Answer
a) $\theta=61.78^{\circ}$
b) total internal reflection
Work Step by Step
(a) The required angle can be determined as follows:
$n_asin\theta_i=n_fsin(90^{\circ}-\theta)$
$\implies sin(90^{\circ}-\theta)=\frac{n_asin\theta_f}{n_f}$
We plug in the known values to obtain:
$\implies sin(90^{\circ}-\theta)=\frac{1.00sin(50^{\circ})}{1.62}$
This simplifies to:
$\theta=61.78^{\circ}$
(b) We know that
$sin\theta_c=\frac{n_a}{n_f}$
$\implies \theta_c=sin^{-1}(\frac{1.00}{1.62})$
$\theta_c=38.1^{\circ}$
Here, the angle of incidence is greater than the critical angle, hence the light ray is totally internally reflected.
