Answer
(a) $2.42cm$
(b) $2.42cm$
Work Step by Step
(a) The effective focal length of the eye can be determined as
$f_{effective}=\frac{1}{refractive \space power}$
We plug in the known values to obtain:
$f_{effective}=\frac{1}{41.4diopters}=2.42cm$
(b) We can find the image distance as
$d_i=(\frac{1}{f}-\frac{1}{d_{\circ}})^{-1}$
$\implies d_i=(\frac{1}{f}-\frac{1}{\infty})^{-1}$
This simplifies to:
$d_i=f=2.42cm$
