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Electromagnetic waves:
Reflection, Transmission
and Interference
Monday October 28, 2002
1
Amplitude Transmission & Reflection
For normal incidence
Amplitude reflection Amplitude transmission
n n 2v 2 2n1
12 2
n n
1
12 0
2 1 v1 v2 n1 n2
Suppose these are plane waves
o i k1 x t o i k 2 x t o i k1 x t i
f1 E e f 2 ET e g1 ER e e
2
Intensity reflection
Amplitude reflection co-efficient
i n2 n1
o
ER i
12 o e e n n
E 2 1
and intensity reflection
1
v11 ERo
2
I n n1
2
R12 2 R 12 2
2
n n
1
2
v11 E o 2 Io 2 1
3
Intensity transmission
Intensity transmission
1
v2 2 ET
o 2
n2 E
2
T12 12 12 21
IT 2 o n2
2
T12 T
Io 1 v E o 2 n1 E
o
n1
1 1
2
and in general
R+T=1
(conservation of energy)
4
Two-source interference
What is the nature of the superposition of radiation
from two coherent sources. The classic example of
this phenomenon is Young’s Double Slit Experiment
Plane wave ()
P
S1 r1
y
r2
a x
S2
L
5
Young’s Double slit experiment
Assumptions
Monochromatic, plane wave
Incident on slits (or pin hole), S1, S2
separated by distance a (centre to centre)
Observed on screen L >> a (L- meters, a –
mm)
Two sources (S1 and S2) are coherent and in
phase (since same wave front produces both
as all times)
Assume slits are very narrow (width b ~ )
so radiation from each slit alone produces
uniform illumination across the screen
6
Young’s double slit experiment
slits at x = 0
The fields at S1 and S2 are
it it
E1 Eo1e E2 Eo 2e
Assume that the slits might have different width
and therefore Eo1 Eo2
7
Young’s double slit experiment
What are the corresponding E-fields at P?
Eo1 i k r1 t
Eo 2 i k r2 t
E1P e E2 P e
r1 r2
Since L >> a ( small) we can put r = |r1| = |r2|
We can also put |k1| = |k2| = 2/ (monochromatic source)
8
Young’s Double slit experiment
The total amplitude at P
EP E1P E2 P
2 1
Intensity at P I P v EP vE oP
2
2
2 *
EP EP E P
* *
E1P E2 P E1P E 2 P
E12P E 22P E1P E *P E 2 P E1*P
2
9
Interference Effects
Are represented by the last two terms
E1P E*P E 2 P E1P
2
*
If the fields are perpendicular
2
EP E1P E 2 P
2 2
then, 2 2 2
v E P v E1P v E 2 P
and, I P I 1P I 2 P
In the absence of interference, the total intensity is a simple sum 10
Interference effects
Interference requires at least parallel components
of E1P and E2P
We will assume the two sources are polarized
parallel to one another (i.e.
E1P E 2 P E1P E 2 P cos E1P E 2 P
11
Interference terms
*
E1P E 2 P ____________________________
Eo1 Eo 2 i kr2 r1
2
e
r
*
E 1 P E 2 P ____________________________
Eo1 Eo 2 i k r2 r1
2
e
r
* *
E 1 P E 2 P E1 P E 2 P _________________
cos 2 1
Eo1 Eo 2
2 2
r
where, 2 1 k r2 r1 12
Intensity – Young’s double slit diffraction
2 1 k r2 r1
Phase difference of beams occurs because of a path difference!
EoP Eo1P Eo 2 P 2Eo1P Eo 2 P cos2 1
2 2 2
I P I1P I 2 P 2 I1P I 2 P cos2 1
13
Young’s Double slit diffraction
I P I1P I 2 P 2 I1P I 2 P cos2 1
I1P = intensity of source 1 (S1) alone
I2P = intensity of source 2 (S2) alone
Thus IP can be greater or less than I1+I2
depending on the values of 2 - 1
In Young’s experiment r1 ~|| r2 ~|| k
r1
Hence
r2
k r2 k r1 k r1 r2
a
Thus r2 – r1 = a sin
r2-r1 14
Intensity maxima and minima
Maxima for,
2 1 0,2 ,4 ... 2m
ka sin 2m or sin m m 0,1,2...
a
If I1P=I2P=Io I MAX I1P I 2 P 2 I1P I 2 P cos2 1 4I o
1
Minima for, 2 1 ,3 ... 2 m
2
1
sin m
2 a
If I1P=I2P=Io I MIN I1P I 2 P 2 I1P I 2 P cos2 1 0
15
Fringe Visibility or Fringe Contrast
To measure the contrast or visibility of these fringes, one
may define a useful quantity, the fringe visibility:
I MAX I MIN
V
I MAX I MIN
16
Co-ordinates on screen
Use sin ≈ tan = y/L
Then ym ax L m
a
1
ym in L m
2 a
These results are seen in the following
Interference pattern
17
Phasor Representation of wave addition
Phasor representation of a wave
E.g. E = Eosint 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
E Eo sin k r t
Now write, as,
E Eo sin t
18
Phasors
Imagine disturbance E1 Eo1P sin t 1
given 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 cos180
2 2
or ,
Carry out addition at t=0
I I1 I 2 2 I1 I 2 cos
where, 2 1 k r2 r1
19
Other forms of two-source interference
Lloyd’s mirror
screen
S
r1
r2
S’
20
Other forms of two source interference
Fresnel Biprism
S1
S
s2
d s
21
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
22
Incidence at an angle
Before slits
Difference in path length
r1
a sin i i
= a sin I in r1
After slits
Difference in path length
r2
= a sin in r2 a sin
Now k(r2-r1) = - k a sin + k a sin i
Thus = ka (sin - sini)
23
Reflection from dielectric layer
n1 n2 n1
Assume phase of wave at
O (x=0, t=0) is 0
Amplitude reflection co-
A
efficient
(n1n2) = 12
’
A’ (n2 n1) ’=21
O’
’ Amplitude transmission
co-efficient
O
(n1n2) = 12
(n2 n1) ’= 21
Path O to O’ introduces a
t 2
x=0 x=t phase change t
k S
2 cos '
2 2
24
Reflection from a dielectric layer
At O:
Incidentamplitude E = Eoe-iωt
Reflected amplitude ER = Eoe-iωt
i k 2 S 2 t
At O’: ' Eo e
Reflected amplitude ' Eo e i k S 2 2 t
Transmitted amplitude
i 2 k 2 S 2 t
At A: ' ' Eo e
Transmitted amplitude ' ' E e i 2 k 2 S 2 t
o
Reflected amplitude
25
Reflection from a dielectric layer
•At A’
E A' ' Eo e i k1S1 t
A
and ΔS1= z sin = 2t tan ’ sin
z = 2t tan ’
Since,
A’
n2 n1
' 0.2 and ' 0.96
n2 n1
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. 26
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’
27
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