---

The above lens diagram shows the visual calculation using geometry for a thin lens. The object is located 50 units away from the lens ($p$), which has a focal length of 30 units ($f$). By drawing three rays from the top of the object, we can figure out where the rays meet and locate the image ($i$).

The thin lens equation is: $$\frac{1}{p}+\frac{1}{i}=\frac{1}{f}$$ Solving for $i$, we obtain: $$i = \frac{pf}{p-f} = \frac{30 \times 50}{20} = 75$$ thus matching the position shown in the diagram.

The magnification is also calculated by comparing: $$m = \frac{-i}{p} = \frac{-75}{50} = -1.5 $$ This value is also predicted by the diagram.