Answer
(a) $-0.60diopters$
(b) $27cm$
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:
$Refractive \space power=\frac{1}{\infty}+\frac{1}{-(1.7-0.020)}=-0.60diopters$
(b) We know that
$d_{\circ}=\frac{1}{f}-\frac{1}{d_i}$
We plug in the known values to obtain:
$d_{\circ}=(-0.60-\frac{1}{-(0.25-0.020)})^{-1}$
$d_{\circ}=27cm$
