1. Electromagnetic waves: Reflection, Refraction and Interference
Friday October 25, 2002

2. Optical cooling v k m Photons
Because of its motion, the atoms “see” an incoming photon with a frequency Doppler-shifted upward by,
Laser frequency (fL) chosen to be just below the resonance frequency of the atom (fo)
fo = fL(1+v/c)

3. 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,
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.

4. Reflection, Transmission and Interference of EM waves

5. Reflection and Transmission at an interface
Normal Incidence – Two media characterized by v1, v2
incident
transmitted
reflected 1 2

6. 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

7. Reflection and Transmission at an interface
Continuity of slope requires,
or,

8. Reflection and Transmission at an interface
Integrating from t = - to t = t
Assuming f1(t = - ) = 0
Then,

9. Amplitude transmission co-efficient ()
Medium 1 to 2
Medium 2 to 1

10. Amplitude reflection co-efficient ()
At a dielectric interface

11. Phase changes on reflection from a dielectric interface
n2 > n1
n2<n1
Less dense to more dense
e.g. air to glass
More dense to less dense
e.g. glass to air
 phase change on reflection
No phase change on reflection

12. Phase changes on transmission through a dielectric interface
Thus there is no phase change on transmission

13. Amplitude Transmission & Reflection
For normal incidence
Amplitude reflection
Amplitude transmission
Suppose these are plane waves

14. Intensity reflection Amplitude reflection co-efficient
and intensity reflection

15. Intensity transmission
and in general
R + T = 1
(conservation of energy)

16. 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 y  x a S2 L

17. 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

18. Young’s double slit experiment
slits at x = 0
The fields at S1 and S2 are
Assume that the slits might have different width and therefore Eo1  Eo2

19. Young’s double slit experiment
What are the corresponding E-fields at P?
Since L >> a ( small) we can put r = |r1| = |r2|
We can also put |k1| = |k2| = 2/ (monochromatic source)

20. Young’s Double slit experiment
The total amplitude at P
Intensity at P

21. Interference Effects Are represented by the last two terms
If the fields are perpendicular
then,
and,
In the absence of interference, the total intensity is a simple sum

22. 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.

23. Interference terms
where,

24. Intensity – Young’s double slit diffraction
Phase difference of beams occurs because of a path difference!

25. Young’s Double slit diffraction
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
Thus r2 – r1 = a sin  r1 r2 a
r2-r1