Fundamentals of Physics Extended (10th Edition)

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

Chapter 17 - Waves-II - Problems - Page 508: 23f

Answer

$x = 41.2~m$

Work Step by Step

We can find the path length difference when the phase difference is $1.50~\lambda$: $1.50~\lambda = (1.50)(2.00~m) = 3.00~m$ We can form a right angle triangle with sides of length $d$, $x$, and $x+3.00$ We can find $x$: $x^2+d^2 = (x+3.00)^2$ $x^2+d^2 = x^2+6.00~x+9.00$ $d^2 = 6.00~x+9.00$ $6.00~x = d^2-9.00$ $x = \frac{d^2-9.00}{6.00}$ $x = \frac{(16.0)^2-9.00}{6.00}$ $x = 41.2~m$
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