JEE Advanced 2026 — Paper 2 — PHY-II solution

JEE Advanced2026Paper 2PHY-II

Consider two isosceles prisms 1 and 2 with prism angles $A_{1}$ and $A_{2}$ and refractive indices $n_{1}$ and $n_{2}$ , respectively, as shown in the figure. The faces $a_{1} b_{1}$ and $a_{2} b_{2}$ are parallel to each other and perpendicular to the mirror $M$ . If a ray of light is incident on the face $a_{1} c_{1}$ and emerges from the face $a_{2} c_{2}$ , then the correct statement(s) is/are:

A
If both the prisms are at minimum deviation condition, then $\frac{n _{2}}{n _{1}} = sin (\frac{A _{1}}{2}) / sin (\frac{A _{2}}{2})$
B
If prism 2 is at minimum deviation condition, then $sin i_{1} = n_{2} sin (\frac{A _{2}}{2})$ is always true.
C
If both the prisms 1 and 2 are thin and are at minimum deviation condition with angles of deviation $δ_{m 1}$ and $δ_{m 2}$ , respectively, then $θ = \frac{δ _{m 1}}{2 ( n _{1} - 1 )} + \frac{δ _{m 2}}{2 ( n _{2} - 1 )}$ .
D
If prism 1 is at minimum deviation condition, then $sin i_{2} = n_{1} sin (\frac{A _{1}}{2})$ is always true.

Solution

Geometry. Because the vertical faces $a_{1} b_{1}$ and $a_{2} b_{2}$ are perpendicular to the mirror, the ray inside prism 1 reflects off the mirror and enters prism 2 symmetrically. The construction forces $e_{1} = i_{2}$ . The minimum-deviation condition on a single prism is $i = e$ for that prism alone.

Option (A). At minimum deviation for prism 1: $sin i_{1} = n_{1} sin (A_{1} /2)$ . At minimum deviation for prism 2: $sin i_{2} = n_{2} sin (A_{2} /2)$ . By the symmetry $i_{1} = i_{2}$ , so

$n_{1} sin (A_{1} /2) = n_{2} sin (A_{2} /2) \Rightarrow \frac{n _{2}}{n _{1}} = \frac{sin ( A _{1} /2 )}{sin ( A _{2} /2 )}$ . Correct.

Option (B). If only prism 2 is at minimum deviation, $i_{2} = e_{2}$ but prism 1 need not be at minimum deviation, so $i_{1} \neq = i_{2}$ in general. Thus $sin i_{1} = n_{2} sin (A_{2} /2)$ is not always true. Incorrect.

Option (D). If prism 1 is at minimum deviation, $i_{1} = e_{1}$ , and by geometry $e_{1} = i_{2}$ , so $sin i_{2} = sin i_{1} = n_{1} sin (A_{1} /2)$ . Correct.

Option (C). For thin prisms at minimum deviation, $δ_{m 1} = (n_{1} - 1) A_{1}$ and $δ_{m 2} = (n_{2} - 1) A_{2}$ . The angle between the two outward faces is $θ = \frac{A _{1}}{2} + \frac{A _{2}}{2}$ , hence

$θ = \frac{1}{2} [\frac{δ _{m 1}}{n _{1} - 1} + \frac{δ _{m 2}}{n _{2} - 1}] = \frac{δ _{m 1}}{2 ( n _{1} - 1 )} + \frac{δ _{m 2}}{2 ( n _{2} - 1 )}$ . Correct.

$∴$ Correct options are (A), (C), (D).

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