Answer
$L=5.15\mu m$
Work Step by Step
We know that
$N_2-N_1=2(\frac{2L}{\lambda})(n_2-n_1)$
This can be rearranged as:
$L=\frac{\lambda(N_2-N_1)}{2(n_2-n_1)}$
We plug in the known values to obtain:
$L=\frac{589(7.0)}{2(1.40-1.00)}=5.15\mu m$
Fundamentals of Physics Extended (10th Edition)
by Halliday, David; Resnick, Robert; Walker, Jearl
Published by
Wiley
ISBN 10:
1-11823-072-8
ISBN 13:
978-1-11823-072-5
Chapter 35 - Interference - Problems - Page 1078: 80
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