A hypermetropic person having near point at a distance of 0.75 m puts on spectacles of power 2.5 D. The near point now is at ____?

A hypermetropic person having near point at a distance of 0.75 m puts on spectacles of power 2.5 D. The near point now is at ____?

Explanation

1/f = 1/v - 1/u

2.5 = 1/0.75 - 1/u

u = -0.26 m

So, the new near point is indeed:

0.26 m