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

The object placed at 25 cm from the correcting lens must produce a virtual image at.
1m or 100 cm.
Object distance = u = - 25 cm, image distance = v = -100 cm.
By lens formula,
`(1)/(f) = (1)/(v) - (1)/(u) = (1)/(-100) -(1)/(-25) = -(1)/(100) + (1)/(25) = +(3)/(100)`
or `f = +(100)/(3) cm = +(1)/(3)m`
Power, `P = (1)/(f) = +(3)/(1) = +3D`.