Question

The angle of minimum deviation \( \delta_m \) for an equilateral glass prism is \( 30^\circ \). The refractive index of the prism is

• A. $1/\sqrt{2}$
• B. $\sqrt{2}$
• C. $2\sqrt{2}$
• D. $1/2\sqrt{2}$

Hint:

The angle of minimum deviation $\deltam$ for an equilateral glass prism is $30

Answer 1

The Correct Option is B

To determine the refractive index of the prism, we use the formula for the angle of minimum deviation in a prism. An equilateral prism has an apex angle \( A \) of 60°.

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

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

Given that \( A = 60^\circ \) and \( \delta_m = 30^\circ \), we substitute these values into the formula:

\(n = \frac{\sin\left(\frac{60^\circ + 30^\circ}{2}\right)}{\sin\left(\frac{60^\circ}{2}\right)}\)

Calculate the angles:

• \(\frac{60^\circ + 30^\circ}{2} = \frac{90^\circ}{2} = 45^\circ\)
• \(\frac{60^\circ}{2} = 30^\circ\)

Substitute to get the refractive index:

\(n = \frac{\sin 45^\circ}{\sin 30^\circ}\)

Calculate the sine values:

• \(\sin 45^\circ = \frac{\sqrt{2}}{2}\)
• \(\sin 30^\circ = \frac{1}{2}\)

Substitute these values into the equation:

\(n = \frac{\frac{\sqrt{2}}{2}}{\frac{1}{2}} = \sqrt{2}\)

Thus, the refractive index of the prism is \(\sqrt{2}\).

This matches option "√2", confirming that it is the correct answer.