Answer
$\theta = 49.9^{\circ}$
Work Step by Step
We can use Equation (33-45) to find the critical angle $\theta_c$ at the interface between materials 2 and 3:
$\theta_c = sin^{-1}~\frac{n_3}{n_2}$
$\theta_c = sin^{-1}~\frac{1.30}{1.50}$
$\theta_c = 60.1^{\circ}$
If the ray is incident at point A at the critical angle $\theta_c$, then the angle of refraction at the interface between materials 1 and 2 is $\theta_2 = 60.1^{\circ}$
We can use Snell's law to find $\theta$:
$n_1~sin~\theta = n_2~sin~\theta_2$
$sin~\theta = \frac{n_2~sin~\theta_2}{n_1}$
$\theta = sin^{-1}~(\frac{n_2~sin~\theta_2}{n_1})$
$\theta = sin^{-1}~(\frac{1.50~sin~60.1^{\circ}}{1.70})$
$\theta = sin^{-1}~(0.7649)$
$\theta = 49.9^{\circ}$
