Saved Bookmarks
| 1. |
A pin is placed 10 cm in front of a convex lens of focal length 20cm made of material having refractive index 1.5. The surface of the 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? |
|
Answer» Solution :As radius of curvature of silvered surface is 22 cm, `f_M=(R )/(2)= (-22)/(2) =- 1 1 cm ` andhence ` 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 thefocallengthof equivalentmirror ` 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 10cmin frontof 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. |
|