Electromagnetic waves:
Interference
Wednesday November 6,
2002
1
Haidinger’s Bands: Fringes of equal inclination
n1
Beam splitter
d
n2
P
x
1

1
f
Focal
plane
Extended
source
PI
P2
Dielectric
slab
2
Fizeau Fringes: fringes of equal thickness
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Now imagine we arrange to keep cos ’ constant
We can do this if we keep ’ small
That is, view near normal incidence
Focus eye near plane of film
Fringes are localized near film since rays
diverge from this region
Now this is still two beam interference, but
whether we have a maximum or minimum will
depend on the value of t
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Fizeau Fringes: fringes of equal thickness
I  I1  I 2  2 I1I 2 cos 
where,



2kt cos  '

2kt  
Then if film varies in thickness we will see fringes as we move our eye.
These are termed Fizeau fringes.
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Fizeau Fringes


 2kt cos  '
 2kt  
Beam splitter
n
Extended source
n2
x
n
5
Wedge between two plates
1 2
glass
glass
y
L
Path difference
= 2y
Phase difference  = 2ky - 
D
air
(phase change for 2, but not for 1)
Maxima 2y = (m + ½) o/n
Minima 2y = mo/n
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Wedge between two plates
Maxima 2y = (m + ½) o/n
y
Minima 2y = mo/n
Look at p and p + 1 maxima
L
D
air
yp+1 – yp = o/2n  Δx
where Δx = distance between adjacent maxima
Now if diameter of object = D
Then L = D
And (D/L) Δx= o/2n or D = oL/2n Δx
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Wedge between two plates
Can be used to test the quality of surfaces
Fringes follow contour of constant y
Thus a flat bottom plate will give straight fringes, otherwise
ripples in the fringes will be seen.
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Newton’s rings
Used to test spherical surfaces
Beam splitter
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Newton’s rings
Maxima when,
y = (m+1/2) o/2n
R- Rcos = y
or,
R2=(R-y)2+r2
 R2(1-2y/R) + r2
i.e.
r2
Gives rings,

= 2yR
R
Rm2=(m+1/2)oR/n
R
y
r
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Reflection from dielectric layer:
Antireflection coatings
Important in instruments such as cameras
where reflections can give rise to spurious
images
 Usually designed for particular wavelength
in this region – i.e. where film or eye are
most sensitive
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Anti-Reflection coatings
2
1
A. Determine thickness of film
n1 < n2 < n3
air
n1
n2
film
n3
glass
Thus both rays (1 and 2) are shifted in phase by  on reflection.
For destructive interference (near normal incidence)
2n2t=(m+1/2)
Determines the thickness of the film
(usually use m=0 for minimum t)
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Anti-Reflection coatings
2
1
air
B. Determine refractive index of film
EA
Near normal incidence
n1
n2
A’
film
Amplitude at A
n3
glass

EA

EA

A
 '  ' Eo
 2n1  2n2   n3  n2 
  



 n1  n2  n2  n3   n3  n2 
n  n2
 3
n3  n2
Since  ’ ~ 1
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Anti-reflection coating
B. Determine refractive index of film
Amplitude at A’
E A'
n2  n1

n2  n1
To get perfect cancellation, we would like EA = E A’
n2  n1n3
n3  n2 n2  n1

n3  n2 n1  n2
should be index of AR film
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Multiple Beam interference
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Thus far in looking at reflectivity from a dielectric
layer we have assumed that the reflectivity is small
The problem then reduces to two beam interference
Now consider a dielectric layer of uniform thickness
d and assume that the reflectivity is large e.g. || >
0.8
This is usually obtained by coating the surface of the
layer with a thin metallic coating – or several
dielectric coatings to give high reflectivity
Or, one can put coatings on glass plates , then
consider space between plates
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Multiple beam interference
Let 12 =  21= ’ 12=  21= ’
’’ Eo
Eo
(’)5’Eo
(’)3’Eo
(’)7’Eo

n1
n2
n1
’
A
B
C
D

’ Eo
(’)2’Eo
(’)4’Eo
(’)6’Eo
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Multiple Beam Interference
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Assume a (for the time being) a monochromatic source
, ’ small ( < 30o) usually
Now || = |’| >> , ’
Thus reflected beams decrease rapidly in amplitude
(from first to second)
But amplitude of adjacent transmitted beam is about the
same amplitude
Amplitude of successfully reflected beams decreases
slowly (from the second)
Thus treat in transmission where contrast should be
somewhat higher
The latter is the configuration of most applications
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