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QJEEAdv24Ph1d1e: JEE Advanced 2024 Physics - Class 12 JEEAdv24Ph1d1e Solution: A glass beaker has a solid, plano-convex base of refractive…

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 cm9 \mathrm{~cm} , while the planar surface (STU) acts as a mirror. This beaker is filled with a liquid of refractive index nn up to the level QPR. If the image of a point object O\mathrm{O} at a height of hh (OT in the figure) is formed onto itself, then, which of the following option(s) is(are) correct?

JEE Advanced 2024 Paper 1 Online Physics - Geometrical Optics Question 4 English

Multiple Choice Options

  • A For n=1.42,h=50 cmn=1.42, h=50 \mathrm{~cm} .
  • B For n=1.35,h=36 cmn=1.35, h=36 \mathrm{~cm} .
  • C For n=1.45,h=65 cmn=1.45, h=65 \mathrm{~cm} .
  • D For n=1.48,h=85 cmn=1.48, h=85 \mathrm{~cm} .

Step-by-Step Verified 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 h below the base.

JEE Advanced 2024 Paper 1 Online Physics - Geometrical Optics Question 4 English Explanation

Since object and image are at same distance from equivalent lens, hence h=2 Fqq\mathrm{h}=2 \mathrm{~F}_{\mathrm{qq}}

amp;1 Feq=(1.6−11)(29)+(n−1)1(−29)amp;1h2=1.29+2(1−n)9amp;2 h=3.2−2n9amp; h=91.6−ncm\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}

(A) for n=1.42, h=50 cm\mathrm{n}=1.42, \mathrm{~h}=50 \mathrm{~cm}

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

(C) for n=1.45, h=60 cm\mathrm{n}=1.45, \mathrm{~h}=60 \mathrm{~cm}

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

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