Transcript Lecture 20: Electromagnetic Waves

Electromagnetic waves:
Reflection, Refraction
and Interference
Friday October 25, 2002
1
Optical cooling
v
k
Photons
m
Because of its motion, the atoms
“see” an incoming photon with a
frequency Doppler-shifted upward
by,
k v
v
 fL
2
c
Laser frequency (fL) chosen to be
just below the resonance frequency
of the atom (fo)
fo = fL(1+v/c)
2
Optical Cooling
Photons of the right frequency will be absorbed by the atom, whose speed
is reduced because of the transfer of the momentum of the photon,
mvi  k L  mv f  0
mv  k L
Emission occurs when the atom falls back to its ground state. However, the
emission is randomly directed
An atom moving in the opposite direction, away from the light source, sees
photons with a frequency, fL(1-v/c), far enough from fo that there can be
little or no absorption, and therefore no momentum gain.
Radiation pressure force : As v decreases, there must be a corresponding
change in the laser frequency…..
One must have three mutually perpendicular laser beams in order to
reduce the speed of the atoms in all directions.
3
Reflection, Transmission
and Interference of EM
waves
4
Reflection and Transmission at an interface
Normal Incidence – Two media characterized by v1, v2
incident
transmitted
f1 x  v1t 
f 2 x  v2t 
reflected
g1 x  v1t 
1
2
5
Reflection and Transmission at an interface

Require continuity of amplitude at interface:
f1 + g1 = f2

Require continuity of slope at interface:
f1’ + g1’ = f2’

Recall u = x – vt
f
f
f '

,
x du
f
f u

 vf '
t
u t
6
Reflection and Transmission at an interface
Continuity of slope requires,

1 f1 1 g1
1 f 2


v1 t v1 t
v2 t
or,
f1 g1 v1 f 2


t
t
v2 t
7
Reflection and Transmission at an interface



Integrating from t = - to t = t
Assuming f1(t = - ) = 0
Then,
v1
f1  g1 
f2
v2
8
Amplitude transmission co-efficient ()
 12
f2
2v2


f1 v1  v2
 21
f2
2v1


f1 v1  v2
Medium 1 to 2
Medium 2 to 1
9
Amplitude reflection co-efficient ()
12
g1 v2  v1


f1 v1  v2
At a dielectric interface
12
v1 
c
n1
v2 
c
n2
 n2  n1 
v2  v1


 
v2  v1
 n2  n1 
10
Phase changes on reflection from a dielectric interface
 n2  n1 

12  
 n2  n1 
n2 > n1
n2<n1
Less dense to more dense
e.g. air to glass
12  12 e
i
 phase change on reflection
More dense to less dense
e.g. glass to air
12  12 ei 0  12
No phase change on reflection
11
Phase changes on transmission through a
dielectric interface
 12
2v 2
2n1


0
v1  v2 n1  n2
Thus there is no phase change on transmission
12
Amplitude Transmission & Reflection
For normal incidence
Amplitude reflection
12
Amplitude transmission
 n2  n1 

 
 n2  n1 
 12 
2v 2
2n1

0
v1  v2 n1  n2
Suppose these are plane waves
o i  k1 x t 
f1  E e
o i  k 2 x t 
f 2  ET e
o i  k1 x t  i
g1  ER e
e
13
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
14
Intensity transmission
Intensity transmission
 
 
1
v2 2 ETo
IT 2
T12 

1
Io
v11 E o
2
2
2
n2  E

n1  E
o
T
o




2
n2
2
T12   12    12 21
n1
and in general
R+T=1
(conservation of energy)
15
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
16
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
17
Young’s double slit experiment


slits at x = 0
The fields at S1 and S2 are

 it
E1  Eo1e

 it
E2  Eo 2e
Assume that the slits might have different width
and therefore Eo1  Eo2
18
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)
19
Young’s Double slit experiment
The total amplitude at P



E P  E1P  E2 P

I P  v E P
Intensity at P
 2  *
EP  EP  E P


* *
 E1P  E2 P  E1 P  E 2 P


2

1
vE o2P
2

 E 12P  E 22P  E 1 P  E *2 P  E 2 P  E 1*P
20
Interference Effects

Are represented by* the last two* terms

If the fields are perpendicular

then,

and,
E1 P  E 2 P  E 2 P  E1 P
v E P
2
 2
2
2
E P  E1 P  E 2 P
2
 v E 1 P  v E 2 P
2
I P  I 1 P  I 2 P
In the absence of interference, the total intensity is a simple sum
21
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
22
Interference terms
 *
E1P  E 2 P  ____________________________
Eo1 Eo 2 i kr2 r1 

e
2
r
* 
E 1 P  E 2 P  ____________________________
Eo1 Eo 2 i kr2  r1 

e
2
r
* 
* 
E 1 P  E 2 P  E 1 P  E 2 P  _________________
2
where,
Eo1 Eo 2
cos 2  1 
2
r
  
 2  1  k  r2  r1 
23
Intensity – Young’s double slit diffraction
  
 2  1  k  r2  r1 
Phase difference of beams occurs because of a path difference!
2
E oP
 E o21P  E o22 P  2E o1P E o 2 P cos2  1 
I P  I1P  I 2 P  2 I1P I 2 P cos2  1 
24
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
Hence
r1
r2
   
k  r2  k  r1  k r1  r2 
Thus r2 – r1 = a sin 
a
r2-r1
25