Question

A parallel beam of light in air makes an angle of 47.5 with the surface of a glass plate having a refractive index of 1.66. (a) what is the angle between the reflected part of the beam and the surface of the glass? (b) what is the angle between the refracted beam and the surface of the glass?

Answer 1

(a) The law of reflection says that the angle of incidence (measured with respect to the perpendicular to the interface between the two mediums) is equal to the angle of reflection (again, measured with respect to the  interface between the two mediums):
$\theta _i = \theta _r$
In our problem, $47.5 ^{\circ}$ is the angle of the incident ray with respect to the surface, so the angle of incidence (with respect to the perpendicular) is
$\theta_i = 90^{\circ} - 47.5 ^{\circ} = 42.5^{\circ}$
And so, the angle of reflection is also $\theta_r = 42.5^{\circ}$, and since this is measured with respect to the perpendicular to the surface, the angle between the reflected ray and the surface is
$\theta= 90^{\circ}-42.5^{\circ}=47.5^{\circ}$

(b) The angle of refraction, $\theta_t$, is given by Snell's law:
$n_i \sin \theta_i = n_t \sin \theta_t$
where $n_i = 1.00$ is the refractive index of air and $n_t = 1.66$ is the refractive index of glass.
Substituting, we find:
$\theta_T = \arcsin ( \frac{n_i}{n_r} \sin \theta_i )=\arcsin ( \frac{1.00}{1.66} \sin (42.5^{\circ}) )=24.0^{\circ}$
However, this is the angle of refraction, measured with respect to the perpendicular to the surface; so, the angle of the refracted ray with respect to the surface is
$\theta=90^{\circ}-24.0^{\circ}=66^{\circ}$