Question

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

Answer 1

For Hypermetropic eye,

v=-1m=-100cm

u=-25cm
Using lens formula,
1/f=1/v -1/u
=1/(-100) - 1/(-25)
=3/100
or f=100 /3 cm=1/3 m
Now,
Power,P=1/f=3 D