Answer
$d=594nm$
Work Step by Step
We know that
$2d=m(\frac{\lambda}{n_{0il}})$
This can be rearranged as:
$d=m(\frac{\lambda}{2n_{0il}})$
We plug in the known values to obtain:
$d=3\times (\frac{475}{2\times1.20})=594nm$
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 1080: 103b
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