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 . 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 :

Two liquids A ($$\mu_A = 1.3$$) and B ($$\mu_B = 1.4$$) form a convex meniscus towards A. An object at 13 cm in A forms an image with magnification $$m = -2$$.

We start by applying the refraction formula at a spherical surface: $$\frac{\mu_B}{v} - \frac{\mu_A}{u} = \frac{\mu_B - \mu_A}{R}$$. Since the meniscus is convex towards A, the center of curvature is on the A side, and taking the direction of light from A to B gives $$u = -13$$ cm by sign convention.

Next, the magnification condition for refraction at a single surface is $$m = \frac{\mu_A \cdot v}{\mu_B \cdot u}$$. Substituting $$m = -2$$ yields $$-2 = \frac{1.3 \times v}{1.4 \times (-13)}$$ and hence $$-2 = \frac{1.3v}{-18.2}$$.

This gives $$v = \frac{-2 \times (-18.2)}{1.3} = \frac{36.4}{1.3} = 28$$ cm.

Now, finding R from the refraction formula, $$\frac{1.4}{28} - \frac{1.3}{-13} = \frac{1.4 - 1.3}{R}$$ leads to $$0.05 + 0.1 = \frac{0.1}{R}$$ and therefore $$0.15 = \frac{0.1}{R}$$.

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

Hence, the correct answer is Option D) $$\frac{2}{3}$$ cm.