Q66 / 75 · JEE Main 2025 (2 Apr Shift 2) · Physics
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 :
- A$\frac{1}{3} \mathrm{~cm}$ and $\frac{1}{3} \mathrm{~cm}$
- Correct answerB$\frac{1}{5} \mathrm{~cm}$ and $\frac{1}{7} \mathrm{~cm}$
- C$\frac{1}{3} \mathrm{~cm}$ and $\frac{1}{7} \mathrm{~cm}$
- D$\frac{1}{6} \mathrm{~cm}$ and $\frac{1}{9} \mathrm{~cm}$
Solution
This will happen when
$\begin{aligned}<br/>& \frac{1}{\mathrm{f}_1}=\frac{1}{\mathrm{f}_2} \\ & (\mu-1)\left(\frac{1}{\mathrm{R}_1}-\frac{1}{-\mathrm{R}_2}\right)=(\mu-1)\left(\frac{2}{\mathrm{R}}\right) \\ & \frac{1}{\mathrm{R}_1}+\frac{1}{\mathrm{R}_2}=\frac{2}{\mathrm{R}}<br/>\end{aligned}$
$\begin{aligned}<br/>& \frac{1}{\mathrm{f}_1}=\frac{1}{\mathrm{f}_2} \\ & (\mu-1)\left(\frac{1}{\mathrm{R}_1}-\frac{1}{-\mathrm{R}_2}\right)=(\mu-1)\left(\frac{2}{\mathrm{R}}\right) \\ & \frac{1}{\mathrm{R}_1}+\frac{1}{\mathrm{R}_2}=\frac{2}{\mathrm{R}}<br/>\end{aligned}$
Hence, the correct answer is option: 'B'.