIT INTERFERENCE

Page 50: 1 INTERFERENCE 2 Topics Two source interference Double-slit interference Coherence Intensity in double slit interference Interference from thin film.

Problem 7

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

Page 51: 1 INTERFERENCE 2 Topics Two source interference Double-slit interference Coherence Intensity in double slit interference Interference from thin film.

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 used?

m

Q

O

ym+1So with small angle approximation:

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

DOUBLE SLIT INTERFERENCE

Page 52: 1 INTERFERENCE 2 Topics Two source interference Double-slit interference Coherence Intensity in double slit interference Interference from thin film.

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 )

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53

CoherenceProblem 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/ timex = c/t = (3.0 x108m/s)/1 x 108s) = 3 m.(b) No.

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54

Thomas Young (1773–1829)

YOUNG’S DOUBLE SLIT INTERFERENCE

• 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, producing the

interference pattern.

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55

INTENSITY IN DOUBLE SLIT INTERFERENCEIn 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 E2Let 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 + )

P

O

E0- amplitude of each wave.

Page 56: 1 INTERFERENCE 2 Topics Two source interference Double-slit interference Coherence Intensity in double slit interference Interference from thin film.

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

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57

INTENSITY IN DOUBLE SLIT INTERFERENCEAdding Wave disturbances: Phasors

The combined electric field can be done algebraically, using a graphical method, which proves to be convenient in more complicated situations.

The sinusoidal wave can be represented graphically by a Phasor of magnitude Eo rotating about the origin counterclockwise with an angular frequency .

w

Phasor diagram

E1(=E0 sinwt)

w

Page 58: 1 INTERFERENCE 2 Topics Two source interference Double-slit interference Coherence Intensity in double slit interference Interference from thin film.

58

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 (t + )

It has the same amplitude and frequency as E1.

Its phase is with respect to E1.

(b)

w

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59

(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 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.

ωt

E1

E2

E

E

E0

E0

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60

ωt

E1

E2

E

E

E0

E0

A B

C

)wtsin(EE

E

E

AC

BC)wtsin(

From the right angled triangle ABC

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61

)wtsin(EE

Where E- The resultant electric fieldEθ- Amplitude of the resultant wave β – the phase difference between amplitude of the resultant wave and first wave.

β

β φ

E0

E0

From the geometry of the right angled triangle

cosE2E

E

)2/E(cos

0

0

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,

2 β= φ or β =φ/2

Amplitude Wave part

Phaseshift

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62

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 Eθ

2 (for resultant wave) and I0 α E0

2 (Intensity of each single wave)

Iθ = 4 E02 cos2 β = 4 E0

2 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 wave.

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63

Intensity distribution in double slit interference

Iθ = 4 I0 cos2 ф/2Intensity Maxima occur where cos2ф/2 =1, or Ф =0 , 2π , 4π,…………..2m π ф= 2mπ (maxima) Or the path difference: d sinθ= m λ m= 0, ±1, ±2……………(maxima)

Iθ = 4 I0 cos2 ф/2Intensity Minima occur where cos2ф/2 =0, or Ф =π , 3π, 5 π …………..(2m+1) π ф= (2m+1)π (minima) Or the path difference: d sinθ= (m+1/2) λ m= 0, ±1, ±2……… (minima)

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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.

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65

The phase difference (φ) between r1 and r2

associated with the path difference S1b (= d sinθ)

)sind

(cosE

E

I

I:ensityintlativeRe

)sind

(cosI4I

cosI4I

aswrittenbecananyatensityintThe

sind

2or

)sind(2sind

2

differencePath

2

differencePhase

2

2

0

2

0

2

0

2

0

INTENSITY IN DOUBLE SLIT INTERFERENCEPHASE AND PATH DIFFERENCE

Page 66: 1 INTERFERENCE 2 Topics Two source interference Double-slit interference Coherence Intensity in double slit interference Interference from thin film.

