# Electromagnetic waves Reflection, Transmission and Interference

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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

v11 ERo
   2
I           n  n1 
2

R12    2                     R  12   2
2
n n   
1
2

v11 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
     it                     it
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               vE 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 kr2 r1 
     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 cos2  1 
2     2      2

I P  I1P  I 2 P  2 I1P I 2 P cos2  1 
13
Young’s Double slit diffraction
I P  I1P  I 2 P  2 I1P I 2 P 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
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 cos2  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 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 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 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   

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 cos180   
2          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

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  - sini)

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
   (n1n2)  = 12
’
A’                                         (n2 n1) ’=21
O’
              ’                 Amplitude transmission
co-efficient
     O
   (n1n2)  =  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  k1S1 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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