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

11 cm

The curved silvered surface will behave as a concave lens of focal length

$f_m =\frac{R}{2}=-\frac{22}{2}=- 11cm =-0.11m$

And hence

P_M = the power of the mirror = $-\frac{1}{f_m}=-\frac{1}{-0.11}=+\frac{1}{0.11}D$

Further as the focal length of lens is 20 cm i.e., 0.20m its power will be given by :

$P_L =\frac{1}{f_L}=\frac{1}{20} D$

Now as in image formation, light after passing through the lens will be reflected back by the curved mirror through the lens again.

$P = P_L + P_M + P_L = 2P_L + P_M$

$P =\frac{2}{0.20}+\frac{1}{0.11}=\frac{210}{11} D$

So the focal length of equivalent mirror

$F =-\frac{1}{P}=\frac{11}{210}m=-\frac{110}{21} cm$

i.e. the silvered lens behaves as a concave mirror of focal length (110/21) cm. So for object at a distance 10 cm in front of it

$\frac{1}{v}+\frac{1}{-10}=\frac{21}{110}$ i.e., v = -11cm

i.e. image will be 11cm infront of the silvered lens and will be real as shown in figure.