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Chapter 9 InterferenceFebruary 16 General considerations of interference

9.1 General considerationsIntroduction: Wave equation Superposition principleInterference: The interaction between two or more waves produces in space a resultant irradiance which is different from the sum of the component irradiances:E = E1+E2, but I I1+ I2.Interferometric devices: wavefront splitting and amplitude splitting.

Poynting vector: is the instantaneous power flow across an unit

area whose normal is parallel to S.0

02

BE

BES

c

For a harmonic, linearly polarized plane wave:

)(cos)cos( ),cos( 2000

200 ωtcωtωt rkBESrkBBrkEE

Irradiance (intensity): The time-averaged energy transport per unit area per unit time.

ce.interferenlight studying in 2

1 assume and , drop thenWe

2

1||

2

1)(cos||

20

20

200000

22000

2

EIc

EccωtcSI

T

TT

E

BErkBE

2

Superposition of two planar, linearly polarized waves (vector model):

)cos()2cos(2

1

)cos()cos(

221122110201

2211020121

rkrkrkrkEE

rkrkEEEE

t

tt

cos)cos(

)cos(2

1

02012211020112

2211020121

EErkrkEE

rkrkEEEE

I

T

2211 rkrkPhase difference

1221

2122

21

2

2122

212121

2

22022

11011

2

2)()(

)cos(),(

)cos(),(

IIII

tt

tt

TTTT

EEEEE

EEEEEEEEE

rkErE

rkErE

interference term

1) When E01 and E02 are perpendicular, I12=0, no interference.2) When E01 and E02 are parallel,

cos2cos 21020112 IIEEI

cos2 2121 IIIII Note that cos averages to 0 in space: energy is conserved.

3

Total constructive interference:

Total destructive interference:

mIIIII 2 ,2 2121max

)12( ,2 2121min mIIIII

When I1=I2=I0, 2cos4)cos1(2 2

00

III

Interference fringes between two plane-waves:Bright fringes:

0constant)()()(

2)(

212121

2121

zkkykkxkk

m

zzyyxx

rkk

Fringes are parallel planes perpendicular to k1-k2 .

cos2 2121 IIIII

k1

k2

k1-k2

Exercise: Prove that the distance between the fringe planes is

Discussion: Fabrication photographic gratings.

.)2/sin(2

d

4

9.2 Conditions for interference1) For producing stable patterns, the two sources must have the same frequency.2) For producing interference patterns, the two fields must have parallel components. 3) For producing clear patterns, the two sources must have similar amplitude.4) For producing interference patterns, coherent sources are required.

Temporal coherence:Time interval tc in which the light resembles a sinusoidal wave. (~10 ns for quasi- monochromatic light sources.)Coherence length: lc= ctc.

Relation to bandwidth (uncertainty principle): tc 1.Useful relations:

Spatial coherence:The correlation of the phase of a light wave between different locations.

=Relative phase predictable

2cctcl cc

5

Read: Ch9: 1-2Homework: Ch9: 2,3 Due: February 27

r1 r2

S1 S2

m=-1 0 +1

P

m = Ma

6

At a far field, I1I2I0.

Maximum:

Minimum:

kmrr /)](2[ 2121

kmrr /)]()12[( 2121

2

)()(cos4 21212

0

rrkII

Hyperboloids

Superposition of two spherical waves:

).(//)( and strengths, sourcesimilar have waves two theSuppose

cos2

)cos()(),(

)cos()(),(

202101

2121

2220222

1110111

rErE

rErE

rErE

IIIII

tkrt

tkrt

r1 r2

P

S1 S2

Interference patterns:1) Viewing screen perpendicular to the S1-S2 axis:

Concentric Rings.2) Viewing screen parallel to the axis: Parallel

fringes (Young’s experiment).3) Maximum order of rings: -M< m <M, where M

=a/. (-a <r1-r2 <a )

February 23 Interference of spherical waves

7

Variation of the interference rings:When reducing a :1) The rings swallow. (What we observe in the Michelson interferometer.)2) The distances between neighbor rings at a fixed viewing point increases.

