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 :

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1 cm

$\frac{1}{3} cm$

$\frac{2}{3} cm$

$\frac{4}{3} cm$

answer is C.

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Detailed Solution

$\begin{array}{l} \frac{n_{2}}{v} - \frac{n_{1}}{u} \frac{n_{2} - n_{1}}{R} \\ \frac{1.4}{v} - \frac{1.3}{- 13} = \frac{0.1}{R} \\ \frac{1.4}{v} = \frac{1 - R}{10 R} \\ \frac{1.4}{v} = \frac{1 - R}{10 R} \\ m = \frac{v / n_{2}}{u / n_{1}} \\ - 2 \times \frac{(- 13)}{1.3} = \frac{10 R}{1 - R} \\ R = \frac{2}{3} cm \end{array}$

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