Question

Question #b5c81

Answer 1

As the lens is being used as magnifying glass the image formed will be virtual and magnified,

The lens formula

`1/v-1/u=1/f`

where

`v->"Image distance"`

`u->"object distance"`

`f->"focal length"`

The magnification given here is 5 times the size of an object and image is formed 20 m from lens. So

`v->"Image distance"=-20m(" -ve sign for virtual image")`

`u->"object distance"="Image distance"/"magnification"=-20/5=-4`

Inserting these in lens equation we have

`1/(-20)-1/(-4)=1/f`

`=>1/f=1/(-20)-1/(-4)=1/4-1/20=(5-1)/20=1/5`

`f=5m`

Hence the focal length of the lens is 5m and the object is to be placed at 4m distance to have 5 times magnified virtual image at a distance 20m from the lens.