JEE Mains · Physics · STD 12 - Ray optics and optical instruments
- JEE Mains 2025
- Hard
- MCQ
In the diagram given below, there are three lenses formed. Considering negligible thickness of each of them as compared to \(\left|R_1\right|\) and \(\left|R_2\right|\), i.e., the radii of curvature for upper and lower surfaces of the glass lens, the power of the combination is :

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Answer & Solution
Correct Answer
(D) \(-\frac{1}{6}\left(\frac{1}{\left|\mathrm{R}_1\right|}-\frac{1}{\left|\mathrm{R}_2\right|}\right)\)
Step-by-step Solution
Detailed explanation

\(\begin{aligned} & \Rightarrow p_{\mathrm{eq}}=\mathrm{p}_1+\mathrm{p}_2+\mathrm{p}_3 \\ & \Rightarrow \mathrm{p}_1=\left(\frac{4}{3}-1\right)\left(\frac{1}{\infty}-\frac{1}{-\left|\mathrm{R}_1\right|}\right) \\ & \Rightarrow \mathrm{p}_1=\left(\frac{1}{3\left|\mathrm{R}_1\right|}\right) \\ & \Rightarrow \mathrm{p}_2=\left(\frac{1}{2}\right)\left(\frac{1}{-\left|\mathrm{R}_1\right|}-\frac{1}{-\left|\mathrm{R}_2\right|}\right) \\ & \Rightarrow \mathrm{p}_2=\frac{1}{2}\left(\frac{1}{\left|\mathrm{R}_2\right|}-\frac{1}{\left|\mathrm{R}_1\right|}\right) \\ & \Rightarrow \mathrm{p}_3=\left(\frac{1}{3}\right)\left(\frac{1}{-\left|\mathrm{R}_2\right|}-\frac{1}{\infty}\right)=-\frac{1}{3\left|\mathrm{R}_2\right|} \\ & \Rightarrow \mathrm{p}_{\mathrm{eq}}=\frac{1}{3}\left(\frac{1}{\left|\mathrm{R}_1\right|}-\frac{1}{\left|\mathrm{R}_2\right|}\right)-\frac{1}{2}\left(\frac{1}{\left|\mathrm{R}_1\right|}-\frac{1}{\left|\mathrm{R}_2\right|}\right) \\ & =-\frac{1}{6}\left(\frac{1}{\left|\mathrm{R}_1\right|}-\frac{1}{\left|\mathrm{R}_2\right|}\right)\end{aligned}\)
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