Q 12-09-054JEE MainJEE Main 2026 (28 Jan, Shift 2)Medium
For a transparent prism, if the angle of minimum deviation is equal to its refracting angle, the refractive index $n$ of the prism satisfies.
Answer: (B) $\sqrt2<n<2$
With $\delta_m = A$:
$$n = \frac{\sin\frac{A+\delta_m}{2}}{\sin\frac A2} = \frac{\sin A}{\sin\frac A2} = 2\cos\frac A2$$
At minimum deviation the angle of incidence is $i = \dfrac{A+\delta_m}{2} = A$, and $i$ cannot exceed $90^\circ$, so $0 < A \le 90^\circ$ (a real grazing case at $90^\circ$ is the limit).
Then $\cos\dfrac A2$ lies between $\cos45^\circ = \dfrac1{\sqrt2}$ and $1$:
$$\sqrt2 < n < 2$$
Solution by Sreeraj P, M.Sc Physics