Question

What power of spectacle lens is needed to allow a farsighted person, whose near point is 1.00 m, to see an object clearly that is 25.0 cm away? Assume the

spectacle (corrective) lens is held 1.50 cm away from the eye by eyeglass frames.

Answer 1

$P=\frac{1}{a}+\frac{1}{b}=\frac{1}{0.25-0.015}-\frac{1}{1-0.015}=4.26-1.02=3.24\ (D)$ . Answer