Verified Solution • 2025–26 Session moderate

QJEEM25Ph4bbd: JEE Main 2025 Physics - Class 12 JEEM25Ph4bbd Solution: 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 ∣R1∣\left|R_1\right| and ∣R2∣\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

JEE Main 2025 (Online) 22nd January Morning Shift Physics - Geometrical Optics Question 15 English

Multiple Choice Options

  • A −16(1∣R1∣−1∣R2∣)-\frac{1}{6}\left(\frac{1}{\left|\mathrm{R}_1\right|}-\frac{1}{\left|\mathrm{R}_2\right|}\right)
  • B −16(1∣R1∣+1∣R2∣)-\frac{1}{6}\left(\frac{1}{\left|\mathrm{R}_1\right|}+\frac{1}{\left|\mathrm{R}_2\right|}\right)
  • C 16(1∣R1∣+1∣R2∣)\frac{1}{6}\left(\frac{1}{\left|\mathrm{R}_1\right|}+\frac{1}{\left|\mathrm{R}_2\right|}\right)
  • D 16(1∣R1∣−1∣R2∣)\frac{1}{6}\left(\frac{1}{\left|\mathrm{R}_1\right|}-\frac{1}{\left|\mathrm{R}_2\right|}\right)

Step-by-Step Verified Solution

JEE Main 2025 (Online) 22nd January Morning Shift Physics - Geometrical Optics Question 15 English Explanation

amp;⇒pcq=p1+p2+p3amp;⇒p1=(43−1)(1∞−1−∣R1∣)amp;⇒p1=(13∣R1∣)amp;⇒p2=(12)(1−∣R1∣−1−∣R2∣)amp;⇒p2=12(1∣R2∣−1∣R1∣)amp;⇒p3=(13)(1−∣R2∣−1∞)=−13∣R2∣amp;⇒peq=13(1∣R1∣−1∣R2∣)−12(1∣R1∣−1∣R2∣)amp;=−16(1∣R1∣−1∣R2∣)\begin{aligned}& \Rightarrow \mathrm{p}_{\mathrm{cq}}=\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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