1 electromagnetic waves: two source interference monday november 4, 2002
TRANSCRIPT
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Electromagnetic waves: Two source Interference
Monday November 4, 2002
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Reflection from dielectric layer Assume phase of wave at O
(x=0, t=0) is 0 Amplitude reflection co-
efficient (n1n2) = 12
(n2 n1) ’=21
Amplitude transmission co-efficient (n1n2) = 12
(n2 n1) ’= 21
Path O to O’ introduces a phase change
nn22nn11 nn11
AA
O’O’
OO
ttx = 0x = 0 x = tx = t
A’A’’’
’’
'cos
2
222
tSk
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Reflection from a dielectric layer At O:
Incident amplitude E = Eoe-iωt
Reflected amplitude ER = Eoe-iωt
At O’: Reflected amplitude Transmitted amplitude
At A: Transmitted amplitude Reflected amplitude
tSkioeE
22' tSki
oeE 22'
tSkioeE
222'' tSki
oeE 222''
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Reflection from a dielectric layer
tSkioA eEE 11'' AA
A’A’
z = 2t tan z = 2t tan ’’
and and ΔΔSS11= z sin = z sin = 2t tan = 2t tan ’ sin ’ sin
•At A’
96.0'2.0'12
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andnn
nn
Since,Since,
The reflected intensities ~ 0.04IThe reflected intensities ~ 0.04Ioo and both beams (A,A’) will have and both beams (A,A’) will have
almost the same intensity.almost the same intensity.Next beam, however, will have ~ |Next beam, however, will have ~ |||33EEoo which is very small which is very small
Thus assume interference at Thus assume interference at , and need only consider the two , and need only consider the two beam problem.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’
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’.
AA
A’A’
z = 2t tan z = 2t tan ’’
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Reflection from a dielectric surface
tSkioA eEE 11''
AA
A’A’
z = 2t tan z = 2t tan ’’
tSkio
iA eEeE 222'
If we assume If we assume ’ ~ 1’ ~ 1
and since and since ’ = |’ = |||
This is just interference between two sources with equal amplitudes This is just interference between two sources with equal amplitudes
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Reflection from a dielectric surface
cos2 2121 IIIII
'cos2
sin'tan2'cos
2
2
2
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1122
tkn
tknt
kn
skSk
o
oo
tSkio
iA eEeE 11
' tSki
oA eEE 222'
where,where,
Since Since kk22 = n = n22kkoo kk11=n=n11kkoo
and nand n11sinsin = n = n22sinsin’’ (Snells Law)(Snells Law)
Thus, Thus,
112212 2 skSk
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Reflection from a dielectric surface
Since ISince I11 ~ I ~ I22 ~ I ~ Ioo
Then, I = 2IThen, I = 2Ioo(1+cos(1+cos))
Constructive interferenceConstructive interference
Destructive interferenceDestructive interference
= = 2m 2m = 2ktcos = 2ktcos’ - ’ - (here k=n(here k=n22kkoo))
2ktcos2ktcos’ = ’ = (2m+1)(2m+1)
ktcosktcos’ = ’ = (m+1/2)(m+1/2)
2n2n22coscos’ = ’ = (m+1/2) (m+1/2)oo
2n2n22coscos’ = ’ = m moo
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Haidinger’s Bands: Fringes of equal inclination
dd
nn22
nn11
Beam splitterBeam splitter
ExtendedExtendedsourcesource
PPII PP22
PP
xx
ff
FocalFocalplaneplane
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DielectricDielectricslabslab
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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
cos2 2121 IIIII
where, where,
kt
kt
2
'cos2
Then if film varies in thickness we will see fringes as we move our eye.Then if film varies in thickness we will see fringes as we move our eye.
These are termed These are termed Fizeau fringesFizeau fringes..
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Fizeau Fringes
Extended sourceExtended source
Beam splitterBeam splitter
xx nn
nn22
nn
kt
kt
2
'cos2
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Wedge between two plates11 22
glassglassglassglass
airair
DDyy
LL
Path difference = 2yPath difference = 2yPhase difference Phase difference = 2ky - = 2ky - (phase change for 2, but not for 1) (phase change for 2, but not for 1)
Maxima 2y = (m + ½) Maxima 2y = (m + ½) oo/n/n
Minima 2y = mMinima 2y = moo/n/n
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Wedge between two plates
Maxima 2y = (m + ½) Maxima 2y = (m + ½) oo/n/n
Minima 2y = mMinima 2y = moo/n/n
Look at p and p + 1 maximaLook at p and p + 1 maxima
yyp+1p+1 – y – ypp = = oo/2n /2n ΔΔxx
where where ΔΔx = distance between adjacent maximax = distance between adjacent maxima
Now if diameter of object = DNow if diameter of object = D
Then LThen L = D = D
And (D/L) And (D/L) ΔΔx= x= oo/2n or /2n or D = D = ooL/2n L/2n ΔΔxx
airair
DDyy
LL
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Wedge between two platesCan be used to test the quality of surfacesCan be used to test the quality of surfaces
Fringes follow contour of constant yFringes follow contour of constant y
Thus a flat bottom plate will give straight fringes, otherwise Thus a flat bottom plate will give straight fringes, otherwise ripples in the fringes will be seen. ripples in the fringes will be seen.