Question

The distance between the eye lens and retina of an eye is 2.6 cm. The eye can see far objects comfortably. The near point is at 31.2 cm. Find the distance by which the focal length of the eye can be adjusted.

Answer 1

$$\frac{1}{25}-\frac{1}{31.2}=\frac{1}{f}\\f=125\ cm$$