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 releative 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`