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

If an object is placd at 25 cm from the correcting lensl it should produce the virtual image at 40 cm. Thus, `u=-25cm, v=-40cm`
`1/f=1/v-1/u`
`=1/(40cm)+1/(25cm)`
or `f=200/3cm=+2/3m`
`P=1/f+1.5D`
The uN/Aided eye can see a maximum distance of 250 cm. Suppose the maximum distance for clear vision is d when the lens is used. Then the object at a distance d is imaged by the lens ast 250 cm. We have
`1/v-1/u=1/f`
`or - 1/(250cm)-1/d=3/(200cm)`
`or d=-53cm`
Thus the person will be able to see up to a maximum distance of 53 cm.