Answer
$i_2=-30cm$
(30cm to the left of the second lens)
Work Step by Step
First we will use the given values in equation 34-9 to find the image distance from the first lens, $i_1$
$\frac{1}{p_1}+\frac{1}{i_1}=\frac{1}{f_1}$
Solving for $i_1$, we have:
$i_1=(\frac{1}{f_1}-\frac{1}{p_1})^{-1}$
$i_1=(\frac{1}{20cm}-\frac{1}{40cm})^{-1}$
$i_1=40cm$
From figure 34-45, we note that $p_2=d-i_1=10cm-(40cm)=-30cm$.
Using equation 34-9 on the second lens, we obtain:
$\frac{1}{p_2}+\frac{1}{i_2}=\frac{1}{f_2}$
$i_2=(\frac{1}{f_2}-\frac{1}{p_2})^{-1}$
$i_2=(\frac{1}{-15cm}-\frac{1}{-30cm})^{-1}$
$i_2=-30cm$ (30cm to the left of the second lens)
