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JEE Advanced 2021 Paper 1 · Physics · Ray and Wave Optics · medium

A thin symmetric double-convex glass lens (refractive index 1.5) has both radii of curvature equal to 20 cm.

One or more options correct

A thin symmetric double-convex glass lens (refractive index 1.51.5) has both radii of curvature equal to 20 cm20\ \mathrm{cm}. Choose the correct statement(s).
  1. A. If the lens is cut into two halves by a plane perpendicular to its principal axis, each half has a focal length of 40 cm40\ \mathrm{cm}.
  2. B. If the lens is cut into two halves by a plane containing the principal axis, each half alone forms an image of an on-axis point object at the same position as the whole lens but with undiminished intensity.
  3. C. The two halves from the cut perpendicular to the axis, placed in contact, have a combined focal length of 10 cm10\ \mathrm{cm}.
  4. D. When the original lens is immersed in a liquid of refractive index 1.61.6, it behaves as a converging lens of focal length 160 cm160\ \mathrm{cm}.
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Answer: A

For the original lens, 1f=(1.5−1)(120+120)=120\frac1f=(1.5-1)\left(\frac1{20}+\frac1{20}\right)=\frac1{20}, so f=20f=20 cm. (A) A cut perpendicular to the axis gives two plano-convex halves, each with one curved surface of radius 2020 cm: 1f′=0.5×120\frac1{f'}=0.5\times\frac1{20}, so f′=40f'=40 cm: correct. (B) Cutting along the axis leaves the focal length unchanged, so the image position is the same, but each half collects only half of the light, so the image intensity is halved: incorrect. (C) Two thin lenses in contact add powers: 140+140=120\frac1{40}+\frac1{40}=\frac1{20}, so the combination has f=20f=20 cm (it is just the original lens), not 1010 cm: incorrect. (D) In liquid of index 1.61.6, 1f=(1.51.6−1)220=−0.00625 cm−1\frac1f=\left(\frac{1.5}{1.6}-1\right)\frac{2}{20}=-0.00625\ \mathrm{cm^{-1}}, so f=−160f=-160 cm: the lens is diverging, so the statement is incorrect.
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