A pin is placed 10cm in front of a convex lens of focal length 20cm, made of a material having refractive index 1.5 . The surface of 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?

Explanation

As radius of curvature of silvered surface is `22`cm ,
so `f_(M) = (R)/(2) = (-22)/(2) = -11 "cm" = -0.11 "m"`
and hence , `M = -(1)/(f_(M)) = -(1)/(-0.11) = (1)/(0.11)D`

Further as the focal length of lens is `20` cm, i.e., `0.20` m its power will be given by :` P_(L) = (1)/(f_(L)) = (1)/(0.20)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) "i.e.," P =(2)/(0.20) + (1)/(0.11) = (210)/(11)D`.
So the focal length of equivalent mirror `F = -(1)/(P) = -(11)/(210) "m" = -(110)/(21)` cm i.e., the silvered lens behave as a concave mirror of focal length `(110//21)` cm. So for object at a distance `10` cm in front of it , `(1)/(v) + (1)/(-10) = -(21)/(110)` 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 Figure.