JEE Mains · Physics · STD 12 - Ray optics and optical instruments
- JEE Mains 2025
- Hard
- MCQ
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 \(\left(R_1 \neq R_2\right)\), without any change in lens power then possible combination of \(R_1\) and \(R_2\) is :
Pick an option to check your answer.
Answer & Solution
Correct Answer
(B) \(\frac{1}{5} \mathrm{~cm}\) and \(\frac{1}{7} \mathrm{~cm}\)
Step-by-step Solution
Detailed explanation
This will happen when
\(\begin{aligned}
& \frac{1}{\mathrm{f}_1}=\frac{1}{\mathrm{f}_2} \\
& (\mu-1)\left(\frac{1}{\mathrm{R}_1}-\frac{1}{-\mathrm{R}_2}\right)=(\mu-1)\left(\frac{2}{\mathrm{R}}\right) \\
& \frac{1}{\mathrm{R}_1}+\frac{1}{\mathrm{R}_2}=\frac{2}{\mathrm{R}}
\end{aligned}\)
\(\begin{aligned}
& \frac{1}{\mathrm{f}_1}=\frac{1}{\mathrm{f}_2} \\
& (\mu-1)\left(\frac{1}{\mathrm{R}_1}-\frac{1}{-\mathrm{R}_2}\right)=(\mu-1)\left(\frac{2}{\mathrm{R}}\right) \\
& \frac{1}{\mathrm{R}_1}+\frac{1}{\mathrm{R}_2}=\frac{2}{\mathrm{R}}
\end{aligned}\)
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