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Reflection, Transmission and Interference of EM
waves
2015/7/7
Phasor Representation of wave addition
• Phasor representation of a wave
• E.g. E = Eosint is represented as a vector of magnitude Eo,
making an angle =t with respect to the y-axis
• Projection onto y-axis for sine and x-axis for cosine
• Now write,
E  Eo sin k  r  t   
as,
E  Eo sin t   
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Phasors
• Imagine disturbance given E1  Eo1P sin t  1 
in the form
E2  Eo 2 P sint   2 
=φ2-φ1
φ2
φ1
E 2  Eo1P   Eo 2 P   2 Eo1P Eo 2 P cos180  
2
2
or,
Carry out addition at t=0 I  I1  I 2  2 I1I 2 cos
  
where,  2  1  k  r2  r1 
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Incidence at an angle
Before slits
Difference in path length
a sin i
= a sin I in r1
i

r1

r2
After slits
Difference in path length
= a sin  in r2
a sin 
Now k(r2-r1) = - k a sin  + k a sin i
Thus  = ka (sin  - sini)
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
Reflection from dielectric layer
n1
n2
n1
– (n1n2)  = 12
– (n2 n1) ’=21
A
• Amplitude transmission coefficient
’
A’


x=0
• Assume phase of wave at O (x=0,
t=0) is 0
• Amplitude reflection co-efficient
’
– (n1n2)  =  12
– (n2 n1) ’=  21
O’
• Path O to O’ introduces a phase
change
O
t
k2 S 2 
x=t
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2
t
2 cos '
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Reflection from a dielectric layer
• At O:
– Incident amplitude
– Reflected amplitude
• At O’:
– Reflected amplitude
– Transmitted amplitude
• At A:
– Transmitted amplitude
– Reflected amplitude
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E = Eoe-iωt
ER = Eoe-iωt
i k2S2 t 
 ' Eoe
i k2S2 t 
  ' Eoe
i 2 k2 S2 t 
 ' ' Eoe
i 2k2S2 t 
 '  ' Eoe
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Reflection from a dielectric layer
•At A’
EA'  ' Eoei k1S1 t 
and
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.
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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’
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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’.
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A

z = 2t tan ’
A’
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Reflection from a dielectric surface
i 2k2S2 t 
i
EA    e  ' Eoe
A
i k1S1 t 

EA'   ' Eoe
z = 2t tan ’
A’
If we assume ’ ~ 1
and since ’ = ||
This is just interference between two sources with
equal amplitudes
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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,       2k S  k s   
2
1
2
2
1
1
Since
k2 = n2ko
and n1sin = n2sin’
Thus,



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k1=n1ko
(Snells Law)
 2k2 S 2  k1s1   
 t 
 2n2 ko 
  n1ko 2t tan ' sin   
 cos ' 
 2n2 kot cos '
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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
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2n2cos’ =  mo
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Haidinger’s Bands: Fringes of equal inclination
d
n1
n2
Beam splitter
P
x
1

1
f
Extended
source
Focal
plane
PI
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P2
Dielectric
slab
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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
•
•
•
•
•
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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
Extended source
n
n2
n
x
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Wedge between two plates
1 2
glass
glass
D
y
air
L
Path difference
= 2y
Phase difference  = 2ky - 
but not for 1)
(phase change for 2,
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
Minima 2y = mo/n
y
Look at p and p + 1 maxima
air
L
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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D
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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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Non-reflective Coatings
Since both paths have the same
phase change at the interfaces,
take only the path differences
into account.
1

2tm    m n
2

t
For destructive
interference
1

2tm    m 
2
n

550nm
t

 94.8nm
4n 41.45
Example:  = 550 nm,
no reflection
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Monochromatic light shines on a pair of identical glass
microscope slides that form a very narrow wedge. The top
surface of the upper slide and the bottom surface of the
lower slide have special coatings on them so that they
reflect no light. The inner two surfaces (A and B) have
nonzero reflectivities.
A top view of the slides looks like
1. I.
2. II.
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