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JEE Main 2025 Jan 22 Shift 1, Physics Q8: Thin Lens and Mirrors

JEE Main2025Jan 22, Shift 1Physics
Q.

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

  1. A

  2. B

  3. C

  4. D

Solution

$ \Rightarrow p_{eq}=p_1+p_2+p_3 $

$ \Rightarrow p_1=\left(\dfrac{4}{3}-1\right)\left(\dfrac{1}{\infty}-\dfrac{1}{-|R_1|}\right) $

$ \Rightarrow p_1=\left(\dfrac{1}{3|R_1|}\right) $

$ \Rightarrow p_2=\left(\dfrac{1}{2}\right)\left(\dfrac{1}{-|R_1|}-\dfrac{1}{-|R_2|}\right) $

$ \Rightarrow p_2=\dfrac{1}{2}\left(\dfrac{1}{|R_2|}-\dfrac{1}{|R_1|}\right) $

$ \Rightarrow p_3=\left(\dfrac{1}{3}\right)\left(\dfrac{1}{-|R_2|}-\dfrac{1}{\infty}\right) =-\dfrac{1}{3|R_2|} $

$ \Rightarrow p_{eq} =\dfrac{1}{3}\left(\dfrac{1}{|R_1|}-\dfrac{1}{|R_2|}\right) -\dfrac{1}{2}\left(\dfrac{1}{|R_1|}-\dfrac{1}{|R_2|}\right) $

$ =-\dfrac{1}{6}\left(\dfrac{1}{|R_1|}-\dfrac{1}{|R_2|}\right) $

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