Answer
$-5.5cm$
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:
$i_1=(\frac{1}{f_1}-\frac{1}{p_1})^{-1}$
$i_1=(\frac{1}{-12cm}-\frac{1}{20cm})^{-1}$
$i_1=-7.5cm$
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 $p_2=d-i_1=10cm-(-7.5cm)=17.5cm$
$\frac{1}{p}+\frac{1}{i}=\frac{1}{f}$
Solving for $i$, we have:
$i_2=(\frac{1}{f_2}-\frac{1}{p_2})^{-1}$
$i_2=(\frac{1}{-8.0cm}-\frac{1}{17.5cm})^{-1}$
$i_2=-5.5cm$
