Question

Make a diagram to show how hypermetropia is corrected. 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.​

Answer 1

Answer:

Near point of defective eye is 1 m and that of normal eye is 25 cm.

Here, u = -25 cm, v = -1m = 100 cm.

Using lens formula

1/f = 1/v – 1/u

1/f = 1/-100 + 1/25 = 3/100

f = 100/3 cm = 1/3m.

P = 1/f(m) = 1/0.33 = +3.0 D.