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.

Make a diagram to show how hypermetropia is corrected. The near point

1.	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.
Answer» <P> Solution : 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`.