Answer
$40^{\circ}$ and $50^{\circ}$, to 2 significant figures.
Work Step by Step
Equation 23–6 gives the critical angle when light passes from a material with a high index of refraction, $n_1$, into a material with a lower index of refraction, $n_2$.
$$sin\theta_C=\frac{n_2}{n_1}=sin 58^{\circ} $$
There are two applicable Brewster’s angles (equation 24–6a). First, assume that light starts in the high index material:
$$tan \theta_p=\frac{n_2}{n_1}= sin 58^{\circ}$$
$$\theta_p=tan^{-1}(sin 58^{\circ})=40.3^{\circ}\approx 40^{\circ}$$
Second, assume that light starts in the low index material:
$$tan\theta_p=\frac{n_1}{n_2}= \frac{1}{sin 58^{\circ}}$$
$$\theta_p=tan^{-1}( \frac{1}{sin 58^{\circ}})=49.7^{\circ}\approx 50^{\circ}$$
The answers are given to 2 significant figures.
