Question

The refractive index of a prism with apex angle \( A \) is \( \cot \frac{A}{2} \). The angle of minimum deviation is:

• A. \( \delta_m = 180^\circ - A \)
• B. \( \delta_m = 180^\circ - 3A \)
• C. \( \delta_m = 180^\circ - 4A \)
• D. \( \delta_m = 180^\circ - 2A \)

Answer 1

The Correct Option is D

To determine the angle of minimum deviation \( \delta_m \) for a prism, we begin with the fundamental equation relating the refractive index \( n \), the apex angle \( A \), and the angle of minimum deviation.

The relation for a prism is given by: 

\(n = \frac{\sin\left(\frac{A + \delta_m}{2}\right)}{\sin\left(\frac{A}{2}\right)}\)

We are given that the refractive index of the prism is \( n = \cot \frac{A}{2} \). Using trigonometric identities, we can rewrite \( \cot \frac{A}{2} \) as:

\(\cot \frac{A}{2} = \frac{\cos \frac{A}{2}}{\sin \frac{A}{2}}\)

Setting these expressions equal, we have:

\(\frac{\sin\left(\frac{A + \delta_m}{2}\right)}{\sin\left(\frac{A}{2}\right)} = \frac{\cos \frac{A}{2}}{\sin \frac{A}{2}}\)

This simplifies to:

\(\sin\left(\frac{A + \delta_m}{2}\right) = \cos \frac{A}{2}\)

Using the trigonometric identity \( \sin(\theta) = \cos(90^\circ - \theta) \), we can equate:

\(\frac{A + \delta_m}{2} = 90^\circ - \frac{A}{2}\)

Solving the equation for \( \delta_m \):

\(\frac{A + \delta_m}{2} + \frac{A}{2} = 90^\circ\)

\(\frac{A + \delta_m + A}{2} = 90^\circ\)

\(A + \delta_m = 180^\circ - A\)

\(\delta_m = 180^\circ - 2A\)

Hence, the angle of minimum deviation is:

The correct answer is: \(\delta_m = 180^\circ - 2A\)

Answer 2

Given:

\[ \mu = \frac{\sin \left( \frac{A + \delta_m}{2} \right)}{\sin \frac{A}{2}} \]

Using the relation:

\[ \cos \frac{A}{2} = \sin \left( \frac{A + \delta_m}{2} \right) \]

We get:

\[ \delta_m = \pi - 2A \]

Therefore:

\[ \delta_m = 180^\circ - 2A \]