JEE Main 2025 Apr 8 Shift 2, Physics Q6: Thin Lens and Mirrors

JEE Main2025Apr 8, Shift 2Physics

A concave-convex lens of refractive index 1.5 and the radii of curvature of its surfaces are 30 cm and 20 cm, respectively. The concave surface is upwards and is filled with a liquid of refractive index 1.3. The focal length of the liquid-glass combination will be

A
$\frac{500}{11}$ cm
B
$\frac{800}{11}$ cm
C
$\frac{700}{11}$ cm
D
$\frac{600}{11}$ cm

Solution

The system consists of two thin lenses in contact: a plano-concave liquid lens (flat top, concave bottom matching the glass) sitting on the concave-convex glass lens.

For the liquid lens ( $μ = 1.3$ ) with flat upper surface ( $R_{1} = \infty$ ) and lower surface matching the glass concave surface ( $R_{2} = - 30$ cm using the sign convention that concave-up means the centre of curvature is below): $\frac{1}{f _{L}} = (1.3 - 1) (\frac{1}{\infty} - \frac{1}{- 30}) = \frac{0.3}{30} = \frac{1}{100}$ .

For the glass lens ( $μ = 1.5$ ), the upper concave surface has $R_{1} = - 30$ cm and the lower convex surface has $R_{2} = - 20$ cm (centre of curvature above the surface): $\frac{1}{f _{G}} = (1.5 - 1) (\frac{1}{- 30} - \frac{1}{- 20}) = 0.5 \cdot \frac{1}{60} = \frac{1}{120}$ .

Combining lenses in contact: $\frac{1}{f} = \frac{1}{100} + \frac{1}{120} = \frac{6 + 5}{600} = \frac{11}{600}$ .

Therefore $f = \frac{600}{11}$ cm, matching option (4).

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JEE Main 2025 Apr 8 Shift 2, Physics Q6: Thin Lens and Mirrors

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