Answer
The height of the image is $~~3.0~mm$
Work Step by Step
We can find the location of the image:
$\frac{1}{s}+\frac{1}{s'} = \frac{1}{f}$
$\frac{1}{s'} = \frac{1}{f} - \frac{1}{s}$
$\frac{1}{s'} = \frac{1}{0.015~m} - \frac{1}{10~m}$
$\frac{1}{s'} = 66.57~m^{-1}$
$s' = 0.015~m$
$s' = 15~mm$
We can find the height of the image:
$h' = M~h$
$h' = (-\frac{s'}{s})(h)$
$h' = (-\frac{0.015~m}{10~m})(2.0~m)$
$h' = -0.0030~m$
$h' = -3.0~mm$
The height of the image is $~~3.0~mm$