Where Eθ =2 E0 cos β: Amplitude of the resultant wave.

‘β’ – the phase difference between resultant wave and first wave. β = ф/2

E = E1+E2 = Eo (sin ωt + sin (ωt+))Using Trigonometric Identity:

2

1sin

2

1cos2sinsin

With = (t + ), = t, get:

)2/wtsin(2/cos2EE

2/)wtwtsin(2/)wtwtcos(2EE

0

0

It can be written as, E= Eθ

sin(wt+β)

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BE-PHYSICS- INTERFERENCE-2010-11

67

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)MIT-MANIPAL

INTENSITY IN DOUBLE SLIT INTERFERENCE

E=E1+E2+E3

+E4

E=Eθ

sin(wt+β)

The phase angle φ between successive phasors is 150.

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68

In any closed n-sided polygon, the sum of the interior angles is (n-2)π.

So, in the five sided polygon, abcdea:

So, 2β=540 - 3(1650)

So, β = 22.50

To find β:

0

0

165dcb

540180)25(

)2n(

dcb2

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69

ax1= ab cosβ

x1x2= bc (cosβ-15)

x2x3= cd (cosβ-15)

x3e= de (cosβ)

x1b

c

x2β-15

x3

β

β-15

To find Eθ:

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70

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)

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71

Alternate method: Verification:

Ey- sum of

verical- com

ponents

Eh- sum of horizontal components

cos wtsin w

Standard equation:E(t)= Eθ sin (wt+β)

2

v

2

h EEEWhere

We can use the phasors to find sum and free to evaluate the phasors at any time t. We choose t=0.

01

h

v1

0

00000v

00000h

5.22)53.3

46.1(tan)

E

E(tan:anglePhase

E83.3E

E46.145sinE30sinE15sinE0sinEE:comonentVertical

E53.345cosE30cosE15cosE0cosEE:ComponentHorizontal

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72

Sample Problem: Intensity in double-slit InterferenceSource 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.

They are in phase and the reach the detector.

Problem: E 41-15

rA

rB

100 m=λ/4 corresponds to π/2 phase

Initially, source A leads source B by 90◦, = 1/4 wavelength (100 m)

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.

DETECTOR

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73

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.

Problem: E 41-16

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74

Tutorial Problem: E 41-18Find the sum of the following quantities (a) graphically, using phasors; and (b) using trigonometry.

y1=10 sin wt, y2=8.0 sin (wt+300)

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.

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75

To find Eθ:

E

sinE

AC

DCsinanglephasethe&

EcosEE2E(E

)sinE()cosEE(

DC)BDAB(E

DCADAC

:ACBtriangleangledrighttheFrom

02

2

020201

2

01

202

20201

222

222

Substitute E01=10, E02=8, & φ=300

Eθ=17.39β= 13.30

The resultant electric field E= 17.39 sin (wt+13.3)

(a) Phasors

βE01

E02

A B D

C

y

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76

Alternate method: Trigonometry

Ey- sum of

verical- com

ponents

Eh- sum of horizontal components

cos wtsin w

Standard equation:E(t)= Eθ sin (wt+β)

2

v

2

h EEEWhere

We can use the phasors to find sum and free to evaluate the phasors at any time t. We choose t=0.

)3.13wtsin(39.17E

3.13)E/E(tan

39.17E

44030sin80sin10E

92.1692.61030cos80cos10E

0

hv

1

v

h

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77

Reflection causes a /2 l phase shift

Reflection causes no phase shift

REFLECTION PHASE SHIFT

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.

If the medium beyond the interface has a lower index of refraction, there is no phase/path length change of the reflected wave.

Incident wave

Reflected wave

Incident wave

Reflected wave

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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

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79

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.

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80

d

n1

n1

n2

Source of light

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 surface (r1) and back

surface (r2) interfere and

enter the eye.

