Consider the figure shown, what will the total magnification after two successive reflections first on M₁ and then on M₂ ?

Two reflections mean, the first image from M₁ to the object for M₂ :

Using mirror formula :

$\frac{1}{\mathrm{v}}+\frac{1}{\mathrm{u}}=\frac{1}{\mathrm{f}}$

First reflection on M₁ :

$\begin{array}{l}\mathrm{u}=-30\mathrm{cm};\mathrm{f}=-20\mathrm{cm}:\\ \frac{1}{-30}+\frac{1}{\mathrm{v}}=\frac{1}{-20}\\ \Rightarrow \mathrm{v}=-60\mathrm{cm}\\ {\mathrm{m}}_1=-\frac{\mathrm{v}}{\mathrm{u}}=-2\end{array}$

The first image forms 60 cms in front of M1, So, the object for M₂ is 20 cms behind the mirror : for second reflection :

$\begin{array}{l}\mathrm{u}=+20\mathrm{cm};\mathrm{f}=+10\mathrm{cm}\\ \frac{1}{\mathrm{v}}+\frac{1}{20}=\frac{1}{10}\\ \Rightarrow \mathrm{v}=20\mathrm{cm}\\ {\mathrm{m}}_2=-\frac{\mathrm{v}}{\mathrm{u}}=-1\end{array}$

Total magnification is given by :

${\mathrm{m}}_{\text{tot }}={\mathrm{m}}_1\times {\mathrm{m}}_2=\left(-2\right)\left(-1\right)=2$