(a) Make a diagram to show how hypermetropia is corrected (b) The near point of a hypermetropic eye is 1 m. 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

Diagram representing the correction of hypermetropia is a follows:

Near point of defective eye is 1 m and that of normal eye is 25 cm.
Here, u = -25 cm, v = -1 m = 100 cm.
Using lens formula
1/f = 1/v-1/u
1/f = 1/-100 + 1/25 = 3/100