Question

A person with presbyopia has a near point of 65 cm. This person wears glasses to correct their vision such that they can comfortably read the screen of their mobile phone when it is held 25 cm from their eyes.

If this person wears their glasses such that they sit 1.4 cm from their eye, what is the optical power of the corrective lenses in their glasses? (to 2 s.f and in D)

Answer 1

The distance between the image and the eyes must be 65 cm. Therefore, the distance between the lens and the image must be 1.4 cm less, or $i=-63.6$ cm. The distance is negative because the image forms on the same side as the object.

The distance between the object and the eye must be 25 cm, or o=23.6 cm from the lens.

The optical power of the corrective lenses in their glasses:

$$P=\frac{1}{o}+\frac{1}{i}=\frac{1}{0.236}+\frac{1}{-0.636}=2.66D.$$