Electromagnetic waves: Interference

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Electromagnetic waves Interference

• Wednesday October 30, 2002

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Two-source interference
What is the nature of the superposition of
radiation from two coherent sources. The classic
example of this phenomenon is Youngs Double Slit
Experiment
Plane wave (?)
P
S1
y
?
x
a
S2
L
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Interference terms
where,
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Intensity Youngs double slit diffraction
Phase difference of beams occurs because of a
path difference!
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Youngs Double slit diffraction

• I1P intensity of source 1 (S1) alone
• I2P intensity of source 2 (S2) alone
• Thus IP can be greater or less than I1I2
  depending on the values of ?2 - ?1
• In Youngs experiment r1 r2 k
• Hence
• Thus r2 r1 a sin ?

r1
r2
a
r2-r1
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Intensity maxima and minima
Maxima for,
If I1PI2PIo
Minima for,
If I1PI2PIo
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Fringe Visibility or Fringe Contrast
To measure the contrast or visibility of these
fringes, one may define a useful quantity, the
fringe visibility
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Co-ordinates on screen

• Use sin? tan ? y/L
• Then
• These results are seen in the following
  Interference pattern

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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,

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Phasors

• Imagine disturbance given in the form

?f2-f1
f2
f1
Carry out addition at t0
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Other forms of two-source interference
Lloyds mirror
screen
S
S
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Other forms of two source interference
Fresnel Biprism
S1
S
s2
d
s
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Other sources of two source interference
Altering path length for r2
r1
r2
n
With dielectric thickness d
kr2 kDd ko(r2-d) nkod ko(r2-d)
kor2 ko(n-1)d
Thus change in path length k(n-1)d
Equivalent to writing, ?2 ?1 ko(n-1)d
Then ? kr2 kor1 ko(r2-r1) ko(n-1)d
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Incidence at an angle
Before slits Difference in path length
a sin ?i
?i
a sin ?I in r1
?
After slits Difference in path length
a sin ?
a sin ? in r2
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

• Assume phase of wave at O (x0, t0) is 0
• Amplitude reflection co-efficient
• (n1?n2) ? ?12
• (n2 ?n1) ??21
• Amplitude transmission co-efficient
• (n1?n2) ? ? 12
• (n2 ?n1) ? ? 21
• Path O to O introduces a phase change

A
?
A
O
?
?
O
?
t
x t
x 0
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Reflection from a dielectric layer

• At O
• Incident amplitude E Eoe-i?t
• Reflected amplitude ER Eoe-i?t?
• At O
• Reflected amplitude
• Transmitted amplitude
• At A
• Transmitted amplitude
• Reflected amplitude

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Reflection from a dielectric layer

• At A

A
and ?S1 z sin ? 2t tan ? sin ?
?
z 2t tan ?
Since,
A
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.

A
?
z 2t tan ?
A
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Reflection from a dielectric surface
A
?
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
where,
Since k2 n2ko k1n1ko
and n1sin? n2sin? (Snells Law)
Thus,
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Reflection from a dielectric surface
Since I1 I2 Io Then, I 2Io(1cos?)
Constructive interference

• ? 2m? 2ktcos? - ? (here kn2ko)
• 2ktcos? ?(2m1)?
• ktcos? ?(m1/2)?
• 2n2cos? ? (m1/2)?o

Destructive interference
2n2cos? ? m?o
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Haidingers Bands Fringes of equal inclination
d
n1
n2
Beam splitter
P
?1
x
?
?1
f
Extended source
Focal plane
Dielectric slab
PI
P2
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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
where,
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
Beam splitter
Extended source
n
n2
n
x
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Wedge between two plates
1
2
glass
D
y
glass
air
L
Path difference 2y Phase difference ?
2ky - ? (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 Minima 2y m?o/n
D
y
air
Look at p and p 1 maxima yp1 yp ?o/2n ?
?x? where ?x distance between adjacent
maxima Now if diameter of object D Then L?
D And (D/L) ?x ?o/2n or D ?oL/2n ?x
L
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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.