Answer
$s' = -20~cm$
$h' = 0.33~cm$
Work Step by Step
We can find the image distance:
$\frac{1}{s}+\frac{1}{s'} = \frac{1}{f}$
$\frac{1}{s'} = \frac{1}{f}-\frac{1}{s}$
$\frac{1}{s'} = \frac{1}{-30~cm}-\frac{1}{60~cm}$
$\frac{1}{s'} = -\frac{1}{20~cm}$
$s' = -20~cm$
We can find the image height:
$h' = \vert M \vert ~h$
$h' = \vert (\frac{s'}{s}) \vert ~(h)$
$h' = \vert (\frac{-20~cm}{60~cm})\vert ~(1.0~cm)$
$h' = 0.33~cm$
