Answer
$+3.1cm$
Work Step by Step
First, we will find the image distance, $i_1$, of the first lens by using equation (34-9) with the given information about the first lens.
$\frac{1}{p}+\frac{1}{i}=\frac{1}{f}$
Solving for $i$, we have:
$i_1=(\frac{1}{f_1}-\frac{1}{p_1})^{-1}$
$i_1=(\frac{1}{9.0cm}-\frac{1}{20cm})^{-1}$
$i_1=16.4cm$
Now we can use equation (34-9) with the information given about the second lens to find the image distance for the image coming out of the second lens, $i_2$. From figure 34-45, we note that $p_2=d-i_1=8cm-(16.4cm)=-8.4cm$
$\frac{1}{p}+\frac{1}{i}=\frac{1}{f}$
Solving for $i$, we have:
$i_2=(\frac{1}{f_2}-\frac{1}{p_2})^{-1}$
$i_2=(\frac{1}{5.0cm}-\frac{1}{-8.4cm})^{-1}$
$i_2=+3.1cm$
