Question

A person wears eye glasses with a power of `- 5.5 D` for distance viewing. His doctor prescribes a correction of `+ 1.5 D` for his near vision. What is the focal length of his distance viewing part of the lens and also for near vision section of the lens ?

Answer 1

(a) For distance viewing,
`P_1 = -5.5 D`
`:. f_1 = (100)/(P_1) = (100)/(-5.5) = -18.73 cm`
(b) As power of near vision part is measured relative to the main part of lens of power `-5.5 D`,
therefore, `P_1 + P_2 = P`
`-5.5 + P_2 = 1.5`
`P_2 = 1.5 + 5.5 = 7.0 D`
`f_2 = (100)/(P_2) = (100)/(7.0) = 14.3 cm`.