A farsighted woman has a near point of 700 cm What power eye

A farsighted woman has a near point of 70.0 cm. What power eyeglasses held 1.90 cm from the eyes will allow her to see objects 25.0 cm away from the eyes?
D

Solution

For woman:

Object distance from lens, do = 25 -2 = 23 cm

Image distance, di = - 70 + 2 = - 68 cm

Applying lens equation,

1/f = 1/di + 1/do

1/f = 1/(-68) + 1/23

f = 34.76 cm = 0.3476 m

P = 1/f = 2.88 D