Answer
far point: $16.0m$
near point: $32.3cm $
Work Step by Step
We can find the far and near points as follows:
$d_i=(\frac{1}{f}-\frac{1}{d_{\circ}})^{-1}=(\frac{1}{f}-\frac{1}{\infty})^{-1}=(-0.0625m^{-1})^{-1}=-16.0m$
Now $far\space point=-d_i+0.0200m=-(-16.0m)+0.0200m=16.0m$
Now we find the near point as
$d_i=(\frac{1}{f}-\frac{1}{d_{\circ}})^{-1}$
$d_i=(1.05dipoters-\frac{1}{0.250-0.0200})^{-1}=-30.3cm$
$near\space point=-d_i+0.0200=-(-30.3cm)+2.00cm=32.3cm$
