p36 056

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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 .


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