A short-sighted person can see distinctly the object situated at a distance of 20 cm from him. What type of lens will he use to see the objects situated at a distance of 100 cm from him ? What will be the power of the lens ?

Explanation

In this case a lens is to be used to form image of the object situated at a distance of 100 cm at 20 cm from the eye i.e ,u =-100 cm , v=-20cm , u and v are both negative .
From the equation `1/v-1/u=1/f` we have ,
`1/(-20)-1/(-100)=1/f or, 1/f=(-5+1)/(100)=-1/(25) or, f= -25 cm `
So, a concave lens of focal length 25 cm is to be used,
Power of the lens , P`=(100)/f=(100)/(-25)=-4D. `
[Short solution : distance of far point ,d =20 cm After using lens distance of far point, d'=100 cm `therefore` Power of lens ,
P`=(100)/(d')-(100)/d=(100)/(100)-(100)/(20)=-4D`
Since power is negative, the lens is a concave lens]