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 ^\prime-2^\prime then the radius of curvature of meniscus is :

Geometrical Optics PYQ — JEE Main 2025 · 28 Jan · Shift 2 · Q66 with Solution

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Correct answer: A

The meniscus is convex towards A, so for light going from $A$ to $B$ , the center of curvature lies on the $B$ side. Hence $R>0$ .

For refraction at a spherical surface,

m = \frac{n_1 v}{n_2 u}

Given $m=-2$ ,

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

So,

-2 = -\frac{1.3v}{18.2}

2 = \frac{1.3v}{18.2}

v = \frac{2\times 18.2}{1.3} = 28\,\text{cm}

Thus image distance is

v = +28\,\text{cm}

The formula is

\frac{n_2}{v} - \frac{n_1}{u} = \frac{n_2-n_1}{R}

Substitute the values:

\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}

Therefore,

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

R = \frac{2}{3}\,\text{cm}

So the correct option is:

A. $\dfrac{2}{3}\,\text{cm}$

Stored correct answer: A

Our derived answer also gives A.