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Chapter 7 : Interference of light. Chapter 7 : Interference of light. in·ter·fer·ence. 1. Life. Hindrance or imposition in the concerns of others. 2. Sports . Obstruction of an opponent, resulting in penalty. .

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slide1

Chapter 7: Interference of light

Chapter 7: Interference of light

slide2

in·ter·fer·ence

1.Life. Hindrance or imposition in the concerns of others.

2.Sports. Obstruction of an opponent, resulting in penalty.

3. Physics.Superposition of two or more waves, resulting in a new wave pattern.

constructive

destructive

slide5

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J.R. Stroop "Studies of interference in serial verbal reactions" Journal of Experimental Psychology 18:643-662 (1935).

slide8

2-beam interference

initial phase (at t=0)

propagation distance from source of disturbance

from superposition principle:

slide9

Measuring interference

  • - Electric fields are rapidly varying (n ~ 1014 Hz)
  • - Quickly averages to 0
  • - Instead of measuring E directly, measure radiant power density
    • = irradiance, Ee (W/m2)
    • = time average of the square of the electric field amplitude
  • - Note: to avoid confusion, Pedotti3 now uses the symbol I instead of Ee
slide10

Irradiance at point P

I =

I1

  • +

I2

  • +

I12

- when E1 and E2 are parallel, maximum interference

- when orthogonal, dot product = 0; no interference

slide11

The interference term I12

dot product of electric fields:

simplify by introducing constant phases:

use trigonometry: 2cosAcosB = cos(A+B) + cos(B-A) and consider again the time average:

w kills it

slide12

The interference term I12

simplify by introducing d:

to yield the interference term of the irradiance:

slide13

Irradiance formula

if E1║E2,

then

-where d is the phase difference

-for parallel electric fields

slide14

Interference

mutually incoherent beams (very short coherence time)

mutually coherent beams (long coherence time)

maximum when cos d = 1

constructiveinterference

d = (2mp)

minimum when cos d = -1

destructive interference

d = (2m+1)p

slide15

Interference fringes

maximum when I1 = I2= I0

1 + 1 = 4 !?!

slide16

Interference in time and space

Young’s experiment

wavefront division

Michelson interferometer

amplitude division

slide18

Double slit experiment with electrons

http://www.youtube.com/watch?v=ZJ-0PBRuthc

slide19

Criteria for light and dark bands

- approximate arc S1Qto be a straight line

- optical path difference D = asinq

conditions for interference:

constructive

destructive

m = 0, 1, 2, 3, …

slide20

Interference from 1 source: reflection

Lloyd’s mirror

part of the wavefront is reflected; part goes direct to the screen

Fresnel’s mirrors

part of the wavefront is reflected off each mirror

slide22

Interference from 1 source: refraction

Fresnel’s biprism

part of the incident light is refracted downward and part upward

slide23

Interference via amplitude division

- thin films

- oil slicks

- soap bubbles

- dielectric coatings

- feathers

- insect wings

- shells

- fish

- …

slide24

Interference intermezzo

Interference intermezzo

slide30

Thin film interference: normal incidence

optical path difference: D = nf(AB + BC) = nf(2t)

slide31

Thin film interference: non-normal incidence

optical path difference: D=nf(AB + BC) –n0(AD)

= 2nf t cosqt

D = ml: constructive interference

D = (m + ½)l: destructive interference

where m = 0,1,2,…

slide32

Keep in mind the phase

“soft”

reflection

“hard”

reflection

Simple version: phase of reflected beam shifted by p if n2 > n1

0 if n1 > n2

Correct version: use Fresnel equations!

slide33

Summary of phase shifts on reflection

external reflection

n1 < n2

TE mode

TM mode

n1

air

n2

glass

internal reflection

n1 > n2

TE mode

TM mode

air

n1

n2

glass

slide35

180o phase change

0o phase change

t

n>1

Colors indicate bubble thickness

How thick here (red band)?

Constructive interference for

2t ~ (m + ½)l

At first red band m = 0

t ~ ¼ (700 nm)

slide36

pop!

Dark, white, and bright bands

Bright: Colored “monochromatic” stripes occur at (1/4)l for visible colors

White: Multiple, overlapping interferences (higher order)

Dark: Super thin; destructive interference for all wavelengths

(no reflected light)

slide37

Multiple beam interference

r, t: external reflection

r’, t’ : internal reflection

Note: thickness t !

  • where d is the phase difference

geometric series

slide38

Multiple beam interference

Introduce Stokes relations: r’=-r and tt’=1-r2and simplify to get:

Irradiance:

slide39

Multiple beam interference

Working through the math, you’ll arrive at:

where Ii is the irradiance of the incident beam

Likewise for transmission leads to:

slide40

This simulation was performed for the two sodium lines described above, with reflectivity                and the separation of the mirrors increasing from 100 microns to 400 microns.

Fabry-Perot interferometer (1897)

d

simulation of two sodium lines:

l1 = 0.5890182 mm

l2 = 0.5896154 mm

mirror reflectivity r = 0.9

mirror separation: 100 - 400 mm

slide41

Fabry-Perot interferometer

see chapter 8

where F is the coefficient of finesse:

slide42

Fabry-Perot interferometer: fringe profiles

Michelson

d

  • -transmission maxima occur when d = 2pm
  • as r approaches 1 (i.e. as Fincreases), the fringes become very narrow
  • see Chapter 8 for more on Fabry-Perot:
      • fringe contrast, FWHM, finesse, free spectral range
slide43

Fringes of equal thickness

Constructive reflection

2d = (m+1/2)λ m=0, 1, 2, 3...

Destructive reflection

2d = mλ m=0, 1, 2, 3...

slide44

Newton’s rings

white-light illumination

pattern depends on contact point: goal is concentric rings

slide45

Oil slick on pavement

Constructive reflection

2d = mλ m=0, 1, 2, 3...

Destructive reflection

2d = (m+1/2)λ m=0, 1, 2, 3...

slide46

Thin film coatings: anti-reflective

Glass: n = 1.5MgF2 coating: n = 1.38To make an AR coating for l = 550 nm, how thick should the MgF2 layer be?

slide48

Multilayer mirrors

  • thin layers with a high refractive index n1,interleaved with thicker layers with a lower refractive index n2
  • path lengths lA and lB differ by exactly one wavelength
  • each film has optical path length D = l/4: all reflected beams in phase
  • ultra-high reflectivity: 99.999% or better over a narrow wavelength range
slide51

Exercises

You are encouraged to solve all problems in the textbook (Pedrotti3).

The following may be covered in the werkcollege on 5 October 2011:

Chapter 7:

1, 2, 7, 9, 15, 16, 24