JEE MAIN - Physics (2025 - 22nd January Morning Shift - No. 17)

Given is a thin convex lens of glass (refractive index $\mu$ ) and each side having radius of curvature $R$. One side is polished for complete reflection. At what distance from the lens, an object be placed on the optic axis so that the image gets formed on the object itself?
$R / \mu$
$R /(2 \mu-1)$
$\mathrm{R} /(2 \mu-3)$
$\mu \mathrm{R}$

Explanation

JEE Main 2025 (Online) 22nd January Morning Shift Physics - Geometrical Optics Question 17 English Explanation

We know, for a combination of a lens and a mirror,

$${P_{comb}} = 2{P_l} + {P_m}$$ ...... (1)

Also using lens maker formula,

$${1 \over {{f_l}}} = (\mu - 1)\left( {{1 \over R} - {1 \over { - R}}} \right) = (\mu - 1){2 \over R}$$

So using eq. (i),

$$ - {1 \over {{f_{comb}}}} = 2\left( {{1 \over {{f_l}}}} \right) - {1 \over {{f_m}}}$$

$$ \Rightarrow - {1 \over {{f_{comb}}}} = {{4(\mu - 1)} \over R} - {1 \over {\left( { - {R \over 2}} \right)}}$$

$$ \Rightarrow - {1 \over {{f_{comb}}}} = {{4(\mu - 1)} \over R} + {2 \over R}$$

$$ \Rightarrow - {1 \over {{f_{comb}}}} = {2 \over R}(2\mu - 2 + 1) = {2 \over R}(2\mu - 1)$$

$$ \Rightarrow {f_{comb}} = {{ - R} \over {2(2\mu - 1)}}$$

$$ \Rightarrow {R_{comb}} = 2{f_{comb}} = {{ - 2R} \over { - 2(2\mu - 1)}} = {{ - R} \over {2\mu - 1}}$$

Hence, distance $$ = {R \over {2\mu - 1}}$$

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