Find the angle of minimum deviation for a prism of angle \( A=60^\circ \) and refractive index \( \mu=\sqrt{2} \).
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → \( 30^\circ \)
At minimum deviation the ray passes symmetrically through the prism, and the prism formula links the refractive index to \( A \) and \( D_m \).
\( \displaystyle \mu=\frac{\sin\left[(A+D_m)/2\right]}{\sin(A/2)} \)
\( \begin{aligned} \sin\frac{A+D_m}{2} &= \mu\sin\frac{A}{2} \\ &= \sqrt{2}\times\sin30^\circ \\ &= \frac{1}{\sqrt{2}} \end{aligned} \)
\( \displaystyle \frac{60^\circ+D_m}{2}=45^\circ \)
\( \Rightarrow\; D_m=30^\circ \)