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ELEC 262 – Problem set 10 solutions
1(a) The coherence time is the inverse of the spectral width, and the coherence length is c
0
divided by the coherence time,
L
c
= c
0
/
∆ν
.
In this case, we compute
∆ν
according to:
()
210
2
1
11
c
∆ν = ν − ν =
λ − λ
, so
1
8
1 632.799 1 632.801
2 10 nm
0.2 m
−
=−
=
×
=
c
L
.
(b) We have seen that the spacing between longitudinal modes in a laser is given by the inverse
of the cavity roundtrip time.
Since the roundtrip length is 0.6 m, we can find the mode spacing
by
8
0
15
1
0
H
z
δν =
τ
=
= ×
RT
RT
cL
.
Divide this into
∆ν
to find the number of modes:
3
∆ν δν =
.
Only three distinct modes can oscillate in this laser.
2. The de Broglie wavelength of the electrons is related to their momentum by
λ=
=
hp hm
v
.
The energy of each electron is 30 keV, which is the same as 4.8
×
10
15
Joules.
We can compute
the velocity
v
because the energy of the wave is
2
1
2
mv
.
Thus we have the velocity as
8
21
0
m
/
s
==
vE
m
, which means that the wavelength is 7.3
×
10
12
meters.
The diffraction
limited spot size for a focused wave is 2
λ
(f/#).
In this case, that works out to 60
λ
, or 0.44
nanometers.
A value of 1.22
λ
(f/#) = 0.27 nanometers is also an acceptable answer here, since
the problem does not specify whether the electron beam is to be treated as a Gaussian beam or a
plane wave.
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 Spring '09
 Mittleman

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