Question

A ray of light passing through a prism (\(\mu = \sqrt{3}\)) suffers minimum deviation. It is found that the angle of incidence is double the angle of refraction within the prism. Then, the angle of prism is ________ (in degrees).

Hint:

At minimum deviation the refraction angle inside is \(r = A/2\); use \(\mu = \sin i/\sin r\) with \(i = 2r\).

Answer 1

Correct Answer: 60

At minimum deviation the ray passes symmetrically, so \(r_1 = r_2 = r = A/2\).
Given \(i = 2r\). Snell's law: \(\mu = \dfrac{\sin i}{\sin r} = \dfrac{\sin 2r}{\sin r} = 2\cos r\).
So \(2\cos r = \sqrt{3} \Rightarrow \cos r = \dfrac{\sqrt{3}}{2} \Rightarrow r = 30^\circ\).
Hence \(A = 2r = 60^\circ\).
Answer: 60