201903310609 Solution to 1980-AL-PHY-I-14

By Snell’s law,

$\begin{aligned} n_1\sin\theta_1 & = n_2\sin\theta_2\\ n_1 \sin 45^\circ & = 1 \times \sin 90^\circ \\ n_1 & = \displaystyle{\frac{1}{\sin 45^\circ}} \\ & = \sqrt{2} \\ \end{aligned}$

And the answer is C.

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