Question #b5c81

1 Answer
P dilip_k
Aug 28, 2016

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.

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