Answer
The negative sign of $s'$ shows that the image is located $30~cm$ behind the mirror.
The height of the image is $1.5~cm$ and a positive $h'$ shows that the image is upright.
Work Step by Step
We can find the image distance:
$\frac{1}{s}+\frac{1}{s'} = \frac{1}{f}$
$\frac{1}{s'} = \frac{1}{f}-\frac{1}{s}$
$\frac{1}{s'} = \frac{1}{60~cm}-\frac{1}{20~cm}$
$\frac{1}{s'} = -\frac{1}{30~cm}$
$s' = -30~cm$
We can find the image height:
$h' = M ~h$
$h' = (\frac{-s'}{s}) ~(h)$
$h' = (\frac{30~cm}{20~cm}) ~(1.0~cm)$
$h' = 1.5~cm$
The negative sign of $s'$ shows that the image is located $30~cm$ behind the mirror.
The height of the image is $1.5~cm$ and a positive $h'$ shows that the image is upright.
