Electromagnetic waves:
Multiple beam Interference
Wed. Nov. 13, 2002
1
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
2
Multiple beam interference
Evaluate
1
 1  1 



* 
2
*
1

r
1

r
1

r

r

r




r 2    '

4


r  r *    ' e i  e i  2  ' cos 
2
Thus,
1
1

*
1  r  1  r
1   '2


 

2
2
1
 2  ' cos 
1
2
1   ' 
2 2
 2  ' 1  cos  
2
3
Multiple beam interference
1
1

*
1 r 1 r
Now,
1
1   ' 
2 2
2

4  '
1 
2



1


'



2
  
sin  
 2 

2
 '2   2  R
and ,
E   E  T 
2
P
o 2
A
2
1  R 
2




1


1  4 R sin 2    
 
 1  R 2
 2 

4
Multiple beam interference
Now recall the definition of the intensity of an electromagnetic wave
 
1
I P  v E Po
2
1
I o  v E Ao
2
2
 
Thus,
2
T2
2

1 R
IP 
Io
4R
2  
1
sin  
2
1  R 
2
is the intensity distribution in the focal plane of the lens.
5
Multiple beam interference
Fringe pattern
   ' 
2n2 k o d
 
 2n1k o d tan  ' sin 
cos  '
  2n2 ko d cos  '
6
Multiple beam interference

Maximum intensity when,
sin 2

2
0
or
  2m
or ,
4n2
o
d cos  '  2m
2n2 d cos '  m
7
Multiple beam interference


Thus maxima are circles in focal plane of lens –
or rings
The maximum intensity
2
I max

 T 

 Io
1 R 
And intensity distribution is,
IP 
I max
4R
2  
1
sin
 
2
1  R 
2
8
Multiple beam interference

Minima intensity when,
sin

Intensity distribution,

If R > 0.9
2

2
1

I min
1  R

2
I max
1  R 
2
Imin<<Imax
9
Multiple beam interference

Let the contrast, or co-efficient
of finesse be defined by,
4R
F
1  R 2
IT


IM
Then the transmitted light is
described by and Airy function,
The same analysis for the
reflected light gives
IR

1
 
1  F sin  
2
2


2

1  F sin  
2
F sin 2 
IM
2
10
Multiple beam interference: Transmission curves
IR
1

I M 1  F sin 2 
 2
IT/II
1.0
F=0.2
R=0.046
F=1
R=0.18
F=200
R=0.87
0.5
0.0
-2
-1
0
1

2
3
4
11
Multiple beam interference: Reflectivity curves
 
 2
F sin 2 
IR
2

I M 1  F sin 2 
1.0
R=0.87
IR/II
F=200
0.5
0.0
-2
-1
0
1

2
F=1
R=0.18
F=0.2
R=0.046
3
4
12
Observation of fringe patterns
Screen or eye
source
f1
f2
Bright circles
13
Typical parameters in an
experiment


Consider a quartz slab (n2=1.5) of thickness d ~
0.5 cm
The condition for constructive interference
requires, 2n d cos '  m
2

For light of wavelength o = 500 nm, incident at
a small angle , i.e. ’ also small, and m is large:
2


2n2 d cos ' 2 1.5 0.5 10 m 1
4
m

 3 10

9
500 10 m
14
Spectroscopy applications: Fabry- Perot Interferometer




Assume we have a
monochromatic light
source and we obtain a
fringe pattern in the focal
plane of a lens
Now plot IT along any
radial direction
Let IMAX=IM
The fringes have a finite
width as we scan
IM
I
m
m-1
m-2
m-3 order
15
Fabry-Perot Interferometer
Full width at half maximum = FWHM, is
defined as the width of the fringe at
I=(½)IM
 Now we need to specify units for our
application
 Let us first find  such that I = ½ IM

16
Fabry-Perot Interferometer

We want,
1
IM
I  IM 
2
1  F sin 2 
 2

This gives,
 2 1
F sin 2 
17
Fabry-Perot Interferometer
IM
m
m-1
I
I= ½ IM
 = 2m
 = 2(m-1)
 = 2(m-Δm)
18
Fabry Perot Interferometer

Thus at I = ½ IM
sin(/2) = sin (m – Δm)   sin Δm

Assume Δm is small and
sin Δm  Δm

Thus
 1 
F 

 m 

2
or
m 
1
 F
FWHM ~ Fraction of an order occupied by fringe
1 R
FWHM  2m 
 R
19
Fabry-Perot Interferometer

The inverse of the FWHM is a measure of the
quality of the instrument. This quality index is
called the finesse
 R
1
Finesse   

1 R
2m

It is the ratio of the separation between the
fringes to the fringe width
20
Fabry-Perot Interferometer
Not that  is determined by the reflectivity
 If R ~ 0.90
 = 30

R ~ 0.95
R ~ 0.97

 = 60
 = 100
In practive, can’t get much better than 100
since the reflectivity is limited by the
flatness of the plates (and other factors of
course)
21
Fabry-Perot Interferometer


Now consider the case of two wavelengths (1, 2)
m
m-2
present in the beam
m-1
Assume 1 2 and 1< 2
2




1
2
1 
2
1
Increase 2, dashed lines shrink
e.g. order m-1 of 2 moves toward mth order of 1
Eventually (m-1) 2=m 1
This defines the free spectral range
22
Fabry-Perot Interferometer



m(2-1)= 2 or mΔ = 
ΔFSR = /m
Now since
2 nd cos  '
2 nd
m 




We have,
 FSR 

2
2 nd
e.g. =500 nm, d = 5mm, n=1

ΔFSR= 25(10-2)mm = 0.25Å
23