A plano-convex lens has thickness 4 cm. When placed on a horizontal table with the curved face in contact with it, the apparent depth of the bottommost point of the lens is found to be 3 cm. If the lens is inverted such that the plane face is in contact with the table, the apparent depth of the center of the plane face of the lens is found to be $\frac{25}{8} \mathrm{cm}$. The focal length of the lens is

When the curved face of the lens is in contact with the table, the virtual image of the bottom-most point O of the lens is formed at $I_1$ due to refraction at the plane surface as shown in Fig (a).

Real depth = 4 cm and apparent depth = 3 cm. Hence

$\mu =\frac{4}{3}$

When the plane face of the lens is in contact with the table, the image of the center O of the plane face of the lens is formed at $I_2$ due to refraction at the curved face as shown in Fig (b). 

For refraction at this face, we have

$$\begin{gathered} \frac{{\mathrm{\mu }}_1}{\mathrm{\nu }}-\frac{{\mathrm{\mu }}_2}{\mathrm{u}}=\frac{{\mathrm{\mu }}_1-{\mathrm{\mu }}_2}{\mathrm{R}} \\ \mathrm{where} {\mathrm{\mu }}_1=1, {\mathrm{\mu }}_2=\mathrm{\mu }, \mathrm{u}=\mathrm{OP}=-4 \mathrm{cm}, \mathrm{and} \mathrm{v}={\mathrm{OI}}_2=-\frac{25}{8} \mathrm{cm}. \\ \mathrm{Putting} \mathrm{these} \mathrm{values}, \mathrm{we} \mathrm{get} \\ -\frac{8}{25}-\frac{\mathrm{\mu }}{-4}=\frac{1-\mathrm{\mu }}{-\mathrm{R}} \\ \mathrm{Putting} \mathrm{\mu } = \frac{4}{3} \mathrm{and} \mathrm{solving} \mathrm{we} \mathrm{get} \mathrm{R} = 25 \mathrm{cm}. \\ \mathrm{Now}, \mathrm{the} \mathrm{focal} \mathrm{length} \mathrm{of} \mathrm{the} \mathrm{plano}-\mathrm{convex} \mathrm{lens} \mathrm{is} \mathrm{given} \mathrm{by} \\ \frac{1}{\mathrm{f}}=\left(\mathrm{\mu }-1\right)\frac{1}{\mathrm{R}}=\left( \frac{4}{3}-1\right)\times \frac{1}{25}=\frac{1}{75} \\ \Rightarrow \mathrm{f}=75 \mathrm{cm}. \end{gathered}$$