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

Explanation

A person suffering from hypermetropia can see distinct objecdts clearly but faces difficulty is seeing nearly objects clearly. It happens because the eye lense focuses the incoming diverget rays beyond the ratina. This defect of vision is corrected by using a convex lens. A convex lens of suitable power converges the incoming light in such a way that the image is formed on the retina, as shown in the following firgure .

The convex lens actually creates a virtual image of a nearby object (N'in hte figure ) at the near point of vision (N) of the perosn suffering from hyermetropia.
The given perons wil be able to clearly see the object kept at 25 cm (near point of the normal eye). if the image of the object is formed 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 lens 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)=33.3 cm=0.33m`
We known,
Powr , `P=(1)/(f(" in meteres"))`
`P=(1)/(0.33)=+30.0D`
A convex lens of power +3.0 D is required to correct the defect.