Question

the far point of a hypermetropic eye is 1m. find the required power of lens used to correct this.Assume that the near point of the eye is 25 cm.

Answer 1

v = - 100 cm

u = -25 cm

Using lens equation,

1/f = 1/v -1/u

1/f = -1/100 - 1/(25)

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

1/f = 3/100

P = 1/f (in m) = 100×3/100 = 3 D