An object of height `h_(0)=1 cm` is moved along principal axis of a convex lens of focal length `f=10 cm`. Figure shows variation of magnitude of height of image with image distance `(v)` . Find `v_(2)-v_(1)` in `cm` .

Answer

10

Explanation

`(h_(i))/(h_(0))=(f-v)/(f) rArrh_(i)=-(v)/(f)h_(0)+h_(0)rArr |h_(i)|=-(v)/(f)h_(0)+h_(0)-inftylevlef`
`|h_(i)|=(v)/(f)h_(0)-h_(0) flevle-infty`
So `h_(2)=h_(2)=1 cm`
From second eq. `v_(2)=2f`
Or When `vrarr0,urarr0&h_(i)rarrh_(0)soh_(2)=h_(0)=1 cm` image of same height is obtained when `v=2f` so `v_(2)=2f`