Answer
$+0.29$
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.
$M=(-\frac{i_1}{p_1})(-\frac{i_2}{p_2})(-\frac{i_3}{p_3})$
$M=(-\frac{-12cm}{4.0cm})(-\frac{-3.33cm}{20cm})(-\frac{-5.2cm}{9.03cm})$
$M=+0.29$
