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JEE Main 20252 AprilEvening Shift
Exam
JEE Main
Year
2025
Date
2 April
Shift
Evening Shift
Subject
Physics
Chapter
Ray Optics

JEE Main 2025 · 2 April · Evening Shift · Physics · Ray Optics

A bi-convex lens has radius of curvature of both the surfaces same...

A bi-convex lens has radius of curvature of both the surfaces same as 1/6 cm1 / 6 \mathrm{~cm}. If this lens is required to be replaced by another convex lens having different radii of curvatures on both sides (R1≠R2)\left(R_1 \neq R_2\right), without any change in lens power then possible combination of R1andR2R_1 and R_2 is :

Options

  1. A13 cm\frac{1}{3} \mathrm{~cm} and 13 cm\frac{1}{3} \mathrm{~cm}
  2. B15 cm\frac{1}{5} \mathrm{~cm} and 17 cm\frac{1}{7} \mathrm{~cm}
  3. C13 cm\frac{1}{3} \mathrm{~cm} and 17 cm\frac{1}{7} \mathrm{~cm}
  4. D16 cm\frac{1}{6} \mathrm{~cm} and 19 cm\frac{1}{9} \mathrm{~cm}

Correct answer

B. 15 cm\frac{1}{5} \mathrm{~cm} and 17 cm\frac{1}{7} \mathrm{~cm}

Step-by-step solution

This will happen when
<br/>1f1=1f2(μ−1)(1R1−1−R2)=(μ−1)(2R)1R1+1R2=2R<br/>\begin{aligned}<br/>& \frac{1}{\mathrm{f}_1}=\frac{1}{\mathrm{f}_2} \\ & (\mu-1)\left(\frac{1}{\mathrm{R}_1}-\frac{1}{-\mathrm{R}_2}\right)=(\mu-1)\left(\frac{2}{\mathrm{R}}\right) \\ & \frac{1}{\mathrm{R}_1}+\frac{1}{\mathrm{R}_2}=\frac{2}{\mathrm{R}}<br/>\end{aligned}
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