Make a diagram to show how hypermetropia is corrected. The near point of a hypermetropia eye is 1m. What is the power of the lens required to correctt this defect ? Assume that the near point of the normal eye is 25 cm.

Explanation

A person suffering from hypermetropia can see distinct objects clearly but faces difficulty in seeing nearby objects clearly. It happens because the eye lens focuses the incoming divergent - rays beyond the retina. This defect of vision is corrected by using a convex lens. A convex lens of suitable power converge the incoming light in such a way that the image is formed on the retina, as shown m the following figure.

The convex lens actually ereates a virtual image of a nearby object (N. in the figure) at the near point of visiori (N) of the person suffering from hypermetropia.
The given person will be able to clearly see the object kept at 25 cm ( near point of the normal ·eye), if the image of the object is fmmed at his. near point, which is given as 1 m.
Object distance, u=-25 cm
Image distance, v=-1m=-100m
Focal length f
using the formula,
`1/v-1/u=1/f`
`-1/(100)-1/(-25)=1/f`
`1/f=-1/(25)-1/(100)`
`1/f=(4-1)/(100)`
`f=(100)/3=33.3cm=0.33m`
we know,
Power, `P=1/(f("in metres"))`
`P=1/(0.33)=+3.0D`
A convex lens of power +3.0 D is required to correct the defect.