Question 146 of 295

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 ' $-2^{\prime}$ then the radius of curvature of meniscus is :

A$\frac{1}{3} \mathrm{~cm}$B$\frac{4}{3} \mathrm{~cm}$C$1\mathrm{~cm}$D$\frac{2}{3} \mathrm{~cm}$

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Answer: (D) $\frac{2}{3} \mathrm{~cm}$

$\begin{aligned} & \frac{\mathrm{n}_2}{\mathrm{v}}-\frac{\mathrm{n}_1}{\mathrm{u}} \frac{\mathrm{n}_2-\mathrm{n}_1}{\mathrm{R}} \\ & \frac{1.4}{\mathrm{v}}-\frac{1.3}{-13}=\frac{0.1}{\mathrm{R}} \\ & \frac{1.4}{\mathrm{v}}=\frac{1-\mathrm{R}}{10 \mathrm{R}} \\ & \frac{1.4}{\mathrm{v}}=\frac{1-\mathrm{R}}{10 \mathrm{R}} \\ & \mathrm{m}=\frac{\mathrm{v} / \mathrm{n}_2}{\mathrm{u} / \mathrm{n}_1} \\ & -2 \times \frac{(-13)}{1.3}=\frac{10 \mathrm{R}}{1-\mathrm{R}} \\ & \mathrm{R}=\frac{2}{3} \mathrm{~cm}\end{aligned}$