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 1075: 18b

Answer

The phase difference is $~~18.2~rad$

Work Step by Step

We can find the path length difference as a multiple of $\lambda$: $\Delta L = d~sin~\theta$ $\Delta L = (4.24\times 10^{-6}~m)~sin~20.0^{\circ}$ $\Delta L = 1.45\times 10^{-6}~m$ We can find the phase difference as a multiple of $\lambda$: $\frac{\Delta L}{\lambda} = \frac{1.45\times 10^{-6}~m}{500\times 10^{-9}~m} = 2.9$ The phase difference is $~~2.9$ We can express the phase difference in radians: $(2.9)(2\pi~rad) = 18.2~rad$
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