INTERFERENCE FROM THIN FILMS

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d

n1

n1

n2

Source of light

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’. It then emerges from the front surface of the film at ‘c’ and also enters the eye, as ray r2.

d- thickness of the film.

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82

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.

d

n1

n1

n2

Source of light

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83

d

n1

n1

n2

Source of light

As r1 and r2 have travelled over

paths of different lengths, have

traversed different media, and

have suffered different kinds of

reflections at ‘a’ and ‘b’.

The phase difference between

two reflected rays r1 and r2

determine whether they

interfere constructively or

destructively.

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84

If a wave travelling from

a medium of index of

refraction ‘n1’ toward a

medium of index of

refraction ‘n2’ undergoes

a phase change 1800

upon reflection when

n2>n1 and undergoes no

phase change if n2<n1.

EQUATIONS FOR THIN –FILM INTERFERENCE

d

n1

n1

n2

Source of light

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85

λ

nn

The wavelength of

light ‘λn’ in a medium

whose index of

refraction is ‘n’

nn

λ - wavelength of the light in free space.

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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 twice

through the film before

reaching the eye.

d

n1

n1

n2

Source of light

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87

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.

d

n1

n1

n2

Source of light

Page 88: 1 INTERFERENCE 2 Topics Two source interference Double-slit interference Coherence Intensity in double slit interference Interference from thin film.

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.

? ?

A general equation for Thin film Interference:

Page 89: 1 INTERFERENCE 2 Topics Two source interference Double-slit interference Coherence Intensity in double slit interference Interference from thin film.

d

Air Air n

1800 phase change

No phase change

Examples: (a) Interference in a thin soap film

For the interference from a

thin soap film of index of

refraction ‘n’ surrounded

by air, we must add the

extra half wavelength for

the front surface reflection,

but not for the back surface .

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90

The total path difference: = 2d + λn/2 =(m+1/2) λn ……… (minima)

where m=0,1,2,3…..

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.

nn

Interference in a thin soap film

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.

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91

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.

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92

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.

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).

wired

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Ray BC reflected from the top –NO phase change.

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.

Back surface

i

r

Page 94: 1 INTERFERENCE 2 Topics Two source interference Double-slit interference Coherence Intensity in double slit interference Interference from thin film.

Constructive InterferenceThe 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………………………………

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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?

No phase change

d

Air Airn=1.33

1800 phase change

No phase change

)ima(maxm2

d2 nn

Solution:The wavelengths which are maximally reflected are constructively interfered.

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96

)21

m(

nm851

)21

m(

nd2

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.

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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=?

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98

Solution:Light strikes the lens at near-normal incidence (θ).

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)

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nm10038.1x4

nm550

n4n2

)21

m(d

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

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100

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

INTERFERENCE FROM THIN FILMS

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101

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).

d=460nm

White light (400 nm-700nm)

Normal incidenceReflected light (λ=?)

R.I=1.0

R.I=1.33

n=1.2 Thin film of oil

water(Both Front & back surface reflections have phase change) .

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102

The total path difference for maxima:Path difference = 2d + λn/2+ λn/2 = mλn , Where (m=1,2,3….)

2d=(m-1) λ/n (λn=λ/n) λ=2nd/(m-1)

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

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103

The wavelengths which are minimally reflected are maximally transmitted, and vice versa. Maximally transmitted wavelengths is the same as finding the minimally reflected wavelengths.

The total path difference for

minima:

Path difference

= 2d + λn/2+ λn/2 =

(m+1/2)λn

where (m=1,2,3….)

substitute : (λn=λ/n)

λ=2nd/(m-1/2)

Maximum Transmitted

light =?

R.I=1.0

R.I=1.33

n=1.2 Thin film of oil

water

Minimally reflectedlight

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104

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)

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105

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 possibleTake m=1

d=λ/2n=215nm

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106

ORIn 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=2π2d=λn

d=λ/2n=572 nm/2 x 1.33=215 nm

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107

Problem: E 41-29A 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?

