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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
HeNe laser
Radio City Rockettes, New York, NY
rood blauw oranje
oranje blauw groen
blauw paars groen
oranje blauw rood
paars oranje rood
groen rood blauw
oranje blauw rood
paars oranje rood
paars
rood
rood
groen
blauw
paars
groen
blauw
J.R. Stroop "Studies of interference in serial verbal reactions" Journal of Experimental Psychology 18:643-662 (1935).
Peacock
Soap bubbles
2-beam interference


E1  E 0 1 cos(ks1  t  1 )


E 2  E 02 cos(ks2  t   2 )
initial phase (at t=0)
propagation distance from
source of disturbance
from superposition principle:
 

EP  E1  E2
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
 
I   0c E  E
Irradiance at point P
I   0c
  0c
  0c
I   0c
I =
2
EP
 
EP  EP
 
 
E1  E2  E1  E2
   
 
E1  E1  E2  E2  2E1  E2


I1
+

I2
- when E1 and E2 are parallel, maximum interference
- when orthogonal, dot product = 0; no interference
+ I12
The interference term I12
 
I12  2 0c E1  E2
dot product of electric fields:
 
 
E1  E2  E01  E02 cos(ks1  t  1 ) cos(ks2  t  2 )
simplify by introducing constant phases:
  ks1  1
  ks2  2
 
 
2E1  E2  2E01  E02 cos(  t ) cos(  t )
use trigonometry: 2cosAcosB = cos(A+B) + cos(B-A)
and consider again the time average:
 
 
2 E1  E2  E01  E02  cos(    2t )  cos(   )
 kills it

The interference term I12
 
 
2 E1  E2  E01  E02  cos(    2t )  cos(   )
 
 E01  E02 cos(   )
 
 E01  E02 cos(k ( s2  s1 )  2  1 )
simplify by introducing d:
d  k ( s2  s1 )  2  1
to yield the interference term of the irradiance:
 
I12   0cE01  E02 cosd

Irradiance formula
I  I1  I 2  I12
 
I1   0 c E1  E1
  0 cE
I1 
2
01
cos (  t )
1
 0 cE 012
2
 
I12   0cE01  E02 cosd
 
I 2   0c E2  E2
2
  0 cE
2
02
cos (   t )
2
1
I 2   0 cE 022
2
if E1║ E2,
 
then E01 E02  E01E02
I12  2 I1I 2 cosd
I  I1  I 2  2 I1I2 cosd
-where d is the phase difference
-for parallel electric fields
Interference
mutually incoherent beams (very short coherence time)
I  I1  I 2
mutually coherent beams (long coherence time)
I  I1  I 2  2 I1I2 cosd
maximum when cos d = 1
I  I1  I 2  2 I1I2
constructive interference
d = (2mp)
minimum when cos d = -1
I  I1  I 2  2 I1I2
destructive interference
d = (2m+1)p
Interference fringes
I  I1  I 2  2 I1I2 cosd
maximum when I1 = I2 = I0
1 + 1 = 4 !?!
Interference in time and space
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  = a sin
conditions for interference:
  m  a sin 
constructive
  m    a sin 
destructive

1
2

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
The Dancing Couple-1663-Jan Steen
Anatomy of a soap bubble
Soap bubble interference
Thin film interference: normal incidence
optical path difference:  = nf(AB + BC) = nf (2t)
Thin film interference: non-normal incidence
optical path difference:  = nf(AB + BC) – n0(AD)
= 2nf t cost
 = m:
 = (m + ½):
constructive interference
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
external reflection
n1 < n2
air
glass
glass
TM mode
TE mode
TM mode
n1
n2
internal reflection
n1 > n2
air
TE mode
n1
n2
Back to the bubbles
Colors indicate bubble thickness
How thick here (red band)?
180o phase change
0o phase change
n>1
t
Constructive interference for
2t ~ (m + ½)
At first red band m = 0
t ~ ¼ (700 nm)
Dark, white, and bright bands
pop!
Bright: Colored “monochromatic” stripes occur at (1/4) for visible colors
White: Multiple, overlapping interferences (higher order)
Dark: Super thin; destructive interference for all wavelengths
(no reflected light)
Multiple beam interference
E1  rE0 e it
E2  tr ' t ' E0 e i (t d )
E3  tr '3 t ' E0 e i (t  2d )
E4  tr '5 t ' E0 e i (t 3d )
...
E N  tr '( 2 N 3) t ' E0 e i[t ( N 1)d ]
r, t : external reflection
r’, t’ : internal reflection
where d is the phase difference
Note: thickness t !
d  k
  2n f t cost
Er  E1r  E2 r  E3r  ...  E Nr  ...
 E0eit {r  tr ' t ' e id  tr '3 t ' e i 2d  ...  tr '[ 2 N 3] t ' e i ( N 1)d  ...}
 E0eit {r  tr ' t ' e id [1  ( r '2 e id )  ( r '2 e id ) 2  ...  ( r '2 e id )( N 2 )  ...]}
geometric series
1  x  x
2
 ...  1/ 1  x 
Multiple beam interference
Er  E0eit {r  tr' t ' eid [1  (r '2 eid )  (r '2 eid )2  ... (r'2 eid )( N 2)  ...]}
1  x  x
2
 ...  1/ 1  x 
 id
r
'
tt
'
e
Er  E0 eit [r 
]
2 id
1 r' e
Introduce Stokes relations: r’=-r
and tt’=1-r2
and simplify to get:
 id

r
(
1

e
)
i t
E r  E0 e 
2  id 
1

r
e 

Irradiance:
I r  Er  Er Er*
2
Er
2
 eit (1  e id )   e it (1  eid ) 
E r 


2 id
2 id
1

r
e
1

r
e



2 2
0
Multiple beam interference
Working through the math, you’ll arrive at:
2r 2 (1  cosd )
I r  Ii
1  r 4  2r 2 cosd
where Ii is the irradiance of the incident beam
Likewise for transmission leads to:
(1  r 2 ) 2
It  Ii
1  r 4  2r 2 cosd
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:
1 = 0.5890182 mm
2 = 0.5896154 mm
mirror reflectivity r = 0.9
mirror separation: 100 - 400 mm
Fabry-Perot interferometer
see chapter 8
It
(1  r 2 ) 2
T 
I i (1  r 2 ) 2  (2r ) 2 sin 2 (d / 2)
T
1
1  F sin 2 (d / 2)
where F is the coefficient of finesse: F 
4r 2
1  r 
2 2
Fabry-Perot interferometer: fringe profiles
Michelson
cos2 d / 2
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
rm2  tm2
R
2tm
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:
MgF2 coating:
n = 1.5
n = 1.38
To make an AR coating for  = 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   /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

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