Fundamentals of Physics Extended (10th Edition)

Published by Wiley
ISBN 10: 1-11823-072-8
ISBN 13: 978-1-11823-072-5

Chapter 35 - Interference - Problems - Page 1074: 2a

Answer

$\frac{L}{\lambda} = \frac{1}{4}$

Work Step by Step

The path length difference is $2L$ Note that initially, the waves are exactly out of phase. To be exactly in phase, the path length difference is $~~(\frac{\lambda}{2}+m~\lambda)$, where $m$ is some integer That is, the path length difference is $~~(\frac{2m+1}{2}~\lambda)$, where $m$ is some integer To find the smallest value of $\frac{L}{\lambda}$, we can let $m = 0$: $2L = (\frac{2m+1}{2}~\lambda)$ $2L = \frac{\lambda}{2}$ $\frac{L}{\lambda} = \frac{1}{4}$
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