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Physics · Geometrical Optics

JEE Main 2025 — 24 January, Morning Shift — Question 58

A thin plano convex lens made of glass of refractive index 1.5 is immersed in a liquid of refractive index 1.2 . When the plane side of the lens is silver coated for complete reflection, the lens immersed in the liquid behaves like a concave mirror of focal length 0.2 m . The radius of curvature of the curved surface of the lens is :-

  1. Option A:

    0.15m

  2. Option B:

    0.10m

    Correct
  3. Option C:

    0.20m

  4. Option D:

    0.25m

Answer: B

Step-by-step solution

1.5v=1.5−1.2R\frac{1.5}{\mathrm{v}}=\frac{1.5-1.2}{\mathrm{R}}

v=1.5R0.3=5R\mathrm{v}=\frac{1.5 \mathrm{R}}{0.3}=5 \mathrm{R}

1.2f−1.55R=1.2−1.5−R\frac{1.2}{\mathrm{f}}-\frac{1.5}{5 \mathrm{R}}=\frac{1.2-1.5}{-\mathrm{R}}

1.2f=0.3R×2⇒f=2R⇒R=0.1\frac{1.2}{\mathrm{f}}=\frac{0.3}{\mathrm{R}} \times 2 \Rightarrow \mathrm{f}=2 \mathrm{R} \Rightarrow \mathrm{R}=0.1

Solution figure

Answer key and solution verified before publishing.

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Exam
JEE Main 2025
Paper
24 January, Morning Shift
Subject
Physics
Chapter
Geometrical Optics
Topic
Lenses and Their Combinations, Silvering of Lens

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