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 675: 39

Answer

a) $n_1=1.41$ b) No c) $n=1.88$

Work Step by Step

a) $n_1=\frac{\sin(90^o)}{\sin(45^o)}=\sqrt{2}=1.41$ b) $\theta=\arcsin(\frac{n_2}{n_1})=\arcsin(\frac{1.33}{1.58})=57.3^o$ Since this is greater than $45^o$, the binoculars will not work underwater c) $n=\frac{1.33}{\sin(45^o)}=1.88$
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