Answer
a)
b) $d_{\circ\space max}=15~cm$
Work Step by Step
(a) We know that when a normal eye relaxes, it is focused on distant objects and its refractive power is at a minimum distance. Since the refractive power and the object distance are inversely proportional to each other, when the distance increases then the refractive power becomes a minimum. Thus, we conclude that the person whose eyes refract more strongly in a relaxed state cannot see distant objects clearly and hence the patient is nearsighted.
(b) We know that
$\frac{1}{d_{\circ\space max}}=\frac{1}{f_{rel}}-\frac{1}{d_i}$
We plug in the known values to obtain:
$\frac{1}{d_{\circ\space max}}=48.5diopters-\frac{1}{0.0240m}$
$\implies \frac{1}{d_{\circ\space max}}=6.84m $
$ d_{\circ\space max}=0.146~m=15~cm$
