JEE Main 2026 — Jan 28, Shift 2 — Physics solution

JEE Main2026Jan 28, Shift 2Physics

For a transparent prism, if the angle of minimum deviation is equal to its refracting angle, the refractive index $n$ of the prism satisfies.

A
$2 < n < 22$
B
$1 < n < 2$
C
$n \geq 2$
D
$2 < n < 2$

Solution

Prism minimum-deviation formula: $μ = \frac{sin ( \frac{δ _{min} + A}{2} )}{sin ( \frac{A}{2} )}$ . Given $δ_{min} = A$ , so $μ = \frac{sin A}{sin ( A /2 )} = 2 cos (A /2)$ .

Also, at minimum deviation each face has incidence angle $i = \frac{δ _{min} + A}{2} = A$ . The grazing-incidence cap requires $i < 9 0^{\circ}$ , so $A < 9 0^{\circ} \Rightarrow A /2 < 4 5^{\circ} \Rightarrow cos (A /2) > \frac{1}{2} \Rightarrow μ > 2$ .

Also $μ = 2 cos (A /2) < 2$ since $cos (A /2) < 1$ . Thus $2 < n < 2$ .

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