Answer
The angle of refraction at point B is $~~60.7^{\circ}$
Work Step by Step
We can use Equation (33-49) to find Brewster's angle $\theta_B$ at the interface between materials 2 and 3:
$\theta_B = tan^{-1}~\frac{n_3}{n_2}$
$\theta_B = tan^{-1}~\frac{1.30}{1.50}$
$\theta_B = 40.9^{\circ}$
If the ray is incident at point A at Brewster's angle $\theta_B$, then the incident angle $\theta_i$ at point B is $~~90.0^{\circ}-40.9^{\circ} = 49.1^{\circ}$
We can use Snell's law to find $\theta_3$, the angle of refraction at point B:
$n_3~sin~\theta_3 = n_2~sin~\theta_i$
$sin~\theta_3 = \frac{n_2~sin~\theta_i}{n_3}$
$\theta_3 = sin^{-1}~(\frac{n_2~sin~\theta_i}{n_3})$
$\theta_3 = sin^{-1}~(\frac{1.50~sin~49.1^{\circ}}{1.30})$
$\theta_3 = sin^{-1}~(0.87214)$
$\theta_3 = 60.7^{\circ}$
The angle of refraction at point B is $~~60.7^{\circ}$
