Answer
$-0.58$
Work Step by Step
Recall that the overall magnification of the system, $M$, is a product of the magnification of each lens $m_1, m_2$, and $m_3$.
We have:
$M=m_1*m_2*m_3$
From equation 34-6, we know that $m=-\frac{i}{p}$; therefore we can obtain :
$M=(-\frac{i_1}{p_1})(-\frac{i_2}{p_2})(-\frac{i_3}{p_3})$
The object distance for the first lens, $p_1$, is given in the table. The other values we know from our work in part A.
$M=(-\frac{i_1}{p_1})(-\frac{i_2}{p_2})(-\frac{i_3}{p_3})$
$M=(-\frac{-4.0cm}{8.0cm})(-\frac{-6.9cm}{12cm})(-\frac{24cm}{12cm})$
$M=-0.58$
