JEE Main 2025 Jan 28 Shift 2, Physics Q6: Refraction at Plane and Spherical surfaces

JEE Main2025Jan 28, Shift 2Physics

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
D
$\frac{4}{3}$ cm

Solution

Given,

$μ_{1} = 1.3, μ_{2} = 1.4, u = - 13 cm, m = - 2$

Using the magnification formula for refraction at a spherical surface,

$m = \frac{μ _{1} v}{μ _{2} u}$

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

$\Rightarrow v = 28 cm$

Now, using the refraction formula,

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

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

$0.05 + 0.10 = \frac{0.1}{R}$

$0.15 = \frac{0.1}{R}$

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

$R = \frac{2}{3} cm$

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JEE Main 2025 Jan 28 Shift 2, Physics Q6: Refraction at Plane and Spherical surfaces

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