MIT-MANIPAL

INTERFERENCE FROM THIN FILMS

d=0.048 x 10-3 m)

λ=680 x10- 9m

θ

x=120 mm

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108

.

14167.1412

1

10x680

)10x048.0(x1x2

2

1nd2m

?m

m2

d2

:whenoccurserferenceintveConstructi

9

3

nn

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BE-PHYSICS- INTERFERENCE-2010-11

109

Problem: E 41-27A 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?

MIT-MANIPAL

INTERFERENCE FROM THIN FILMS

Thin film n=1.25

Glass plate 1.50

Put: λn=λ/n

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110

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

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111

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?Glass=1.5Film= 1.25

Air=1

Page 112: 1 INTERFERENCE 2 Topics Two source interference Double-slit interference Coherence Intensity in double slit interference Interference from thin film.

112

Both the reflected rays r1(front surface

reflection) and r2(back surface reflection)

have additional path difference (λ/2).

)ima(min)2

1m(

22d2 n

nn

nm12425.1x4

nm620

n4n2

)21

m(d

m=1

Page 113: 1 INTERFERENCE 2 Topics Two source interference Double-slit interference Coherence Intensity in double slit interference Interference from thin film.

113

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?

No phase change

d=410 nm

Air Airn=1.50

1800 phase change

Page 114: 1 INTERFERENCE 2 Topics Two source interference Double-slit interference Coherence Intensity in double slit interference Interference from thin film.

114

)ima(maxm2

d2 nn

)21

m(

nm1230

)21

m(

)nm410)(5.1)(2(

)21

m(

nd2

The result is only in the visible range when m = 3, so λ= 492 nm.

Page 115: 1 INTERFERENCE 2 Topics Two source interference Double-slit interference Coherence Intensity in double slit interference Interference from thin film.

115

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?

Glass=n=1.5Film=n= 2.0Air=n=1

Page 116: 1 INTERFERENCE 2 Topics Two source interference Double-slit interference Coherence Intensity in double slit interference Interference from thin film.

116

)ima(maxm2

d2 nn

nm702x4

nm560

n4n2

)21

m(d

Page 117: 1 INTERFERENCE 2 Topics Two source interference Double-slit interference Coherence Intensity in double slit interference Interference from thin film.

117

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.No phase change

d=0.00121 mm

Air Airn=1.33

1800 phase change

Page 118: 1 INTERFERENCE 2 Topics Two source interference Double-slit interference Coherence Intensity in double slit interference Interference from thin film.

118

6.00 m)10xm)/(58521x102(1.33)(1.2

1nd2m

)ima(maxm2

d2

9-6-

nn

So the interference is NOT dark.

5.5 m)10xm)/(58521x102(1.33)(1.nd2

m

)ima(min)2/1m(2

d2

9-6-

nn

m should be an integer.

So the interference is bright.

Page 119: 1 INTERFERENCE 2 Topics Two source interference Double-slit interference Coherence Intensity in double slit interference Interference from thin film.

119

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?

No phase change

d=?

Air Airn=1.33

1800 phase change

Page 120: 1 INTERFERENCE 2 Topics Two source interference Double-slit interference Coherence Intensity in double slit interference Interference from thin film.

120

With First source (λ=600nm):

)1...(..........m600300nd2

m6002

600nd2

)ima(maxm2

d2 nn

2)450600(2

600mgetWe

:)2(by)1(eqnDividing

)2(..........450)2

1m(225nd2

)ima(min)2

1m(

2d2 n

n

With second source (λ=450nm):

Use equation (1) or (2) to solve for d:

From equation (2)d= 2 x 450/2 x1.33=338 nm

Page 121: 1 INTERFERENCE 2 Topics Two source interference Double-slit interference Coherence Intensity in double slit interference Interference from thin film.

