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JEE Advanced · Physics · 24. Ray Optics

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. A For \(n=1.42, h=50 \mathrm{~cm}\).
  2. B For \(n=1.35, h=36 \mathrm{~cm}\).
  3. C For \(n=1.45, h=65 \mathrm{~cm}\).
  4. D For \(n=1.48, h=85 \mathrm{~cm}\).
Verified Solution

Answer & Solution

Correct Answer

(B) For \(n=1.35, h=36 \mathrm{~cm}\).

Step-by-step Solution

Detailed explanation

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 \(\mathrm{n}=1.48, \mathrm{~h}=75 \mathrm{~cm}\)
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