Physics Technology Update (4th Edition)

Published by Pearson
ISBN 10: 0-32190-308-0
ISBN 13: 978-0-32190-308-2

Chapter 28 - Physical Optics: Interference and Diffraction - Problems and Conceptual Exercises - Page 1006: 45

Answer

$36^{\circ}$

Work Step by Step

We can find the required angle as follows: $\lambda=\frac{v}{f}$ $\implies \lambda=\frac{340}{1300}=0.2615m$ We know that $W sin\theta=m\lambda$ This can be rearranged as: $\theta=sin^{-1}(\frac{m\lambda}{W})$ We plug in the known values to obtain: $\theta=sin^{-1}[\frac{(1)(0.2615m)}{0.84m}]$ $\theta=18^{\circ}$ Now the angle between the first two minima is given as $\alpha=2\theta$ $\implies \alpha=2(18^{\circ})$ $\alpha=36^{\circ}$
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