A convex lens throws on a screen 12 m from the lens a magnified image of the object. If the magnification is 15, find the focal length of the lens.

Magnification m = 15

Image distance = v

$$\begin{gathered} \text{ Magnification }=+\frac{\mathrm{v}}{\mathrm{u}} \\ \text{ From lens formula }\frac{1}{\mathrm{f}}=\frac{1}{\mathrm{v}}-\frac{1}{\mathrm{u}} \\ \text{ using Cartesian sign }\frac{1}{\mathrm{f}}=\frac{1}{\mathrm{v}}-\frac{1}{-\mathrm{u}} \\ \Rightarrow \frac{1}{\mathrm{f}}=\frac{1}{\mathrm{v}}+\frac{1}{\mathrm{u}} \\ \therefore \frac{1}{\mathrm{u}}=\frac{1}{\mathrm{f}}-\frac{1}{\mathrm{v}} \\ \Rightarrow \frac{1}{\mathrm{u}}=\frac{\mathrm{v}-\mathrm{f}}{\mathrm{vf}} \\ \Rightarrow \mathrm{u}=\frac{\mathrm{vf}}{\mathrm{v}-\mathrm{f}} \\ \text{ Substituting (2) in (1) } \\ \Rightarrow \mathrm{m}=\frac{\mathrm{v}(\mathrm{v}-\mathrm{f})}{\mathrm{vf}} \\ \therefore 15=\frac{12-\mathrm{f}}{\mathrm{f}} \\ \Rightarrow 15 \mathrm{f}=12-\mathrm{f} \\ \Rightarrow \mathrm{f}=\frac{12}{16}=0.75 \mathrm{m} \end{gathered}$$