Water (with refractive index =(4)/(3) ) in a tank is 18 ~cm deep....

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 $R=\propto$