Document info 18.

InterferenceTuesday, 11/28/2006

Physics 158Peter Beyersdorf

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18.

Class Outline

Review of Exam

Superposition and Interference

Interference Conditions

Wavefront Splitting interferometers

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18.

InterferenceWhen the electric fields from two waves overlap the principle of superposition tells us

But it is the intensity that we can detect. Neglecting the constants 1/2, ε0, and c:

or

with

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!Etot = !E1 + !E2

Itot = !Etot · !Etot = E21 + 2E1 · E2 + E2

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Itot = I1 + I12 + I2

I1 =!E2

1

"

I2 =!E2

2

"I1 =

!E2

1

"I12 = 2

!!E1 · !E2

"

interference term

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Interference Term

The time average

in phasor notation is

For incoherent waves

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!!E1 · !E2

"=

1T

# T

0

!E1(t) · !E2(t)dt

!!E1 · !E2

"= 0

!!E1 · !E2

"= E1(") · E!

2 (")

18.

Interference Term

Show that

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!!E1 · !E2

"= E1(")E!

2 (")

…

![E(!)] =E(!) + E!(!)

2

the time average of this is zero since it fluctuates at about twice the

optical frequency

This term is slowly varying so its time average is its instantaneous value

!!E1 · !E2

"= !

#E1("1)E!

2 ("2)2

$

=#

E1("1)ei!1t + E!1 ("1)e"i!1t

2· E2("2)ei!2t + E!

2 ("2)e"i!2t

2

$

=#

E1("1) · E2("2)ei(!1+!2)t + E!1 ("1) · E!

2 ("2)e"i(!1+!2)t

2+

E1("1) · E!2 ("2)ei(!1"!2)t + E!

1 ("1) · E2("2)e"i(!1"!2)t

2

$

= "E1("1) · E2("2) cos ("1 + "2)t + ! [E1("1) · E!2 ("2)]#

= E1("1) · E!2 ("2)

18.

Interference Conditions

Let Δφ be the phase difference between two overlapping beams

in general Δφ will be a function of time and position, then when there is constructive interference I=Imax:

for Δφ=0, ±2π, ±4π…

and when there is destructive interference I=Imin

for Δφ=±π, ±3π… 6

I12 = 2!

I1I2 cos !!

Imax = I1 + I2 + 2!

I1I2

Imin = I1 + I2 ! 2!

I1I2

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Fringe Visibility

If the two beams have unequal intensity (or if they are not fully coherent) the minimum intensity will not be zero and the fringe visibility or fringe contrast will be

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V =Imax ! Imin

Imax + Imin

high visibility fringes low visibility fringes

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Summing Multiple Sources

If you have two sources of light, each with an intensity of I0, what is the intensity of the wave that you get when you add them together in-phase,

if they are incoherent?

If they are coherent?

consider

Does this violate conservation of energy?

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I = I1 + I2 + 2!

I1I2 cos !!

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Fresnel-Arago Laws

Interference of two orthogonal polarization states does not produce visible fringes

Interference of two parallel, coherent polarization states will produce visible fringes

Interference of two incoherent state will not produce visible fringes

Orthogonal polarization states of natural light have no mutual coherence

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18.

Coherence

A wave split and recombined will interfere with itself it the path length difference between the beams is less than the coherence length. Otherwise the interference will rapidly fluctuate through constructive and destructive and will average itself out

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+ +

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Coherence

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Lc=5λ-10λ<ΔL<10λ

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Coherence

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Lc=1λ-10λ<ΔL<10λ

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Wavefront Splitting Interferometers

If a beam is spatially coherent then the wavefront can be split and recombined to give visible interference fringes

Consider Young’s double slit

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A•Bx

Ld

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Young’s Double Slit

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A

•B

θ

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A

•B

θθθ

d

dsinθ

Δφ=(2π/λ)d sinθ

A

•B

θθθ

d

dsinθ

Δφ=(2π/λ)d sinθbright fringes at (2π/λ)dx/L=2mπ

xL

dark fringes at (2π/λ)dx/L=(2m+1)π

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Fresnel’s Double Mirror

Consider the interference pattern of Fresnel’s double mirror

Find the image of the point source in each mirror

Treat interference pattern as that from two point sources (i.e. the two images of the points source)

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18.

Summary

Intensity of two interfering beams depends on intensity of the beams and the relative phase shift between the beams

Only coherent beams of the same polarization produce observable interference fringes

Fringe visibility measures contrast of a fringe pattern

Spatially coherent waves can be split spatially and made to interfere - the resulting interference pattern can be used to measure the separation of the virtual sources 16