An equibiconvex lens has a focal length of 20 cm. A ball pin of length 5 cm is placed on one side of the lens, such that the mid point of the pin is at a distance of 30 cm from the centre of the lens. Calculate the size of the image of the pin and its magnification f = 20 cm l = 5 cm.

Explanation

`(1)/(f)=(1)/(-u)+(1)/(v)`
Given : `-u_(A)=27.5 cm`
`(1)/(f)=(1)/(-u_(A))+(1)/(v_(A))`
`(1)/(20)=(1)/(27.5)+(1)/(v_(A))`
`(1)/(20)-(1)/(27.5)=(1)/(v_(A))`
`v_(A)= 73.33 cm`
`(1)/(f) = (1)/(-u_(B))+(1)/(v_(B))`
`-u_(B)=32.5 cm`
`(1)/(20)-(1)/(32.5)=(1)/(v_(B))`
`(1)/(v_(B))=(12.5)/(650)`
`v_(B)=52 cm`
`v_(A)-v_(B)=21.33cm`
magnification `m=(v_(1))/(v_(0))=(21.33)/(5)=4.266`