Answer
$-4.0cm$
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:
$i_1=(\frac{1}{f_1}-\frac{1}{p_1})^{-1}$
$i_1=(\frac{1}{-6.0cm}-\frac{1}{4.0cm})^{-1}$
$i_1=-2.4cm$
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 34-17, we note that $p_2=d_{12}-i_1=9.6cm-(-2.4cm)=12cm$
$\frac{1}{p}+\frac{1}{i}=\frac{1}{f}$
Solving for $i$, we have:
$i_2=(\frac{1}{f_2}-\frac{1}{p_2})^{-1}$
$i_2=(\frac{1}{6.0cm}-\frac{1}{12cm})^{-1}$
$i_2=12cm$
Now we can use equation (34-9) with the information given about the third lens to find the image distance for the final image (the one coming from the third lens), $i_3$. From figure 34-17, we note that $p_3=d_{23}-i_2=14cm-(12cm)=2cm$
$\frac{1}{p}+\frac{1}{i}=\frac{1}{f}$
Solving for $i_3$, we have:
$i_3=(\frac{1}{f_3}-\frac{1}{p_3})^{-1}$
$i_3=(\frac{1}{4.0cm}-\frac{1}{2cm})^{-1}$
$i_3=-4.0cm$
