Answer
(a) $48.8^{\circ}$
(b) No
Work Step by Step
(a) We know that
$\theta_w=sin^{-1}[\frac{n_a}{n_w}sin\theta_a]$
We plug in the known values to obtain:
$\theta_w=sin^{-1}[\frac{1.00}{1.33}sin(90^{\circ})]$
$\theta_w=48.8^{\circ}$
(b) We know that when $\theta$ decreases then the angle of incidence on the oil-air interface decreases as well. Thus, there will be no total internal reflection.
