Two transparent media having refractive indices 1.0 and 1.5 separated by a spherical refracting surface of radius of curvature 30 cm. The centre of curvature of surface is towards denser medium and a point object is placed on the principal axis in rarer medium at a distance of 15 cm from the pole of the surface. The distance of image from the pole of the surface is …… cm.

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answer is 30.

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Detailed Solution

$\begin{array}{l} \frac{μ_{2}}{v} - \frac{μ_{1}}{u} = \frac{μ_{2} - μ_{1}}{R} \\ \Rightarrow \frac{1.5}{v} + \frac{1}{15} = \frac{1.5 - 1}{30} = \frac{1}{60} \\ \Rightarrow \frac{1.5}{v} = \frac{1}{60} - \frac{1}{15} = \frac{- 3}{60} \\ \Rightarrow v = - 30 cm \end{array}$