Water (with refractive index $=\frac{4}{3}$ ) in a tank is $18 \mathrm{~cm}$ deep. Oil of refractive index $\frac{7}{4}$ lies on water making a convex surface of radius of curvature $R=6 \mathrm{~cm}$ as shown. Consider oil to act as a thin lens. An object $S$ is placed $24 \mathrm{~cm}$ above water surface. The location of its image is at $x \mathrm{~cm}$ above the bottom of the tank. Then, $x$ is

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Aicharya · Accepted Answer

Two refractions will take place, first from spherical surface and the other from the plane surface. So, applying $ \frac{\mu_2}{v}-\frac{\mu_1}{u}=\frac{\mu_2-\mu_1}{R} $ two times with proper sign convention. Ray of light is travelling downwards. Therefore, downward direction is taken as positive direction. $ \begin{gathered} \frac{7 / 4}{v}-\frac{1.0}{-24}=\frac{7 / 4-1.0}{+6} \\ \frac{4 / 3}{(18-x)}-\frac{7 / 4}{v}=\frac{4 / 3-7 / 4}{\propto} \end{gathered} $ Solving these equations, we get $ x=2 \mathrm{~cm} $ $\therefore$ Answer is 2 . Analysis of Question (i) Question is moderately difficult from calculation point of view, otherwise it is simple. (ii) $\frac{\mu_2}{v}-\frac{\mu_1}{u}=\frac{\mu_2-\mu_1}{R}$ can be applied for plane surface also with $R=\propto$ ;

Water (with refractive index $=\frac{4}{3}$ ) in a tank is $18 \mathrm{~cm}$ deep. Oil of refractive index…

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