Given a thin convex lens (refractive index μ₂), kept in a liquid… | JEE Main 2025 (23 Jan, Shift 1)

Q22JEE Main 2025 · 23 JAN · SHIFT 1Difficult

Combination of Lenses

Given a thin convex lens (refractive index $\mu_2$ ), kept in a liquid (refractive index $\mu_1, \mu_1 < \mu_2$ ) having radii of curvatures $|R_1|$ and $|R_2|$ . Its second surface is silver polished. Where should an object be placed on the optic axis so that a real and inverted image is formed at the same place?

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Answer: (d)

Image falls on the object when rays refracted at surface 1 hit the silvered surface normally, i.e. head for its centre $C_2$ , at $|R_2|$ on the object side.
Refraction at surface 1: $\frac{\mu_2}{v} - \frac{\mu_1}{u} = \frac{\mu_2-\mu_1}{R}$ with $u = -x$ , $v = -|R_2|$ , $R = +|R_1|$ .
$\frac{\mu_1}{x} = \frac{\mu_2-\mu_1}{|R_1|} + \frac{\mu_2}{|R_2|} = \frac{\mu_2(|R_1|+|R_2|) - \mu_1|R_2|}{|R_1||R_2|}$
$x = \frac{\mu_1|R_1||R_2|}{\mu_2(|R_1|+|R_2|) - \mu_1|R_2|}$
So the answer is (d).

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