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 d2x

B10B1 B2 B3 B4 B5 B6 B7 B8 B9

m1=1 m2=10

Page 122: 1 INTERFERENCE 2 Topics Two source interference Double-slit interference Coherence Intensity in double slit interference Interference from thin film.

122

m 1.89 (1.50)nm)/2 9(630 /2n 9)dd(

gives)2(eqn)2(Eqn

)2..(..........)2

1m(nd2&

)1.....()2

1-(m2nd Then,

10m be bandbright last the&

1m ofvalueminimum bandbright first for theLet

)2

1m(nd2m

2nd2m

2d2

:whenoccurserferenceintveConstructi

12

22

11

nn

nn

Number of bright bands in ‘x’ mm length. 10 bright & 9 dark bands. Film thickness over the length d2-d1=?

Page 123: 1 INTERFERENCE 2 Topics Two source interference Double-slit interference Coherence Intensity in double slit interference Interference from thin film.

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.

INTERFERENCE FROM THIN FILMS

Page 124: 1 INTERFERENCE 2 Topics Two source interference Double-slit interference Coherence Intensity in double slit interference Interference from thin film.

124

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.

Page 125: 1 INTERFERENCE 2 Topics Two source interference Double-slit interference Coherence Intensity in double slit interference Interference from thin film.

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-to-glass 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 .

Page 126: 1 INTERFERENCE 2 Topics Two source interference Double-slit interference Coherence Intensity in double slit interference Interference from thin film.

126

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.

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).

Air-Glass interfacePhase change (π)

Page 127: 1 INTERFERENCE 2 Topics Two source interference Double-slit interference Coherence Intensity in double slit interference Interference from thin film.

The condition for constructive interference remains unchanged.Path difference:

)1......(..........)2

1m(d2

1nfilmair

10cos,0rincidencenormalFor

m2

d2

Page 128: 1 INTERFERENCE 2 Topics Two source interference Double-slit interference Coherence Intensity in double slit interference Interference from thin film.

128

Consider the section AB of the lens, wherein it forms an air film of thickness d. Let mth order bright ring of radius ‘r‘ forms here.

Let C be the center of curvature of the plano- convex lens.

R(=CB=OC) be the radius of curvature of the plano-convex lens.

The radius (r) of the bright ring:

O

C

BA D

R

Page 129: 1 INTERFERENCE 2 Topics Two source interference Double-slit interference Coherence Intensity in double slit interference Interference from thin film.

129R2

r.....

R

r

2

11RRd

or,termstwoonlykeeping

,theorembinomialthebybracketsquaretheandexpcanwe

,1R/rIf

22

We can write: d= OD=OC-CD

O

C

BAD

R

2

12

22

R

r1RR

rRRd

Page 130: 1 INTERFERENCE 2 Topics Two source interference Double-slit interference Coherence Intensity in double slit interference Interference from thin film.

Substituting ‘d’ value in equation (1) and solving for ‘r’:

r2D

)2

1m(Rr

…….(For bright ring)

The radii/diameters of the bright rings.

gives)1nairfor()2

1m()

R2

r(2

)1......(..........)2

1m(d2

2

Page 131: 1 INTERFERENCE 2 Topics Two source interference Double-slit interference Coherence Intensity in double slit interference Interference from thin film.

131

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.

Page 132: 1 INTERFERENCE 2 Topics Two source interference Double-slit interference Coherence Intensity in double slit interference Interference from thin film.

132

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 nmRadius of curvature R= 5 mdiameter of the ring=d=20mmTherefore, radius r= 10 mm=0.01 m(a) m=? (b) n=1.33, m=?

Page 133: 1 INTERFERENCE 2 Topics Two source interference Double-slit interference Coherence Intensity in double slit interference Interference from thin film.

