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65.
(a) If both mirrors are perfectly reﬂecting, there is a node at each end of the crystal. With one end
partially silvered, there is a node very close to that end. We assume nodes at both ends, so there are
an integer number of halfwavelengths in the length of the crystal. The wavelength in the crystal
is
λ
c
=
λ/n
,where
λ
is the wavelength in a vacuum and
n
is the index of refraction of ruby. Thus
N
(
λ/
2
n
)=
L
,where
N
is the number of standing wave nodes, so
N
=
2
nL
λ
=
2(1
.
75)(0
.
0600 m)
694
×
10
−
9
m
=3
.
03
×
10
5
.
(b) Since
λ
=
c/f
,where
f
is the frequency,
N
=2
nLf/c
and ∆
N
=(2
nL/c
)∆
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 Fall '08
 SPRUNGER
 Physics

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