## Physics for Scientists and Engineers: A Strategic Approach with Modern Physics (4th Edition)

$s' = -8.57~cm$ $h' = 1.14~cm$
We can find the image distance: $\frac{1}{s}+\frac{1}{s'} = \frac{1}{f}$ $\frac{1}{s'} = \frac{1}{f}-\frac{1}{s}$ $\frac{1}{s'} = \frac{1}{-20~cm}-\frac{1}{15~cm}$ $\frac{1}{s'} = -\frac{7}{60~cm}$ $s' = -8.57~cm$ We can find the image height: $h' = \vert M \vert ~h$ $h' = \vert (\frac{s'}{s}) \vert ~(h)$ $h' = \vert (\frac{-8.57~cm}{15~cm})\vert ~(2.0~cm)$ $h' = 1.14~cm$