A glass beaker has a solid, plano-convex base of refractive index 1.60, as shown in the figure. The radius…

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 $9 \mathrm{~cm}$, 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 $\mathrm{O}$ at a height of $h$ (OT in the figure) is formed onto itself, then, which of the following option(s) is(are) correct?
  1. For $n=1.42, h=50 \mathrm{~cm}$.
  2. For $n=1.35, h=36 \mathrm{~cm}$.
  3. For $n=1.45, h=65 \mathrm{~cm}$.
  4. For $n=1.48, h=85 \mathrm{~cm}$.

Solution

Since STU is a plane mirror, we can take mirror image of the whole situation about it and final image can be assumed to be at a distance $\mathrm{h}$ below the base. Since object and image are at same distance from equivalent lens, hence $\mathrm{h}=2 \mathrm{~F}_{\mathrm{eq}}$ $\begin{aligned} & \frac{1}{\mathrm{~F}_{\mathrm{eq}}}=\left(\frac{1.6-1}{1}\right)\left(\frac{2}{9}\right)+\frac{(\mathrm{n}-1)}{1}\left(\frac{-2}{9}\right) \\ & \frac{1}{\frac{\mathrm{h}}{2}}=\frac{1.2}{9}+\frac{2(1-\mathrm{n})}{9} \\ & \frac{2}{\mathrm{~h}}=\frac{3.2-2 \mathrm{n}}{9} \\ & \mathrm{~h}=\frac{9}{1.6-\mathrm{n}} \mathrm{cm}\end{aligned}$ (1) for $\mathrm{n}=1.42, \mathrm{~h}=50 \mathrm{~cm}$ (2) for $\mathrm{n}=1.35, \mathrm{~h}=36 \mathrm{~cm}$ (3) for $\mathrm{n}=1.45, \mathrm{~h}=60 \mathrm{~cm}$ (4) for $\mathrm{n}=1.48, \mathrm{~h}=75 \mathrm{~cm}$

Asked in: JEE Advanced 2024 (Paper 1)

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