A pin is placed 10 cm in front of a convex lens of focal length 20cm made of material having refractive
index 1.5. The surface of the lens farther away from the pin is silvered and has a radius of curvature 22cm. Determine the position of the final image. Is the image real or virtual?

As radius of curvature of silvered surface is 22 cm, ${\mathrm{f}}_{\mathrm{M}}=\frac{\mathrm{R}}{2}=\frac{-22}{2}=-11\mathrm{cm}$
and hence, ${\mathrm{P}}_{\mathrm{M}}=-\frac{1}{{\mathrm{f}}_{\mathrm{M}}}=-\frac{1}{-0.11}=\frac{1}{0.11}\mathrm{D}$
Further as the focal length of lens is 20 cm,
i.e., 0.20 m, its power will be given by: ${\mathrm{P}}_{\mathrm{L}}=-\frac{1}{{\mathrm{f}}_{\mathrm{L}}}=\frac{1}{0.20}\mathrm{D}$
Now as in image formation, light after passing through the lens will be reflected back by the curved mirror through the lens again.
$\mathrm{P}={\mathrm{P}}_{\mathrm{L}}+{\mathrm{P}}_{\mathrm{M}}+{\mathrm{P}}_{\mathrm{L}}=2{\mathrm{P}}_{\mathrm{L}}+{\mathrm{P}}_{\mathrm{M}}\text{ i.e., }\mathrm{P}=\frac{2}{0.20}+\frac{1}{0.11}=\frac{210}{11}\mathrm{D}$
So the focal length of equivalent mirror $\mathrm{F}=-\frac{1}{\mathrm{P}}=-\frac{11}{210}\mathrm{m}=\frac{110}{21}\mathrm{cm}$
i.e., the silvered lens behaves as a concave mirror of focal length (110/21) cm.
So for object at a distance 10cm in front of it, $\frac{1}{\mathrm{P}}=-\frac{11}{210}\mathrm{m}=-\frac{110}{21}\mathrm{cm}$
i.e., v = –11 cm i.e., image will be 11 cm in front of the silvered lens and will be real as shown in Fig.