Electromagnetic waves: Reflection, Refraction and Interference

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Electromagnetic waves Reflection, Refraction and
Interference

• Friday October 25, 2002

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Optical cooling
Photons
v
k
m
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(1v/c)
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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.
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Reflection, Transmission and Interference of EM
waves
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Reflection and Transmission at an interface
Normal Incidence Two media characterized by v1,
v2
incident
transmitted
reflected
1
2
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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

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

• Integrating from t -? to t t
• Assuming f1(t -? ) 0
• Then,

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Amplitude transmission co-efficient (?)
Medium 1 to 2
Medium 2 to 1
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Amplitude reflection co-efficient (?)
At a dielectric interface
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Phase changes on reflection from a dielectric
interface
n2 gt n1
n2ltn1
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
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Phase changes on transmission through a
dielectric interface
Thus there is no phase change on transmission
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Amplitude Transmission Reflection
For normal incidence
Amplitude reflection
Amplitude transmission
Suppose these are plane waves
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Intensity reflection
Amplitude reflection co-efficient
and intensity reflection
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Intensity transmission
Intensity transmission
and in general
R T 1
(conservation of energy)
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Two-source interference
What is the nature of the superposition of
radiation from two coherent sources. The classic
example of this phenomenon is Youngs Double Slit
Experiment
Plane wave (?)
P
S1
y
?
x
a
S2
L
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Youngs 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 gtgt 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

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Youngs 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
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Youngs double slit experiment
What are the corresponding E-fields at P?
Since L gtgt a (? small) we can put r r1
r2 We can also put k1 k2 2?/?
(monochromatic source)
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Youngs Double slit experiment
The total amplitude at P
Intensity at P
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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
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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.

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Interference terms
where,
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Intensity Youngs double slit diffraction
Phase difference of beams occurs because of a
path difference!
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Youngs 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 I1I2
  depending on the values of ?2 - ?1
• In Youngs experiment r1 r2 k
• Hence
• Thus r2 r1 a sin ?

r1
r2
a
r2-r1