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 .