Transcript Propagation of waves - Dalhousie University

Electromagnetic waves:
Reflection, Transmission
and Interference
Monday October 28, 2002
1
Amplitude Transmission & Reflection
For normal incidence
Amplitude reflection
Amplitude transmission
n n 
1

12   2
 n2  n1 
 12 
2v2
2n1

0
v1  v2 n1  n2
Suppose these are plane waves
o i k1xt 
f1  E e
o i k2 xt 
f 2  ET e
o i k1xt  i
g1  ER e
e
2
Intensity reflection
Amplitude reflection co-efficient
12
ER i
i  n2  n1 

 o e  e 
E
 n2  n1 
o
and intensity reflection
R12


1
v1 1 E Ro
 2
1
v1 1 E o
2


2
2
 n2  n1 
I
2

 R  12



Io
 n2  n1 
2
3
Intensity transmission
Intensity transmission
 
 
1
o 2
2
v2 2 ET
o
IT 2
n2  ET 
T12 


I o 1 v  E o 2 n1  E o 
1 1
2
n2
2
T12   12    12 21
n1
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
a

r1

r2

y
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

 it
E1  Eo1e

 it
E2  Eo2e
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?

E1P

Eo1 i kr1 t 
  e
r1

E2 P

Eo 2 i kr2 t 
  e
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

I P  v E P
Intensity at P
 2  *
EP  EP  E P


* *
 E1P  E2 P  E1P  E 2 P


2

1
vE oP2
2

 E12P  E 22P  E1P  E *2 P  E 2 P  E1*P
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Interference Effects

Are represented by the last two terms

If the fields are perpendicular

then,

and,
E1P  E*2 P  E2 P  E1*P
2
 2
2
2
E P  E1 P  E 2 P
2
v E P  v E1P  v E 2 P
2
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
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Interference terms
 *
E1P  E 2 P  __________
__________
________
Eo1Eo 2 i kr2 r1 

e
2
r
* 
E1 P  E 2 P  __________
__________
________
Eo1 Eo 2 i kr2 r1 

e
2
r
* 
* 
E1P  E 2 P  E1P  E 2 P  __________
_______
Eo1 Eo 2
2
cos2  1 
2
r
where,
  
2  1  k  r2  r1 
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Intensity – Young’s double slit diffraction
  
2  1  k  r2  r1 
Phase difference of beams occurs because of a path difference!
Eo2P  Eo21P  Eo22 P  2Eo1P Eo 2 P cos2  1 
I P  I1P  I 2P  2 I1P I 2P cos2  1 
13
Young’s Double slit diffraction
I P  I1P  I 2P  2 I1P I 2P cos2  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
Hence
r1
r2
   
k  r2  k  r1  k r1  r2 
Thus r2 – r1 = a sin 
a
r2-r1
14
Intensity maxima and minima
Maxima for,
  2  1  0,2 ,4 ...  2m
ka sin   2m
If I1P=I2P=Io
or
sin   m

a
m  0,1,2...
I MAX  I1P  I 2P  2 I1P I 2 P cos2  1   4I o
1

  2  1   ,3 ...  2 m  
2

1

sin    m  
2 a

Minima for,
If I1P=I2P=Io
I MIN  I1P  I 2P  2 I1P I 2 P cos2  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 ymax  L m  
ymin

 a
1

 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 = 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
E  Eo sin k  r  t   
Now write,
as,
E  Eo sin t   
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Phasors

E1  Eo1P sin t  1 
Imagine disturbance
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 cos180  
2
Carry out addition at t=0
2
or,
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

r1
S
screen

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
a sin i
= a sin I in r1
After slits
Difference in path length
= a sin  in r2
i

r1

r2

a sin 
Now k(r2-r1) = - k a sin  + k a sin i
Thus  = ka (sin  - sini)
23
Reflection from dielectric layer
n1
n2
n1


A

’
A’


Assume phase of wave at
O (x=0, t=0) is 0
Amplitude reflection coefficient
’
O’


O
Amplitude transmission
co-efficient


x=0
t

x=t
(n1n2)  = 12
(n2 n1) ’=21
(n1n2)  =  12
(n2 n1) ’=  21
Path O to O’ introduces a
2 t
phase
change
k2 S 2 
2 cos '
24
Reflection from a dielectric layer

At O:
 Incident
amplitude
 Reflected amplitude

At O’:
 Reflected
E = Eoe-iωt
ER = Eoe-iωt
 ' Eoei k2S2 t 
i k2S2 t 
amplitude
 Transmitted amplitude

  ' Eoe
 ' ' Eoei 2k2S2 t 
 Transmitted amplitude '  ' E ei 2k2S2 t 
o
At A:
 Reflected
amplitude
25
Reflection from a dielectric layer
•At A’
i k1S1 t 
and
A
EA'  ' Eoe
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.
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’
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