Q64 / 75 · JEE Main 2025 (2 Apr Shift 2) · Physics
A ray of light suffers minimum deviation when incident on a prism having angle of the prism equal to $60^{\circ}$. The refractive index of the prism material is $\sqrt{2}$. The angle of incidence (in degrees) is ______.
Correct answer: 45
Solution
$\begin{aligned} & \mu=\frac{\sin \left(\frac{\mathrm{A}+\delta_{\mathrm{m}}}{2}\right)}{\sin \left(\frac{\mathrm{A}}{2}\right)}, \text { since } \mathrm{A}=60^{\circ} \quad \therefore \delta \mathrm{m}=30^{\circ} \\ & \delta_{\mathrm{m}}=2 \mathrm{i}-\mathrm{A}[\text { as } \mathrm{i}=\mathrm{e}] \\ & \Rightarrow \mathrm{i}=45^{\circ}\end{aligned}$
Hence, the correct answer is 45.