Answer
(a) $\theta_1 = 2~cos^{-1}~(\frac{n}{2})$
(b) $\theta_1 = 82.8^{\circ}$
Work Step by Step
(a) We can use Snell's Law:
$n_1~sin~\theta_1 = n_2~sin~\theta_2$
$(1)~sin~\theta_1 = n~sin~\frac{\theta_1}{2}$
$sin~(2\cdot \frac{\theta_1}{2}) = n~sin~\frac{\theta_1}{2}$
$2~sin~\frac{\theta_1}{2}~cos~\frac{\theta_1}{2} = n~sin~\frac{\theta_1}{2}$
$2~cos~\frac{\theta_1}{2} = n$
$cos~\frac{\theta_1}{2} = \frac{n}{2}$
$\frac{\theta_1}{2} = cos^{-1}~(\frac{n}{2})$
$\theta_1 = 2~cos^{-1}~(\frac{n}{2})$
(b) We can evaluate our expression for light incident on glass:
$\theta_1 = 2~cos^{-1}~(\frac{n}{2})$
$\theta_1 = 2~cos^{-1}~(\frac{1.50}{2})$
$\theta_1 = (2)~(41.4^{\circ})$
$\theta_1 = 82.8^{\circ}$
