An object 4 cm tall is placed 15 cm from a diverging lens of focal length 30 cm. What is the height of the image formed?

Using the lens formula $\frac{1}{v} - \frac{1}{u} = \frac{1}{f}$ with f = −30 cm and u = −15 cm: $\frac{1}{v} = \frac{1}{-30} + \frac{1}{-15} = \frac{-1-2}{30} = \frac{-3}{30}$, so v = −10 cm. Magnification $m = \frac{v}{u} = \frac{-10}{-15} = \frac{2}{3}$. Image height = $\frac{2}{3} \times 4 = 2.67$ cm.