Question

The angle of a prism is 60\(^\circ\) and its refractive index is \(\sqrt{2}\). The angle of minimum deviation suffered by a ray of light in passing through it is

• A. about 20\(^\circ\)
• B. 30\(^\circ\)
• C. 60\(^\circ\)
• D. 45\(^\circ\)

Hint:

For minimum deviation in a prism: \(n = \dfrac{\sin\frac{A+D_m}{2}}{\sin\frac{A}{2}}\). Rearrange to find \(D_m\).

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
At minimum deviation: \(n = \dfrac{\sin\left(\dfrac{A+D_m}{2}\right)}{\sin\left(\dfrac{A}{2}\right)}\).
Step 2: Detailed Explanation:
\[ \sqrt{2} = \frac{\sin\left(\frac{60 + D_m}{2}\right)}{\sin 30^\circ} = \frac{\sin\left(\frac{60 + D_m}{2}\right)}{0.5} \] \[ \sin\left(\frac{60 + D_m}{2}\right) = \frac{\sqrt{2}}{2} = \sin 45^\circ \] \[ \frac{60 + D_m}{2} = 45^\circ \Rightarrow D_m = 30^\circ \]
Step 3: Final Answer:
Minimum deviation \(D_m = \mathbf{30^\circ}\).