A short sighted person is wearing specs of power `-3.5 D`. His doctor prescribes a correction of `+ 2.5 D` for his near vision. What is focal length of his distance viewing part and near vision.

Answer

`28.5cm` ; `16.7 cm`

Explanation

For distance viewing part , `P_(1) = - 3.5 D`
`:. f_(1) = (100)/(P_(1)) = (100)/(-3.5)cm = - 28.5 cm`
For near vision part, `P_(1) + P_(2) = P`
`P_(2) = P - P_(1) = 2.5 D - (3.5 D) = 6.0 D`
`f_(2) = (100)/(P_(2)) = (100)/(6.0) = 16.7 cm`