A 55 year old man can focus objects clearly from $100 \mathrm{~cm}$ to $300 \mathrm{~cm}$. Representing the eye as a simple lens $2 \mathrm{~cm}$ from the retina, what is the focal length of the lens at the far point (focussed at $300 \mathrm{~cm}$)?
$\frac{1}{f_{\text {far }}}=\frac{1}{u_{\text {far }}}+\frac{1}{v}, \quad u_{\text {far }}=300 \mathrm{~cm}, v=2 \mathrm{~cm}$.

Solving the equation yields $f_{\text {far }}=1.987 \mathrm{~cm}$.