Physics for Scientists and Engineers: A Strategic Approach with Modern Physics (4th Edition)

Published by Pearson
ISBN 10: 0133942651
ISBN 13: 978-0-13394-265-1

Chapter 35 - Optical Instruments - Exercises and Problems - Page 1017: 20

Answer

The angular spread of the light in the glass is $~~0.38^{\circ}$

Work Step by Step

We can find the angle $\theta_r$ of the red light after it enters the glass: $n_r~sin~\theta_r = n_1~sin~\theta_1$ $sin~\theta_r = \frac{n_1~sin~\theta_1}{n_r}$ $\theta_r = sin^{-1}~(\frac{n_1~sin~\theta_1}{n_r})$ $\theta_r = sin^{-1}~(\frac{1.00~sin~30^{\circ}}{1.52})$ $\theta_r = sin^{-1}~(0.328947)$ $\theta_r = 19.20^{\circ}$ We can find the angle $\theta_r$ of the violet light after it enters the glass: $n_v~sin~\theta_v = n_1~sin~\theta_1$ $sin~\theta_v = \frac{n_1~sin~\theta_1}{n_v}$ $\theta_v = sin^{-1}~(\frac{n_1~sin~\theta_1}{n_v})$ $\theta_v = sin^{-1}~(\frac{1.00~sin~30^{\circ}}{1.55})$ $\theta_v = sin^{-1}~(0.32258)$ $\theta_v = 18.82^{\circ}$ The angular spread of the light in the glass is $~~0.38^{\circ}$
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