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,
`f_M =(R )/(2) = (-22)/(2) =- 1 1 cm ` and hence
` 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 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)/( 0.20 ) +(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 behaves 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., image will be 11 cm in front of the silvered lens and will be real as shown in Fig.