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JEE MAIN - Physics (2025 - 22nd January Morning Shift - No. 5)

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

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

$-\frac{1}{6}\left(\frac{1}{\left|\mathrm{R}_1\right|}-\frac{1}{\left|\mathrm{R}_2\right|}\right)$
$-\frac{1}{6}\left(\frac{1}{\left|\mathrm{R}_1\right|}+\frac{1}{\left|\mathrm{R}_2\right|}\right)$
$\frac{1}{6}\left(\frac{1}{\left|\mathrm{R}_1\right|}+\frac{1}{\left|\mathrm{R}_2\right|}\right)$
$\frac{1}{6}\left(\frac{1}{\left|\mathrm{R}_1\right|}-\frac{1}{\left|\mathrm{R}_2\right|}\right)$

Explanation

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

Using lens maker formula, $${1 \over f} = \left( {{{{\mu _2}} \over {{\mu _1}}} - 1} \right)\left( {{1 \over {{R_1}}} - {1 \over {{R_2}}}} \right)$$

Note : here, we will take air as medium for all three lenses so $${\mu _1} = 1$$

$${P_1} = {1 \over {{f_1}}} = \left( {{4 \over 3} - 1} \right)\left( {{1 \over \infty } - {1 \over { - {R_1}}}} \right)$$

$${P_1} = {1 \over 3}\left( {{1 \over {{R_1}}}} \right)$$ ..... (1)

$${P_2} = {1 \over {{f_2}}} = \left( {{3 \over 2} - 1} \right)\left( {{1 \over { - {R_1}}} - {1 \over { - {R_2}}}} \right)$$

$$ \Rightarrow {P_2} = {{ - 1} \over 2}\left( {{1 \over {{R_1}}} - {1 \over {{R_2}}}} \right)$$ .... (2)

$${P_3} = {1 \over {{f_3}}} = \left( {{4 \over 3} - 1} \right)\left( {{1 \over { - {R_2}}} - {1 \over \infty }} \right)$$

$$ \Rightarrow {P_3} = {{ - 1} \over 3}{1 \over {{R_2}}}$$ ...... (3)

So, $${P_{eq}} = {1 \over {3{R_1}}} - {1 \over 2}\left( {{1 \over {{R_1}}} - {1 \over {{R_2}}}} \right) - {1 \over {3{R_2}}}$$

$$ = {1 \over 3}\left( {{1 \over {{R_1}}} - {1 \over {{R_2}}}} \right) - {1 \over 2}\left( {{1 \over {{R_1}}} - {1 \over {{R_2}}}} \right)$$

$$ = \left( {{1 \over 3} - {1 \over 2}} \right)\left( {{1 \over {{R_1}}} - {1 \over {{R_2}}}} \right)$$

$${P_{eq}} = {{ - 1} \over 6}\left( {{1 \over {{R_1}}} - {1 \over {{R_2}}}} \right)$$

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