133

452

1

)m5(x)nm443(

)01.0( m

be now willrings ofnumber theso

)n

( nm 443 (1.33)/nm) (589

h to wavelengt thechangeseectively in water apparatus thePutting(b)

observed. rings ofnumber theis342

1

10x589x5

01.0

2

1

R

rm

)2

1m(Rr)a(

2

n

9

22

Page 134: 1 INTERFERENCE 2 Topics Two source interference Double-slit interference Coherence Intensity in double slit interference Interference from thin film.

134

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 = ?

Page 135: 1 INTERFERENCE 2 Topics Two source interference Double-slit interference Coherence Intensity in double slit interference Interference from thin film.

135

251271

421

r

rn

3eqn2eqnDivide

32

110Rr

22

110Rr

2

1mRr

1n

nei

nknowWe

2

2

liquid

air

liquid

air

liquidliquid

airair

liquid

air

liquid

liquid

air

liquidn

..

.)(

:)(/)(

).....()(

).(..........)()(

)......(..........?

..:

Page 136: 1 INTERFERENCE 2 Topics Two source interference Double-slit interference Coherence Intensity in double slit interference Interference from thin film.

136

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 nmRadius of the nth ring=rn

=0.162cm

Radius of the (n+20)th

=rn+20=0.368cm m=? Radius of curvature R= ?

Page 137: 1 INTERFERENCE 2 Topics Two source interference Double-slit interference Coherence Intensity in double slit interference Interference from thin film.

137

1.00m nm) 0.5)(546- (5.308cm) (0.162

1/2)-(n /r RThen

5.308. m of value theusing Rfor solve

308.5)cm162.0(

)cm368.0(

)5.0n(

)5.19n(

getwe,anotherbyoneDivide

)2....(..........R)2

1-20(n)cm368.0(

)1........(..........R)2

1n(cm) (0.162

)2

1m(Rr:ringbrightFor

2

2

2

Page 138: 1 INTERFERENCE 2 Topics Two source interference Double-slit interference Coherence Intensity in double slit interference Interference from thin film.

BE-PHYSICS- INTERFERENCE-2010-11

138

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.

MIT-MANIPAL

INTERFERENCE FROM THIN FILMS

d

n

Page 139: 1 INTERFERENCE 2 Topics Two source interference Double-slit interference Coherence Intensity in double slit interference Interference from thin film.

139

PurposeInterferometers 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.

MICHELSON’S INTERFEROMETER

Page 140: 1 INTERFERENCE 2 Topics Two source interference Double-slit interference Coherence Intensity in double slit interference Interference from thin film.

BE-PHYSICS- INTERFERENCE-2010-11

140

MICHELSON’S INTERFEROMETER

Light from an extended monochromatic

source P falls on a half-silvered mirror

M.

The incident beam is divided into

reflected and transmitted beams of

equal intensity (Division of amplitude).

These two beams travel almost in

perpendicular directions and will be

reflected normally from movable

mirror (M2) and fixed mirror (M1).MIT-MANIPAL

B

A

G

Page 141: 1 INTERFERENCE 2 Topics Two source interference Double-slit interference Coherence Intensity in double slit interference Interference from thin film.

BE-PHYSICS- INTERFERENCE-2010-11 141

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).

MIT-MANIPAL

MICHELSON’S INTERFEROMETER

Page 142: 1 INTERFERENCE 2 Topics Two source interference Double-slit interference Coherence Intensity in double slit interference Interference from thin film.

BE-PHYSICS- INTERFERENCE-2010-11 142

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.

MIT-MANIPAL

MICHELSON’S INTERFEROMETER

Page 143: 1 INTERFERENCE 2 Topics Two source interference Double-slit interference Coherence Intensity in double slit interference Interference from thin film.

BE-PHYSICS- INTERFERENCE-2010-11 143

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.

MIT-MANIPAL

MICHELSON’S INTERFEROMETER

Page 144: 1 INTERFERENCE 2 Topics Two source interference Double-slit interference Coherence Intensity in double slit interference Interference from thin film.

144

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.