Superposition of a plane wave and a spherical wave:Move S1 to infinity.

r1 r2

S1 S2

m=-1 0 +1

P

m = Ma

8

(*Reading) Interference pattern and holography:1) Two coherent waves produce an interference pattern.2) The interference pattern is photographically recorded, which makes a hologram.3) Now remove one of the waves (called the object wave), and illuminate the

hologram with the other wave (called the reference wave). The removed wave is then magically reproduced in space!

)()()()()()()()()()(

:)( withdilluminateonly when wavedTransmitte

)()( :hologram theofvity Transmissi

)()()()()()()()()( :Fringes

2*222

**222

rrrrrrrrrr

r

rr

rrrrrrrrr

BABAABATAC

A

IT

BABABABAI

A B

Record

A

Reconstruction

9

(*Reading) Size of the interference rings:

.11

2 ,

211

. ,

11

2

1 )()(

., ringorder th theof radius theSuppose

2

2222221

uv

nx

x

n

uv

mMnMvu

mxvu

vuxvxurr

vuxm

nn

mmm

m

Note:1) The size of the nth ring from the center:

2) S1, S2 and the viewing screen form a lens system:

3) Distance between rings:

4) For interference between a plane wave and a spherical wave:

. nxn

.2

11 ,

111 21

n

x

uvf

fuvn

.2

,2 n

v

dn

dxnvx n

n

r1

r2

P

S1 S2

xm

uv

(xn)

.11

2

uvndn

dxn

10

(*Reading) Proof of the change of the rings (Method I, complicated but most accurate):

r1

r2

P

S1 S2

R

x

D

mmxRxDmxRDrr 222221 ),()(),(

1) Calculate dR/dx for a given m:

.)( ,When 0)(

)(

0)(

)(

0)(

21

1

2222

2222

xRxrrR

rxD

dx

dR

RxDdxdR

RxD

RDdxdR

R

RxDRDdx

d

2) Calculate dR/dm at a given R:

separation Ring ,)(

)(

)(

21

21

2222

2222

rrR

rr

dm

dR

RxDdmdR

R

RDdmdR

R

RxDRDdm

d

)],()([

),()(),(

21

21

xRrRrR

xRrRrxR

At a given R, .),(2 xRrx

The ring separation at (R, x) is given by

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(*Reading) Proof of the change of the rings (Method II, easier):

r1

r2

P

S1 S2

R

x

D

(*Reading) Proof of the change of the rings (Method III, simplified):

mxrr cos21

1) Calculate d/dx for a given m:

)( ,When

0tan

10

)cos(

xx

xdx

d

dx

xd

2) Calculate d/dm at a given :

separation Ringsin

sincos

xdm

ddm

dxmx

The ring separation at (, x) is given by

sin),(

xx

At a given , .),( xx

The ring separation at (m, x) is given by

222),(

mxxm

At a given m, .),( xmx

R

12

Read: Ch9: 1-2No homework

a

s

13

February 25 Wavefront-splitting Interferometers

9.3 Wavefront-splitting interferometers9.3.1 Young’s experiment (1801) Young’s originality: Obtaining spatially coherent light from a pinhole, and splitting the wavefront of the spatially coherent light with two pinholes.

Optical path difference:r2 – r1 = a sin θ = ay/s ( If s>> a, s>> y )

Positions of bright fringes:

m = 0, ±1, ±2, … is the order number.

a

smy

am

m

m

ma msin

Distance between fringes:a

sy

Intensity distribution of the fringes:

s

ayI

rrkIII

2

0212

02

0 cos42

)(cos4

2cos4

ay/s

14

Effects of finite coherent length:

• Light from each slit has a coherent length lc. For sunlight (blackbody) lc3.• The waves from two slits can only interfere if r2 – r1 <lc.• The contrast of the fringes degrades when the

amount of the overlap between uncorrelated wavegroups increases.

