Question

at what distance from a concave mirror of focal length 10cm should an object 2 cm long be placed in order to get An erect image 6cmtall?

Answer 1

Let object distance be $\textcolor {red}{v}$. Let image distance be $\textcolor{red}{u}$. Since it is a concave mirror, using natural sign conventions, we have u as positive and v as negative. Also, magnification is -3. Hence $\frac{-\mathrm{v}}{\mathrm{u}}=-3$ and we get $\frac{\mathrm{v}}{\mathrm{u}}=3$ , that is, $\mathrm{v}=3\mathrm{u}$.
Using mirror formula:
$\frac{1}{\mathrm{u}}- \frac{1}{3\mathrm{u}}=\frac{1}{f}\\ \mathrm{Since} \: f=10, \\ \frac{2}{3\mathrm{u}}=\frac{1}{10}\\ \frac{20}{3}=\mathrm{u}$