Page 145: 1 INTERFERENCE 2 Topics Two source interference Double-slit interference Coherence Intensity in double slit interference Interference from thin film.

145

d2

N N)d(2

Michelson interferometer

equation

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.

Page 146: 1 INTERFERENCE 2 Topics Two source interference Double-slit interference Coherence Intensity in double slit interference Interference from thin film.

146

Applications

Determination of wavelength

The fact that whenever the movable mirror moves by

a fringe originates or vanishes at the center is used

to determine from the equation 2d = N , where d is

the distance moved and N, the number of fringes

originated or vanished.

,2

Page 147: 1 INTERFERENCE 2 Topics Two source interference Double-slit interference Coherence Intensity in double slit interference Interference from thin film.

147

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

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.

d

Page 148: 1 INTERFERENCE 2 Topics Two source interference Double-slit interference Coherence Intensity in double slit interference Interference from thin film.

BE-PHYSICS- INTERFERENCE-2010-11

148

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?

MIT-MANIPAL

MICHELSON’S INTERFEROMETER

The number of fringes is the same as the number of half wavelengths in 1.0000 cm.

Nλ= 2dN= 2 d/λ=2(1.0000 x10-2 m)/(589 x

10-9m)= 33,956 fringes

Page 149: 1 INTERFERENCE 2 Topics Two source interference Double-slit interference Coherence Intensity in double slit interference Interference from thin film.

BE-PHYSICS- INTERFERENCE-2010-11

149

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?

MIT-MANIPAL

MICHELSON’S INTERFEROMETER

Nλ= 2dλ= 2 d/N=2(0.233 mm)/792= 588 nm

= 588 nm

Page 150: 1 INTERFERENCE 2 Topics Two source interference Double-slit interference Coherence Intensity in double slit interference Interference from thin film.

BE-PHYSICS- INTERFERENCE-2010-11

150

Problem: E41-40An 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.MIT-MANIPAL

MICHELSON’S INTERFEROMETER

Page 151: 1 INTERFERENCE 2 Topics Two source interference Double-slit interference Coherence Intensity in double slit interference Interference from thin film.

151

Answer: The change in the optical path length is

2(nd – d)2d(n-1)=mλ

n=mλ/2d+1=1.00030

Page 152: 1 INTERFERENCE 2 Topics Two source interference Double-slit interference Coherence Intensity in double slit interference Interference from thin film.

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) = md = mλ/ 2(n-1) d = 7 x (589 x10-9 m)/ 2(1.41-1) = 4.9 x 10-6 m

Page 153: 1 INTERFERENCE 2 Topics Two source interference Double-slit interference Coherence Intensity in double slit interference Interference from thin film.

BE-PHYSICS- INTERFERENCE-2010-11

153MIT-MANIPAL

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 slit-separation d. [5] Explain the term coherence. [2] Obtain an expression for the intensity of light in double-slit interference using phasor-diagram. [5]

Page 154: 1 INTERFERENCE 2 Topics Two source interference Double-slit interference Coherence Intensity in double slit interference Interference from thin film.

BE-PHYSICS- INTERFERENCE-2010-11

154MIT-MANIPAL

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]

Page 155: 1 INTERFERENCE 2 Topics Two source interference Double-slit interference Coherence Intensity in double slit interference Interference from thin film.

BE-PHYSICS- INTERFERENCE-2010-11

155MIT-MANIPAL

INTERFERENCE – ANSWERS TO PROBLEMS

SP41-1: 0.12degree, 2.5mmE41-1: 0.22radian, 12 degreesE41-5: 650nmE41-9: 600nmSP41-2: E(t)= 3.83 Eo sin(t+22.5o )E41-15: 0 degreeE41-19: 1.21, 2.22, 8.13mSP41-3: 567nm (Yellow-green)SP41-4: 100nmE41-23: 552nm, 442nmE41-27: 840nmE41-29: 141E41-33: 34, 45SP41-6: 33956 fringesE41-39: 588nm


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