Q 12-09-043JEE MainJEE Main 2026 (5 Apr, Shift 1)Medium
A ray of light passing through an equilateral prism is having velocity $2.12 \times 10^8$ m/s in the prism material, then the minimum angle of deviation is ______ degrees.
Answer: (B) $30$
$\mu = \dfrac{c}{v} = \dfrac{3 \times 10^8}{2.12 \times 10^8} \approx 1.414 = \sqrt{2}$.
$\sqrt{2} = \dfrac{\sin\frac{60^\circ + \delta_m}{2}}{\sin 30^\circ} \Rightarrow \sin\dfrac{60^\circ + \delta_m}{2} = \dfrac{1}{\sqrt{2}} \Rightarrow \delta_m = 30^\circ$.
Solution by Sreeraj P, M.Sc Physics