Q 12-09-167JEE MainJEE Main 2021 (25 Jul, Shift 2)Medium
A prism of refractive index $\mu$ and angle of prism $A$ is placed in the position of minimum angle of deviation. If minimum angle of deviation is also $A$, then in terms of refractive index,
Answer: (A) $2\cos^{-1}\left(\frac{\mu}{2}\right)$
$\mu = \dfrac{\sin\left(\frac{A + \delta_m}{2}\right)}{\sin\frac{A}{2}} = \dfrac{\sin A}{\sin\frac{A}{2}} = 2\cos\dfrac{A}{2}$ (with $\delta_m = A$).
$$A = 2\cos^{-1}\left(\frac{\mu}{2}\right)$$
Solution by Sreeraj P, M.Sc Physics