Answer
far point: $2.27m$
near point: $42.0cm$
Work Step by Step
We can find the far and near points as follows:
$d_i=(\frac{1}{f}-\frac{1}{d_{\circ}})^{-1}=(\frac{1}{f}-\frac{1}{\infty})^{-1}=(-0.445m^{-1})^{-1}=-2.25m$
Now $far\space point=-d_i+0.0200m=-(-2.25m)+0.0200m=2.27m$
Now we find the near point as
$d_i=(\frac{1}{f}-\frac{1}{d_{\circ}})^{-1}$
$d_i=(1.85dipoters-\frac{1}{0.250-0.0200})^{-1}=-0.400m$
$near\space point=-d_i+0.0200=-(-40.0cm)+2.00cm=42.0cm$
