Physics · Geometrical Optics

JEE Main 2025 — 2 April, Evening Shift — Question 54

A bi-convex lens has radius of curvature of both the surfaces same as $1 / 6 \mathrm{~cm}$ . If this lens is required to be replaced by another convex lens having different radii of curvatures on both sides ( $R_{1} \neq R_{2}$ ), without any change in lens power then possible combination of $R_{1}$ and $R_{2}$ is

Option A:
$\frac{1}{5} \mathrm{~cm}$ and $\frac{1}{7} \mathrm{~cm}$
Correct
Option B:
$\frac{1}{3} \mathrm{~cm}$ and $\frac{1}{7} \mathrm{~cm}$
Option C:
$\frac{1}{6} \mathrm{~cm}$ and $\frac{1}{9} \mathrm{~cm}$
Option D:
$\frac{1}{3} \mathrm{~cm}$ and $\frac{1}{3} \mathrm{~cm}$

Answer: A

Step-by-step solution

$\frac{1}{f}=(\mu-1)\left(\frac{2}{R}\right)=(\mu-1) 12$

$\frac{1}{f^{\prime}}=(\mu-1)\left(\frac{1}{R_{1}}+\frac{1}{R_{2}}\right)$

$f^{\prime}=f$ $12=\frac{1}{R_{1}}+\frac{1}{R_{2}}$

Answer key and solution verified before publishing.

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Exam: JEE Main 2025
Paper: 2 April, Evening Shift
Subject: Physics
Chapter: Geometrical Optics
Topic: Lenses and Their Combinations, Silvering of Lens

Answer: A

Step-by-step solution

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