Physics: Principles with Applications (7th Edition)

Published by Pearson
ISBN 10: 0-32162-592-7
ISBN 13: 978-0-32162-592-2

Chapter 23 - Light: Geometric Optics - Problems - Page 674: 33

Answer

$81.9^{\circ}$.

Work Step by Step

Apply Snell’s law. $$n_{air}sin \theta_1=n_{glass}sin \theta_2$$ We are told that the angle of reflection is twice the angle of refraction. We also know, from the law of reflection, that the angle of incidence equals the angle of reflection. Therefore, the angle of incidence is twice the angle of refraction. $$\theta_1=2\theta_2$$ $$n_{air}sin2\theta_2=n_{glass}sin\theta_2$$ $$(1.00)sin2\theta_2=(1.51)sin\theta_2$$ Apply a “double-angle” trigonometric identity, $ sin2\theta_2= 2 sin\theta_2cos \theta_2$. $$(1.00) 2 sin\theta_2cos \theta_2=(1.51)sin\theta_2$$ $$cos \theta_2=\frac{1.51}{2}$$ $$\theta_2=40.97^{\circ}$$ Now find the angle of incidence. $$\theta_1=2\theta_2=81.9^{\circ}$$
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