Question

1. An object is 29 cm in front of a convex lens with a focal length of 10 cm. Using ray tracing and thin lens equation, determine whether the image is real or virtual, upright or inverted, reduced or magnified.

Answer 1

29 cm 10 cm F 2F OLL 2F 15.3 cm 키

Thin lens equation :- (1/v) - (1/u) = 1/f

where v = lens-to-image distance ,

u = -29 cm , lens-to-object distance ( Cartesian sign convention is used ) ,

f = 10 cm , focal length of lens

1/v = ( 1/ 10 ) - (1/29)   , we get v = 15.3 cm

magnification :- v/u = 15.3/(-29) = - 0.53

Image is real because if we place a screen at distance 15.3 cm from lens at right side of lens if the object is at left side , then image is formed on screen.

Image is inverted as seen from ray diagram . From thin lens equation we get v = 15.3 , while the object distance u = -29. Hence magnification is -ve and image is inverted.

Image is diminished . In the ray diagram , we see that object is placed at left side of lens beyond 2F point.

Hence diminished image is formed between F and 2F. Using thin lens equation we get magnitude of magnification less than one. This also confirm that image is diminished