56. We denote the two wavelengths as
λ
and
λ
0
, respectively. We apply Eq. 3640 to both wavelengths and
take the diference:
N
0
−
N
=
2
L
λ
0
−
2
L
λ
=2
L
µ
1
λ
0
−
1
λ
¶
.
We now require
N
0
−
N
= 1 and solve For
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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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