The far point of a student is 3.5 m away. Find the power of the lens that corrects her eye.

Short sight (myopia): far objects blur; a concave lens with $f = -(\text{far point})$. Long sight (hypermetropia): near objects blur; a convex lens that makes a page at $u = -25\ \text{cm}$ appear at $v = -(\text{near point})$, from $\dfrac{1}{f} = \dfrac{1}{v} - \dfrac{1}{u}$. Power $P = \dfrac{1}{f}$ with $f$ in metres. Old age (presbyopia) may need bifocals.