P

r 2 – r 1

B A

A'B'

lc

Other wavefront-splitting interferometers

Fresnel’s double mirror: Slits S1 and S2 act as virtual coherent sources. They are images of slit S in the two mirrors.

= r2 – r1

PM1

M2

S1

S2

S

a

r1

r2

s

y

Shield

Space between fringes:a

sy

15

Fresnel’s double prism: Interference between light refracted from the upper and the lower prisms. The prisms produce two virtual coherent source S1 and S2.

Question: Where are S1 and S2?

S1

S2

Sa

Lloyd’s mirror: Interference between light from source S and its image S ' in the mirror.Glancing incidence causes a phase shift of , therefore the fringes are complementary to those of Young’s.

s

ayII

2

0 sin4

S

S'

as

y

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Read: Ch9: 3Homework: Ch9: 6,14,17,19Due: March 6

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9.4 Amplitude-splitting interferometers9.4.1 Dielectric films – double-beam interferenceApplication: Optical coating reduces or enhances the reflection of optical devices through thin-film interference effects.

Fringes of equal inclinationConsider the first two reflections (other reflections are weak if the reflectance is not large).Optical path length difference:

February 27 Amplitude-splitting Interferometers-1

dA

B

C

D

P

S

i

tnf

n1

n2

tf

t

tf

t

f

itt

f

it

f

f

dn

dndn

dndn

ACndn

ADnBCABn

cos2

cos

sin2

cos

2

sintan2cos

2

sincos

2

)(

2

1

1

1

tf dn cos2

18

coscos

coscos

coscos

coscos

ti

it

itti

tiitp

ti

ti

ttii

ttiis

nn

nn

nn

nnr

nn

nn

nn

nnr

1) External reflection, ni<nt .There is a phase shift of between the incident electric field and the reflected electric field.

2) Internal reflection, ni>nt .There is no phase shift between the incident electric field and the reflected electric field.

Phase shift in reflections when the angle of incidence is small: Eis

kii r

t

ni

Ers

kr

Ets

kt

nt

Erp

Etp

Eip

tf

tf dd

nk cos

4cos

4

19

Phase difference:Assume nf >n1, nf >n2. There is a phase shift of between external and internal reflections (when the incident angle is not large).

Bright spot: 4

)12(cos ft md

Dark spot: 4

2cos ft md

Extended source: All rays inclined at the same angle arrive at the same point.Fringes of equal inclination:Arcs centered on the perpendicular from the eye to the film.

Haidinger fringes: The fringes of equal inclination viewed at nearly normal incidence.Concentric circular bands.

t

f dn

cos2

sin

2cos

2

2

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Read: Ch9: 4-5Homework: Ch9: 27,29,32,34Due: March 6

Thickness: d = xInterference maximum:

2nf dm = (m + ½)

Distance between fringes: x = f /2

.4

)12(

,4

)12(

fm

fm

mx

md

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March 2 Amplitude-splitting Interferometers-2

Fringes of equal thicknessFizeau fringes: Contours from a non-uniform film when viewed at nearly normal incidence.

x

n1

n2

nfd

Compare: For the interference between two plane waves, the distance between

the fringe planes is .)2/sin(2

d

22

Interference colorsWhen a white light source is used in an interference experiment, each spectral component will produce its own interference pattern in space. The overlapped interference patterns from all spectral components result in the appearance of a special sequence of colors.

Calculated interference colors

Crossed polarizers

Parallel polarizers

Examples:•Soap bubbles•Oil films•Thin film coatings•Crystals between polarizers

Applications:•Identify minerals•Control coatings•Inspect strains

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Newton’s rings: Interference pattern from an air film between two glass surfaces.

xd

Thickness:

Bright rings: 2nf dm = (m + ½)

Dark rings:

R

xddRdx

2)2(

22

.2

1

)0,1,2( ,4

)12(

Rmx

mmd

fm

fm

. Rmx fm

222 sin

2cos x

R f

Discussions:Superposition of a plane wave and a spherical wave.Fabricating zone plates.

