Question

When a light ray incident on an equilateral prism, the angle of minimum deviation is found to be half of the angle of prism. The refractive index of the material of the prism is:

• A. 1.5
• B. 2
• C. \( \sqrt{3} \)
• D. \( \sqrt{2} \)

Hint:

For an equilateral prism, the angle of minimum deviation is half the angle of the prism. Use this to calculate the refractive index.

Answer 1

The Correct Option is D

Let the angle of prism be \( A \) and the angle of minimum deviation be \( D \). Given: \[ D = \frac{A}{2} \] For a prism, the refractive index \( n \) is given by the formula: \[ n = \frac{\sin \left( \frac{A + D}{2} \right)}{\sin \left( \frac{A}{2} \right)} \] Substitute \( D = \frac{A}{2} \) into the formula: \[ n = \frac{\sin \left( \frac{A + \frac{A}{2}}{2} \right)}{\sin \left( \frac{A}{2} \right)} = \frac{\sin \left( \frac{3A}{4} \right)}{\sin \left( \frac{A}{2} \right)} \] For an equilateral prism, the angle of prism \( A = 60^\circ \). Thus: \[ n = \frac{\sin \left( \frac{3 \times 60}{4} \right)}{\sin \left( \frac{60}{2} \right)} = \frac{\sin 45^\circ}{\sin 30^\circ} \] \[ n = \frac{\frac{\sqrt{2}}{2}}{\frac{1}{2}} = \sqrt{2} \] Thus, the refractive index of the material of the prism is \( \sqrt{2} \).