Electromagnetic waves:
Two source Interference
Monday November 4, 2002
1
Reflection from dielectric layer

n1
n2
n1

Assume phase of wave at
O (x=0, t=0) is 0
Amplitude reflection coefficient


A

’
A’


x=0
’
O’


t
Amplitude transmission
co-efficient

O
(n1n2)  = 12
(n2 n1) ’=21
(n1n2)  =  12
(n2 n1) ’=  21
Path O to O’ introduces a
phase change
2
x=t
t
k 2 S 2 
2 cos '
2
Reflection from a dielectric layer

At O:
 Incident
amplitude
 Reflected amplitude

At O’:
i  k 2 S 2 t 
amplitude   ' Eo e
i  k 2 S 2 t 


'
E
e
 Transmitted amplitude
o
 Reflected

E = Eoe-iωt
ER = Eoe-iωt
At A:
 Transmitted
amplitude
 Reflected amplitude
i  2 k 2 S 2 t 
  ' ' Eo e
  '  ' Eo e i 2 k 2 S 2 t 
3
Reflection from a dielectric layer
•At A’
and
E A'   ' Eo e i k1S1 t 
A
ΔS1= z sin  = 2t tan ’ sin 
Since,
  ' 

z = 2t tan ’
A’
n2  n1
 0.2
n2  n1
and
 '  0.96
The reflected intensities ~ 0.04Io and both beams (A,A’) will have
almost the same intensity.
Next beam, however, will have ~ ||3Eo which is very small
Thus assume interference at , and need only consider the two
beam problem.
4
Transmission through a dielectric layer
At O’: Amplitude ~ ’Eo ~ 0.96 Eo
 At O”: Amplitude ~ ’(’)2Eo ~ 0.04 Eo
 Thus amplitude at O” is very small

O”
O’
5
Reflection from a dielectric layer




Interference pattern should
be observed at infinity
By using a lens the pattern
can be formed in the focal
plane (for fringes localized
at )
Path length from A, A’ to
screen is the same for
both rays
Thus need to find phase
difference between two
rays at A, A’.
A

z = 2t tan ’
A’
6
Reflection from a dielectric surface
i
EA    e  ' Eoe
i 2 k2 S2 t 
A
E A '   ' Eo e

i  k1S1 t 
z = 2t tan ’
A’
If we assume ’ ~ 1
and since ’ = ||
This is just interference between two sources with equal amplitudes
7
Reflection from a dielectric surface
EA     ' Eoei 2k2S2 t 
EA'   ei Eoei k1S1 t 
I  I1  I 2  2 I1I 2 cos 
where,
  2  1  2k2S2  k1s1   
Since
k 2 = n2 k o
and n1sin = n2sin’
Thus,



k1=n1ko
(Snells Law)
 2k 2 S 2  k1s1   
 t 
 2n2 ko 
  n1ko 2t tan  ' sin   
 cos ' 
 2n2 kot cos '
8
Reflection from a dielectric surface
Since I1 ~ I2 ~ Io
Then, I = 2Io(1+cos)
Constructive interference
=  2m = 2ktcos’ - 
(here k=n2ko)
2ktcos’ = (2m+1)
ktcos’ = (m+1/2)
2n2cos’ =  (m+1/2)o
Destructive interference
2n2cos’ =  mo
9
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
10
Fizeau Fringes: fringes of equal thickness






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
11
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.
12
Fizeau Fringes


 2kt cos  '
 2kt  
Beam splitter
n
Extended source
n2
x
n
13
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
14
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
15
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.
16