Question

Minimum deviation for an equilateral prism is \(30^\circ\). The refractive index of the prism material is:

• A. \( \sqrt{2} \)
• B. \( \sqrt{\frac{3}{2}} \)
• C. \(2 \)
• D. \(4 \)

Answer 1

The Correct Option is A

Concept: For a prism at the condition of minimum deviation, the refractive index \( \mu \) of the prism is given by \[ \mu = \frac{\sin\left(\frac{A + \delta_m}{2}\right)}{\sin\left(\frac{A}{2}\right)} \] where \begin{itemize} \item \(A\) = angle of the prism \item \( \delta_m \) = angle of minimum deviation \end{itemize} For an equilateral prism, \(A = 60^\circ\). Step 1: {\color{red}Substitute the given values in the prism formula.} Given: \[ A = 60^\circ, \qquad \delta_m = 30^\circ \] \[ \mu = \frac{\sin\left(\frac{60^\circ + 30^\circ}{2}\right)}{\sin\left(\frac{60^\circ}{2}\right)} \] \[ \mu = \frac{\sin(45^\circ)}{\sin(30^\circ)} \] Step 2: {\color{red}Use the trigonometric values.} \[ \sin 45^\circ = \frac{\sqrt{2}}{2}, \qquad \sin 30^\circ = \frac{1}{2} \] \[ \mu = \frac{\frac{\sqrt{2}}{2}}{\frac{1}{2}} \] \[ \mu = \sqrt{2} \] Hence, the refractive index of the prism is \[ \boxed{\sqrt{2}} \]