Answer
$+10cm$
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:
$i_1=(\frac{1}{f_1}-\frac{1}{p_1})^{-1}$
$i_1=(\frac{1}{15cm}-\frac{1}{10cm})^{-1}$
$i_1=-30cm$
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 $p_2=d-i-1=10cm-(-30cm)=40cm$
$\frac{1}{p}+\frac{1}{i}=\frac{1}{f}$
Solving for $i$, we have:
$i_2=(\frac{1}{f_2}-\frac{1}{p_2})^{-1}$
$i_2=(\frac{1}{8.0cm}-\frac{1}{40cm})^{-1}$
$i_2=+10cm$
