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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This note was uploaded on 11/12/2011 for the course PHYS 2001 taught by Professor Sprunger during the Fall '08 term at LSU.
 Fall '08
 SPRUNGER
 Physics

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