A person having long sight cannot see thins distinctly at a distance less than 40 cm . If he wants to see things situated at 25 cm from him, what should be the power of his spectacles ?

Explanation

Let the focal lengths of the lens of the spectacles =f. Here the power of the spectacles will be such that if an object is situated at a distance of 25 cm its virtual image will be formed at a distance of 40 cm from the eye i.e u =25 cm , v=40cm, both u and v are negative .
From the equation `1/v-1/u=1/f` we have,
`1/(-40)-1/(-25)=1/f or,1/f=(8-5)/(200)=3/(200) or, f=(200)/3 cm `
`therefore` power of the lens of the spectacles,
`P=(100)/f=100xx3/(200)=1.5m^(-1)=1.5 D`
[Short solution: distance of near point, d =40 cm `therefore` Power of the lens, `P=4 -(100)/(40)=4-2.5=1.5m^(-1)` As power is positive , the lens is a convex lens. ]