Answer
a. $47.5^{\circ}$.
b. $66.0^{\circ}$.
Work Step by Step
a. The incident angle, measured with respect to the normal (not with the surface) is $90^{\circ}-47.5^{\circ}=42.5^{\circ}$. Apply the law of reflection. The reflected ray makes the same angle with the normal, so the angle it makes with the surface is $90^{\circ}-42.5^{\circ}=47.5^{\circ}$.
b. Use Snell’s Law, where the angles are measured from the normal to the interface.
$$n_a sin\theta_a= n_b sin\theta_b$$
$$\theta_b=sin^{-1}(\frac{n_a}{n_b}sin\theta_1) $$
$$=sin^{-1}(\frac{1.00}{1.66}sin42.5^{\circ}) =24.0^{\circ}$$
The angle the refracted beam makes with the surface is $90^{\circ}-24.0^{\circ}=66.0^{\circ}$.
