JEE Main · 2025 · 28 Jan 2025 · Shift 2 · Geometrical Optics

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In a long glass tube, mixture of two liquids A and B with refractive indices 1.3 and 1.4 respectively, forms a convex refractive meniscus towards A . If an object placed at 13 cm from the vertex of the meniscus in A forms an image with a magnification of ' -2 ' then the radius of curvature of meniscus is :

(A)1 cm
(B) $\frac{1}{3} cm$
(C) $\frac{2}{3} cm$ ✓ Correct
(D) $\frac{4}{3} cm$

Answer: (C)

Solution

Refraction at a single spherical surface separating media of refractive index

n_{1}

(object side, A) and

n_{2}

(image side, B) obeys

\frac{n _{2}}{v} - \frac{n _{1}}{u} = \frac{n _{2} - n _{1}}{R}

and the magnification is

m = \frac{n _{1} v}{n _{2} u}

Given

n_{1} = 1.3

n_{2} = 1.4

u = - 13 cm

m = - 2

- 2 = \frac{1.3 v}{1.4 \times ( - 13 )}

v = \frac{- 2 \times 1.4 \times ( - 13 )}{1.3} = \frac{36.4}{1.3} = 28 cm

Substitute into the refraction equation:

\frac{1.4}{28} - \frac{1.3}{- 13} = \frac{1.4 - 1.3}{R}

0.05 + 0.1 = \frac{0.1}{R}

0.15 = \frac{0.1}{R}

R = \frac{0.1}{0.15} = \frac{2}{3} cm

JEE Main · 2025 · 28 Jan 2025 · Shift 2 · Single correct option

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JEE Main · 2025 · 28 Jan 2025 · Shift 2 · Geometrical Optics

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