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 22cm, so
`f_(M)=(R)/(2)=(-22)/(2)=-11cm=-0.11m`
And hence, `P_(M)=-(1)/(f_(M))-(1)/(-0.11)=(1)/(0.11)D`
Further as the focal length of lens is 20cm, i.e., 0.20m, 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 behaves as a concave mirror of focal lneght `(110//210)` cm. So for object at a distance 10cm in front of it,
`(1)/(v)+(1)/(-10)=-(21)/(110)i.e., v=-11cm`
i.e., image will be 11 cm in front of th esilvered lens and will be real as shown in the figure.