Question #b5c81

1 Answer
Aug 28, 2016

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

The lens formula

1v1u=1f

where

vImage distance

uobject distance

ffocal length

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

vImage distance=20m( -ve sign for virtual image)

uobject distance=Image distancemagnification=205=4

Inserting these in lens equation we have

12014=1f

1f=12014=14120=5120=15

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.