A thin equi-convex lens of refractive index $\frac{3}{2}$  is placed on a horizontal plane mirror as shown in the figure. The space between the lens and the mirror is then filled with water of refractive index $\frac{4}{3}$. It is found that when a point object is placed 15cm above the lens on its principal axis, the object coincides with its own image. On repeating with another liquid, the object and the image again coincide at a distance 25cm from the lens. Calculate the refractive index of the liquid.

Let R be the radius of curvature of both the surfaces of the equi – convex lens, then in the first case, the situation is shown in figure.

Let $f_1$ be the focal length of equi – convex lens of refractive index  ${\mu }_1$ and $f_2$ be the focal length of Plano–concave lens (made of water) of refractive index  ${\mu }_2$. The focal length of the combined lens 
 system is given by
$$\begin{gathered} \frac{1}{F}=\frac{1}{f_1}+\frac{1}{f_2}=({\mu }_1-1)(\frac{1}{R}-\frac{1}{-R})+({\mu }_2-1)(\frac{1}{-R}-\frac{1}{\infty }) \\ \Rightarrow \frac{1}{F}=(\frac{3}{2}-1)\left(\frac{2}{R}\right)+(\frac{4}{3}-1)(-\frac{1}{R})=\frac{1}{R}-\frac{1}{3R}=\frac{2}{3R} \\ \Rightarrow F=\frac{3R}{2} \end{gathered}$$

Now, image coincides with the object when ray of light retraces its path i.e., it is falls normally on the plane mirror. This is possible only when object lies at center of curvature of the lens system.

$$\begin{gathered} \Rightarrow F=15cm \\ \Rightarrow \frac{3R}{2}=15cm \\ \Rightarrow R=10cm \end{gathered}$$

In the second case, let $\mu$ be the refractive index of the liquid filled between lens and mirror and let $F^1$ be the focal length of new lens system. Then,
$$\begin{gathered} \frac{1}{F^1}=({\mu }_1-1)(\frac{1}{R}-\frac{1}{-R})+(\mu -1)(\frac{1}{-R}-\frac{1}{\infty }) \\ \Rightarrow \frac{1}{F^1}=(\frac{3}{2}-1)\left(\frac{2}{R}\right)-\frac{(\mu -1)}{R}=\frac{1}{R}-\frac{\mu -1}{R}=\frac{(2-\mu )}{R} \\ \Rightarrow F^1=\frac{R}{2-\mu }=\frac{10}{2-\mu } \{\because R=10cm\} \end{gathered}$$

Now, the image coincides with the object when it is placed at 25 cm distance.

$$\begin{gathered} \Rightarrow F^1=25 \\ \Rightarrow \frac{10}{2-\mu }=25 \\ \Rightarrow 50-25\mu =10 \\ \Rightarrow 25\mu =40 \\ \Rightarrow \mu =\frac{40}{25}=1.6 \\ \Rightarrow \mu =1.6 \end{gathered}$$