The near and far points of a person are at `40 cm` and `250cm` respectively. Find the power of the lens he/she should use while reading at `25cm`. With this lens on the eye, what maximum distance is clearly visible?

Explanation

`N.P. = 40 cm`
`(1)/(-(N.P.)) -(1)/(-("distance of object")) = (1)/(f)`
`(1)/(-40)-(1)/(-25) = (1)/(f) rArr (1)/(f)=(1)/(25) - (1)/(40) = (3)/(200)`
`f = (200)/(3)cm`
`P= (100)/(f(cm)) = (100)/(200//3) =+15.D`
The eye without glasses can see up to `250cm`. Let maximum distance for clear vision is `x`. Thus the object at a distance `x` is imaged by the lens at `250cm`
`(1)/(v)-(1)/(u)=(1)/(f)`
`(1)/(-250)-(1)/(-x) = (1)/(-200//3) rArr (1)/(x) = (3)/(200)+(1)/(250) = (15+4)/(1000)`
`x = (1000)/(19) = 52.6 cm `