Answer
The height of the final image is $-1.2cm$, or in magnitude, $1.2cm$
Work Step by Step
First we can use the given values and our answer to part A in equation 34-6 to find the magnification of each lens.
$m_1=-i_1/p_1$
$m_1=-(-6.0cm)/10cm$
$m_1=0.60$
$m_2=-i_2/p_2$
$m_2=-36cm/18cm$
$m_2=-2.0$
Recall that the magnification of the system, $M$, is the product of the magnification of each lens.
$M=m_1m_2$
$M=(0.60)(-2.0)$
$M=-1.2$
Now we can multiply the magnification, ($-1.2$), by the actual object height, ($1.0cm$), to find the final image height.
$height=-1.2*1.0cm=-1.2cm$
The height is -1.2cm, or 1.2cm inverted.
