JEE Advanced 2024 — Paper 1 — PHY-II solution

JEE Advanced2024Paper 1PHY-II

A glass beaker has a solid, plano-convex base of refractive index $1.60$ , as shown in the figure. The radius of curvature of the convex surface (SPU) is $9$ cm, while the planar surface (STU) acts as a mirror. This beaker is filled with a liquid of refractive index $n$ up to the level QPR. If the image of a point object O at a height of $h$ (OT in the figure) is formed onto itself, then, which of the following option(s) is (are) correct?

A
For $n = 1.42$ , $h = 50$ cm
B
For $n = 1.35$ , $h = 36$ cm
C
For $n = 1.45$ , $h = 65$ cm
D
For $n = 1.48$ , $h = 85$ cm

Solution

For the image of the on-axis object O to form on itself, light must retrace its path. This means rays from O hit the planar mirror normally after refracting through the liquid-glass interface. The combined system acts as an equivalent mirror, and the object must sit at twice the equivalent focal length, i.e. $h = 2∣ f_{eq} ∣$ .

Using the equivalent-mirror formula, the system consists of two refractions at the curved liquid/glass interface (going down) and a plane mirror. The power adds as:

$\frac{1}{f _{eq}} = 2 \cdot \frac{n - 1}{- 9} + 2 \cdot \frac{0.6}{9}$ (sign convention with $R = - 9$ cm for the curved interface as encountered by descending light treated symmetrically)

Simplifying gives $\frac{1}{f _{eq}} = \frac{2 ( 0.6 - ( n - 1 ))}{9} = \frac{2 ( 1.6 - n )}{9}$ .

Hence $h = 2∣ f_{eq} ∣ = \frac{9}{1.6 - n}$ .

$n = 1.42 \Rightarrow h = 9/0.18 = 50$ cm. (A) correct.

$n = 1.35 \Rightarrow h = 9/0.25 = 36$ cm. (B) correct.

$n = 1.45 \Rightarrow h = 9/0.15 = 60$ cm (not 65). (C) incorrect.

$n = 1.48 \Rightarrow h = 9/0.12 = 75$ cm (not 85). (D) incorrect.

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