In the diagram given below, there are three lenses formed. Considering negligible thickness of each of them as compared to |R_1| and |R_2| , i.e., the radii of curvature for upper and lower surfaces of the glass lens

Correct answer: D. -(1)/(6)( 1 |R_1| - 1 |R_2| ) aligned & p_ eq =p_1+p_2+p_3 & p_1=((4)/(3)-1)((1)/(∞)- 1 -|R_1| ) & p_1=( 1 3|R_1| ) & p_2=((1)/(2))( 1 -|R_1| - 1 -|R_2| ) & p_2=(1)/(2)( 1 |R_2| - 1 |R_1| ) & p_3=((1)/(3))( 1 -|R_2| -(1)/(∞))=- 1 3|R_2| & p_ eq =(1)/(3)( 1 |R_1| - 1 |R_2| )-(1)/(2)( 1 |R_1| - 1 |R_2| ) & =-(1)/(6)( 1 |R_1| - 1 |R_2| ) aligned

JEE Main 202522 JanuaryMorning Shift

Exam: JEE Main
Year: 2025
Date: 22 January
Shift: Morning Shift
Subject: Physics
Chapter: Ray Optics

JEE Main 2025 · 22 January · Morning Shift · Physics · Ray Optics

In the diagram given below, there are three lenses formed....

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 In the diagram given below, there are three lenses formed. Considering negligible thickness of each of them as compared

In the diagram given below, there are three lenses formed. Considering negligible thickness of each of them as compared

Options

A $\frac{1}{6}\left(\frac{1}{\left|\mathrm{R}_1\right|}-\frac{1}{\left|\mathrm{R}_2\right|}\right)$
B $-\frac{1}{6}\left(\frac{1}{\left|\mathrm{R}_1\right|}+\frac{1}{\left|\mathrm{R}_2\right|}\right)$
C $\frac{1}{6}\left(\frac{1}{\left|R_1\right|}+\frac{1}{\left|R_2\right|}\right)$
D $-\frac{1}{6}\left(\frac{1}{\left|\mathrm{R}_1\right|}-\frac{1}{\left|\mathrm{R}_2\right|}\right)$

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

\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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