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 cm`

and here `P_(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 throught the lens will be reflected back by the curved mirror through the lens again
`P= P_(L) + P_(M) + P_(L) -= 2P_(L) + P_(L)`
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) M`
i.e., the silvered lens behave as a concave mirror of focal length `(11//21)` cm so for object at a distance 10 cm in fromt of it.
`(1)/(v)+ (1)/(-10) = - (21)/( 110) `
i.e., image will be 11 cm fromt of the silvered lens and will be real as shown in Figure.