An object 4 cm tall is placed 32 cm in front of a convex lens of focal length 12 cm…

Step 1. The image. $\dfrac{1}{v} = \dfrac{1}{12\ \text{cm}} - \dfrac{1}{32\ \text{cm}} = \dfrac{8 - 3}{96\ \text{cm}} = \dfrac{5}{96\ \text{cm}}$, so $v = +19.2\ \text{cm}$.

Step 2. The magnification. $m = \dfrac{19.2\ \text{cm}}{-32\ \text{cm}} = -0.6$.

Step 3. The height. $h' = -0.6 \times 4\ \text{cm} = -2.4\ \text{cm}$.