24

Beam splitter

Movable mirror

(©WIU OptoLab)

S S1 S2

M1 M2

2dcos

P

cos2d

cos

4d

Optical path length difference:

Phase difference:

Dark fringes: md m cos2

9.4.2 Mirrored interferometersMichelson interferometer1) Compensation plate: Negates dispersion from the beam splitter2) Collimated source: Fringes of equal thickness3) Point source: Interference of spherical waves4) Extended source: Fringes of equal inclination

Application: Accurate length measurement.

from the splitter.

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Read: Ch9: 4-5No homework

26

9.6 Multiple beam interferenceInterference between multiple reflection and multiple refractions:

td

S

i

nf

0E

rE0 ''0 ttrE ''30 ttrE ''50 ttrE

'0ttE2

0 ''rttE 40 ''rttE

2sin

12

1

2sin

12

1

2sin

12

1

1

cos21

)1(

1

1

1

)1(

'1

'

...)'''1('

1' ,'

cos4

cos2

22

2

22

2

22

2

24

222

2

2

2

20

20

362420

2

rr

rr

I

I

I

I

rr

rr

r

er

r

I

I

er

rE

er

ttE

erererttEE

rttrr

dn

kdn

i

t

i

r

ii

t

ii

iiit

tf

tf

March 4, 6 Multiple beam interference

Reduces to fringes of equal inclination when r is small.

27

)2/(sin1

1

1

2

2

2

2

FI

I

r

rF

i

t

it II /

/

F = 0.2 (r2 = 0.046)F = 1 (r2 = 0.17)F = 200 (r2 = 0.87)

2

21

2

r

rFCoefficient of finesse:

Airy function:)2/(sin1

)2/(sin

)2/(sin1

1

2

2

2

F

F

I

I

FI

I

i

r

i

t

)2/(sin1

1 )(

2

FA

t

f

dn

cos4

28

9.6.1 The Fabry-Perot interferometer1) High resolving power2) Prototype of laser cavity

S P

d

) large(For 21

arcsin2 2

1

)2/(sin1

1 )( 2/12

FFFF

A

Half-width of transmission:

Finesse:

F

42 2/1

2

2 FF

2

21

2

cos4

r

rF

dn t

f

)(A

29

Applications:I. How a laser wavelength is selected.1) The laser itself is a Fabry-Perot cavity. Two highly reflective mirrors select an

output of a certain wavelength with an extremely narrow bandwidth, which is called a mode.

2) The mode separation in a laser cavity (=c/2l) can be narrower than the gain profile of the laser medium, therefore output of multiple modes is possible.

3) Laser modes can be longitudinal or transverse.

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II. How a scanning spectrum analyzer works.1) A scanning spectrum analyzer can be used to measure the modes of a laser.2) The analyzer is a Fabry-Perot interferometer whose length can be slightly varied by

piezoelectric device.3) A saw-tooth voltage is used to repeatedly change the cavity length, and the output

signal is displayed on an oscilloscope.

scanning scanning

31

i r

t

ni

nt

1

t

rr

ttr'

rt

r2

tt'

rrtrrt

rttttr

' 0'

1' 1' 22

32

Read: Ch9: 6Homework: Ch9: 37,40,42,45,47Due: March 13

Hint to P9.45:You are asked for the refractive index and the thickness of the film, so two conditions should be figured out. In order to completely eliminate the overall reflection, the light reflected from the front surface and from the rear surface of the film must have exactly the same amplitude, as well as having exactly a phase difference of . Suppose , then nf can be found from the Fresnel equations if we require the two reflections exactly cancel each other. You may also prove that won’t work for this problem.

gf nn 1

gf nn

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“You can’t judge a book by its cover.”