Chapter 7: Interference of light Chapter 7: Interference of light

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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. 3. Physics. Superposition of two or more waves, resulting in a new wave pattern. constructive destructive 2

HeNe laser

Radio City Rockettes, New York, NY

rood blauw oranje paars oranje blauw groen rood blauw paars groen rood oranje blauw rood groen paars oranje rood blauw groen rood blauw paars oranje blauw rood groen paars oranje rood blauw J.R. Stroop "Studies of interference in serial verbal reactions" Journal of Experimental Psychology 18:643-662 (1935). 5

Peacock

Soap bubbles

from superposition principle: 2-beam interference initial phase (at t=0) propagation distance from source of disturbance from superposition principle:

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

Irradiance at point P I = I1 + I2 + I12 - when E1 and E2 are parallel, maximum interference - when orthogonal, dot product = 0; no interference

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

The interference term I12 simplify by introducing d: to yield the interference term of the irradiance:

Irradiance formula if E1║ E2, then -where d is the phase difference -for parallel electric fields

constructive interference destructive interference mutually incoherent beams (very short coherence time) mutually coherent beams (long coherence time) maximum when cos d = 1 constructive interference d = (2mp) minimum when cos d = -1 destructive interference d = (2m+1)p

Interference fringes maximum when I1 = I2 = I0 1 + 1 = 4 !?!

Interference in time and space Michelson interferometer Young’s experiment wavefront division Michelson interferometer amplitude division

The double slit experiment (first performed in 1803)

Double slit experiment with electrons http://www.youtube.com/watch?v=ZJ-0PBRuthc

Criteria for light and dark bands - approximate arc S1Q to be a straight line - optical path difference D = a sinq conditions for interference: constructive destructive m = 0, 1, 2, 3, …

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

Fresnel’s mirrors as solar collectors

Interference from 1 source: refraction Fresnel’s biprism part of the incident light is refracted downward and part upward

Interference via amplitude division - thin films - oil slicks - soap bubbles - dielectric coatings - feathers - insect wings - shells - fish - …

Interference intermezzo Interference intermezzo

The Dancing Couple-1663-Jan Steen

Anatomy of a soap bubble

Soap bubble interference

Thin film interference: normal incidence optical path difference: D = nf(AB + BC) = nf (2t)

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,…

Keep in mind the phase “hard” reflection “soft” reflection Simple version: phase of reflected beam shifted by p if n2 > n1 0 if n1 > n2 Correct version: use Fresnel equations!

Summary of phase shifts on reflection air glass external reflection n1 < n2 TE mode TM mode n1 n2 air glass internal reflection n1 > n2 TE mode TM mode n1 n2

Back to the bubbles

Colors indicate bubble thickness How thick here (red band)? t n>1 180o phase change 0o phase change Constructive interference for 2t ~ (m + ½)l At first red band m = 0 t ~ ¼ (700 nm)

Dark, white, and bright bands pop! 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)

Multiple beam interference r, t : external reflection r’, t’ : internal reflection Note: thickness t ! where d is the phase difference geometric series

Multiple beam interference Introduce Stokes relations: r’=-r and tt’=1-r2 and simplify to get: Irradiance:

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:

Fabry-Perot interferometer (1897) 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

Fabry-Perot interferometer see chapter 8 where F is the coefficient of finesse:

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

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...

Newton’s rings white-light illumination pattern depends on contact point: goal is concentric rings

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...

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

Broadband anti-reflective films

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

Anodized titanium

Natural multi-layer reflectors

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