Figure shows an arrangement of an equiconvex lens (f = 20 cm in air) and a concave mirror (R = 80 cm). A point object 'O' is placed on the principal axis at a distance 40 cm from the lens such that the final image is also formed at the position of object. Find d.  

 

For the lens in air, ${\mathrm{f}}_0=\frac{{\mathrm{R}}_1}{2\left({\mathrm{\mu }}_2-1\right)}$
$20=\frac{{\mathrm{R}}_1}{2(1.5-1)}\Rightarrow {\mathrm{R}}_1=20\mathrm{cm}$
 

This is the radius of curvature of each spherical surface of lens $\left[{\mathrm{R}}_1={\mathrm{R}}_2={\mathrm{R}}_1\right]$ relation between object and image distances
for a thin lens is given by 
$$\begin{gathered} \frac{{\mu }_3}{\mathrm{V}}-\frac{{\mu }_1}{\mathrm{u}}=\frac{{\mu }_2-{\mu }_1}{{\mathrm{R}}_1}-\frac{{\mu }_2-{\mu }_3}{R_2} \\ \text{ here }{\mu }_3=2,{\mu }_1=1.2,{\mu }_2=1.5,\mathrm{u} \\ =-40\mathrm{cm},{\mathrm{R}}_1=+20\mathrm{cm};{\mathrm{R}}_2=-20\mathrm{cm},\mathrm{V}=? \\ \frac{2}{\mathrm{V}}-\frac{1.2}{-40}=\frac{1.5-1.2}{20}-\frac{-2+1.5}{-20}\Rightarrow \mathrm{V}=-50\mathrm{cm} \end{gathered}$$

The lens forms a virtual image I₁ at a distance of 50 cm from it. This acts as real object to the concave mirror. 

If PI₁ = R = 80cm, the rays after reflection from mirror retrace their path and the final image is formed at ‘O’ itself. 

Hence from figure $\mathrm{d}={\mathrm{PI}}_1-50=80-50=30\mathrm{cm}$