Answer
$A. 1.9mm$.
Work Step by Step
We know that
$d_i=\frac{d_{\circ}f}{d_{\circ} -f}$
We plug in the known values to obtain:
$d_i=\frac{50\times 3}{50-3}cm$
$\implies d_i=3.19cm$
Thus, the distance between the IOL and the retina should be $3.19cm$
Now $3.19cm-3cm=0.19cm=1.9mm$
We conclude that IOL should be moved $1.9mm$ to change the focus of the eye from an object at infinity to an object at a distance of $50cm$ and the correct answer is $A. 1.9mm$.
