JEE Main 2025 Jan 24 Shift 1, Physics Q12: Thin Lens and Mirrors

JEE Main2025Jan 24, Shift 1Physics

A thin plano convex lens made of glass of refractive index 1.5 is immersed in a liquid of refractive index 1.2. When the plane side of the lens is silver coated for complete reflection, the lens immersed in the liquid behaves like a concave mirror of focal length $0.2$ m. The radius of curvature of the curved surface of the lens is :-

A
$0.15$ m
B
$0.10$ m
C
$0.20$ m
D
$0.25$ m

Solution

A silvered lens immersed in a medium behaves as an equivalent mirror with power

$P_{e q} = 2 P_{l e n s} + P_{mi r r or}$

The plane silvered face has $P_{mi r r or} = 0$ (plane mirror). So $P_{e q} = 2 P_{l e n s}$ , i.e.

$\frac{1}{f _{e q}} = \frac{2}{f _{l e n s}}$

For the plano-convex lens immersed in the liquid, lensmaker's formula with relative refractive index $μ_{r e l} = \frac{1.5}{1.2} = 1.25$ :

$\frac{1}{f _{l e n s}} = (μ_{r e l} - 1) (\frac{1}{R _{1}} - \frac{1}{R _{2}}) = (1.25 - 1) (\frac{1}{R} - 0) = \frac{0.25}{R}$

(Curved face first, $R_{1} = R$ ; plane face $R_{2} = \infty$ .)

So $\frac{1}{f _{e q}} = \frac{2 \cdot 0.25}{R} = \frac{0.5}{R}$ , giving $R = 0.5 f_{e q} = 0.5 \times 0.2 = 0.10$ m.

$∴$ Option (2).

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JEE Main 2025 Jan 24 Shift 1, Physics Q12: Thin Lens and Mirrors

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