56. We denote the two wavelengths as λ and λ
, respectively. We apply Eq. 36-40 to both wavelengths and
take the difference:
N
− N =
2L
λ
−
2L
λ
= 2L
1
λ
−
1
λ
.
We now require N
− N = 1 and solve for L:
L
=
1
2
1
λ
−
1
λ
−1
=
1
2
1
589.10 nm
−
1
589.59 nm
−1
=
3.54
× 10
5
nm = 354 µm .