A 60° prism has a refractive index of 1.52. Calculate the angle of incidence for the initial light ray that is then refracted inside the prism and just undergoes total internal reflection at the opposite face of the prism.
A) 41.1°
B) 18.9°
C) 29.4°
D) 30.6°
E) 60.6°

Question

Quizplus · Accepted Answer

When light enters the prism at an angle of incidence $i$, it bends towards the normal by an angle of refraction $r$, given by Snell’s Law:

$n_{1}sin\:i=n_{2}sin\:r$

where $n_{1}$ is the refractive index of the medium outside the prism (usually air), and $n_{2}$ is the refractive index of the prism.

At the second face of the prism, the ray meets the boundary between the prism and air at an angle of incidence of $r$, and is refracted back into the prism at an angle of refraction $i$, again given by Snell’s Law. Total internal reflection occurs when there is no angle of refraction possible in the lower refractive index medium.

For the critical angle $	heta_{c}$, the angle of refraction $r$ is $90°$. Applying Snell’s Law gives:

$i=sin^{-1}\:\left(\frac{n_{2}}{n_{1}}ight)sin\:90°=sin^{-1}\left(\frac{n_{2}}{n_{1}}ight)$

Therefore:

$i=sin^{-1}\left(\frac{1.52}{1}\:sin\:60°\:ight)\approx29.4°$

Choice C is the correct answer.

A 60° Prism Has a Refractive Index of 1