Answer
(a) $45.8diopters$
(b) decreases
Work Step by Step
(a) We know that
$Refractive \space power=\frac{1}{f}=\frac{1}{d_{\circ}}+\frac{1}{d_i}$
We plug in the known values to obtain:
$Refractive \space power=\frac{1}{0.24}+\frac{1}{0.0240}=45.8 diopters$
(b) We know that with the increase of $d_{\circ}$, the refractive power of the eyes decreases as the image distance is held constant.
