Let R be the radius of curvature of both the surfaces of the equi-convex lens. In the first case:
Let f₁ be the focal length of planoconcave lens of index ${\mu }_1$. The focal length of the combined lens system will be given by:
               $\frac{1}{\text{F}}=\frac{1}{{\text{f}}_1}+\frac{1}{{\text{f}}_2}$



         $=\left({\mu }_1-1\right)\left(\frac{1}{\text{R}}-\frac{1}{-\text{R}}\right)+\left({\mu }_2-1\right)\left(\frac{1}{-\text{R}}-\frac{1}{\infty }\right)$

         $=\left(\frac{3}{2}-1\right)\left(\frac{2}{\text{R}}\right)+\left(\frac{4}{3}-1\right)\left(-\frac{1}{\text{R}}\right)$

         $=\frac{1}{\text{R}}-\frac{1}{3\text{R}}=\frac{2}{3\text{R}} \text{ of }\text{F}=\frac{3\text{R}}{2}$


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

Hence,  F 15 cm      ( Distance of object = 15 cm).

or         $\frac{3\text{R}}{2}=15 \text{cm}\text{ or }\text{R}=10\text{cm}$

In the second case, let $\mu$be the refractive index of the liquid filled between lens and mirror and let F' be the focal length of new lens system. Then,



$\frac{1 }{{\text{F}}^{\prime }}=\left({\mu }_1-1\right)\left(\frac{1}{\text{R}}-\frac{1}{\text{R}-}\right)+\left(\mu -1\right)\left(\frac{1}{-R}-\frac{1}{\infty }\right)$

or,   $\frac{1}{{\text{F}}^{\prime }}=\left(\frac{3}{2}-1\right)\left(\frac{2}{\text{R}}\right)-\frac{\left(\mu -1\right)}{\text{R}}$

or,   $\frac{1}{\text{R}}-\frac{\mu -1}{\text{R}}=\frac{\left(2-\mu \right)}{\text{R}}$

 ${\text{F}}^{\prime }=\frac{\text{R}}{2-\mu }=\frac{10}{2-\mu }\left(\text{R}=10 \text{cm}\right)$


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

Hence,  ${\text{F}}^{\prime }=25$

or,    $\frac{10}{2-\mu }=25 \text{ or }50-25 \mu =10 \text{ or } 25\mu =40$

 $\; \therefore \mu =\frac{40}{25}=1·6 \text{ or }\mu =1·6$