Answer
$0.60m$
Work Step by Step
We can use the given values in equation 34-9 to find the distance between the lens and the image, $i$.
$\frac{1}{p}+\frac{1}{i}=\frac{1}{f}$
Solving for $i$, we have:
$i=(\frac{1}{f}-\frac{1}{p})^{-1}$
$i=(\frac{1}{0.50m}-\frac{1}{1.0m})^{-1}$
$i=1.0m$
Therefore the image formed by the mirror is 1.0m away plus the distance between the mirror and the lens, 2.0m, for a total of 3.0m.
We can use the given values in equation 34-9 to find the distance between the lens and the final image after it's reflected by the mirror, $i_f$.
$\frac{1}{p_f}+\frac{1}{i_f}=\frac{1}{f_f}$
Solving for $i$, we have:
$i_f=(\frac{1}{f_f}-\frac{1}{p_f})^{-1}$
$i_f=(\frac{1}{0.50m}-\frac{1}{3.0m})^{-1}$
$i_f=0.60m$
