Answer
$\theta_1=74.1^o$
Work Step by Step
$n_1\sin(\theta_1)=n_2\sin(\theta_2)$
$\theta_2=90-\frac{\phi}{2}$
$\theta_1=\arcsin(\frac{n_2\sin(\theta_2)}{n_1})$
$=\arcsin(1.58\sin(90-(90-\frac{75^o}{2})))=74.1^o$
Physics: Principles with Applications (7th Edition)
by Giancoli, Douglas C.
Published by
Pearson
ISBN 10:
0-32162-592-7
ISBN 13:
978-0-32162-592-2
Chapter 23 - Light: Geometric Optics - General Problems - Page 676: 77
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