JEE Advanced 2025 — Paper 2 — PHY-II solution

JEE Advanced2025Paper 2PHY-II

Two identical concave mirrors each of focal length $f$ are facing each other as shown in the schematic diagram. The focal length $f$ is much larger than the size of the mirrors. A glass slab of thickness $t$ and refractive index $n_{0}$ is kept equidistant from the mirrors and perpendicular to their common principal axis. A monochromatic point light source $S$ is embedded at the center of the slab on the principal axis, as shown in the schematic diagram. For the image to be formed on $S$ itself, which of the following distances between the two mirrors is/are correct:

A
$4 f + (1 - \frac{1}{n _{0}}) t$
B
$2 f + (1 - \frac{1}{n _{0}}) t$
C
$4 f + (n_{0} - 1) t$
D
$2 f + (n_{0} - 1) t$

Solution

Let $S$ denote the mirror separation. The source sits at the midpoint of the slab; from $S$ to the edge of the slab is $t /2$ , after which there is air of length $(S - t) /2$ up to the mirror. Because the slab is on the path, the slab of thickness $t$ in air looks like an apparent path $t / n_{0}$ when viewed from outside.

So the apparent object distance of $S$ from either mirror, measured through the slab, is

$u_{app} = \frac{S - t}{2} + \frac{t}{2 n _{0}}$

For the image to land back on $S$ , the light must retrace its path; this happens either when the object is at the centre of curvature ( $u_{app} = 2 f$ ) or at the focus ( $u_{app} = f$ , giving parallel rays that reflect off the second mirror's focal plane and converge back).

Condition 1: $u_{app} = 2 f$ gives

$\frac{S - t}{2} + \frac{t}{2 n _{0}} = 2 f \Rightarrow S = 4 f + (1 - \frac{1}{n _{0}}) t$

Condition 2: $u_{app} = f$ gives

$S = 2 f + (1 - \frac{1}{n _{0}}) t$

$∴$ Options (A) and (B) are correct.

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