ELEC 262

Problem Set 4
solutions
1. (a) The model developed in lecture predicts that the metal should become transparent for
frequencies equal to (and above) the plasma frequency.
The plot shown in lecture illustrates an
abrupt drop in the reflectivity (and therefore an increase in the transmission) at a photon energy
of about 16 electron volts (eV).
This is a good approximation of the photon energy
corresponding to the plasma frequency.
Using
h
= 4.13
×
10
15
eV sec, we find that
15
3.87 10 Hz
ν
==×
Eh
.
(b)
λ
= 2 nm corresponds to a frequency that is well above the plasma frequency:
ν
= c/
λ
= 1.5
×
10
17
Hz.
So the refractive index is very close to (but slightly
less
than) one:
2
2
1
0.9997
ν
=−
=
ν
P
n
(c) We compute the derivative
22
ν
ν
=
ν
ν −ν
P
P
dn
d
.
For
ν
>
ν
P
, this is always positive.
Thus the
material exhibits normal
dispersion.
2. The power radiated by an accelerating charge is given by the Larmor formula:
()
3
0
a
6c
=
q
Pt
πε
In this case, the acceleration is simply given by the centripetal acceleration of a circulating
electron: a = v
2
/R, where v is the velocity of the electron and R is the radius of the circle.
Putting
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 Spring '09
 Mittleman
 Light, Frequency, Snell's Law, refractive index, small angle approximation, Snell’s law

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