Question

The near point of a person suffering from hypermetropia is at 50 cm from his eye. What is the power of the lens needed to correct this defect? (near point of normal eye is 25 cm)

Answer 1

v = - 50 cm

u = -25 cm

Using lens equation,

1/f = 1/v -1/u

1/f = -1/50 - 1/(25)

1/f = -1/50 + 1/25

1/f = 1/50

P = 1/f (in m) = 100/50 = 2 D