Question

Light of wavelength 500nm traveling in air is incident along the normal on Face A of a 60-60-60 glass prism of refractive index 1.52, as shown in the figure.

a) The refracted ray is incident to Face B. Calculate the angle of incidence of the ray, the angle of refraction of the refracted ray (if any) and the angle of reflection of the reflected ray. Draw the rays.

b) The reflected ray at Face B is incident on Face C. Calculate the angle of incidence of the ray, the angle of refraction of the refracted ray (if any) and the angle of reflection of the reflected ray. Draw the rays.

c) If the index of refraction of the prism was 1.10, calculate the angle of refraction on Face B. Please draw and label the ray.

Answer 1

The figure is below. The incidence angle is 60 degree. The angle for total reflection in the prism is

r =arcsin(1/n) =arcsin(1/1.52) =41.14 degree

Since the angle of incidence =60 deg is greater that 41.14 degree there will be total reflection on face B.

For face C the angle of incidence is 0 degree (normal incidence, ray is perpendicular to face C) so that the ray will go outside unrefracted in the air. (refraction angle =0 deg)

In case C) if n=1.10 then the refraction angle is

sin(r)/sin(60) =n

r =arcsin(1.10*sin(60)) =72.29 degree.

Answer 2