Question

The exit surface of a prism with refractive index $n$ is coated with a material having refractive index $\dfrac{n}{2}$. When this prism is set for minimum angle of deviation, it exactly meets the condition of critical angle. The prism angle is ___.

• A. $30^\circ$
• B. $60^\circ$
• C. $15^\circ$
• D. $45^\circ$

Hint:

At minimum deviation, internal refraction angles in a prism are equal and symmetric.

Answer 1

The Correct Option is B

To determine the prism angle where the exit surface is coated with a material of refractive index \(\frac{n}{2}\) and the prism is set at the minimum angle of deviation satisfying the condition of critical angle, we need to follow these logical steps:

1. The critical angle condition for the exit surface can be represented as: \(\theta_c = \sin^{-1}\left(\frac{n_2}{n_1}\right)\), where \(n_2 = \frac{n}{2}\) and \(n_1 = n\).

This gives us: \(\theta_c = \sin^{-1}\left(\frac{\frac{n}{2}}{n}\right) = \sin^{-1}\left(\frac{1}{2}\right)\).

1. We know that \(\sin^{-1}\left(\frac{1}{2}\right) = 30^\circ\). Thus, the critical angle \(\theta_c\) is \(30^\circ\).
2. In the condition of the minimum angle of deviation for prisms:

The formula connecting angle of the prism \(A\), refractive index \(n\), and the minimum deviation \(\delta_m\) is:

1. \(n = \frac{\sin\left(\frac{A + \delta_m}{2}\right)}{\sin\left(\frac{A}{2}\right)}\).
2. When the light travels through a prism at minimum deviation:

The emergent ray exits at the critical angle. Thus, for an isosceles prism, each angle \(r_2 = \theta_c = 30^\circ\) and \(r_2 = \frac{A}{2} = 30^\circ\).

1. Solving \(r_2 = \frac{A}{2} = 30^\circ\) gives: \(A = 60^\circ\).

Hence, the prism angle is \(60^\circ\).