Chapter 23 - Ray Optics - Exercises and Problems - Page 690: 34

$s' = 30~cm$ $h' = 0.50~cm$

We can find the image distance: $\frac{1}{s}+\frac{1}{s'} = \frac{1}{f}$ $\frac{1}{s'} = \frac{1}{f}-\frac{1}{s}$ $\frac{1}{s'} = \frac{1}{20~cm}-\frac{1}{60~cm}$ $\frac{1}{s'} = \frac{1}{30~cm}$ $s' = 30~cm$ We can find the image height: $h' = \vert M \vert ~h$ $h' = (\frac{s'}{s}) ~(h)$ $h' = (\frac{30~cm}{60~cm}) ~(1.0~cm)$ $h' = 0.50~cm$