The focal length of a mirror is given by $\frac{1}{v}-\frac{1}{u}=\frac{2}{\text{⨍ }}.$ If equal errors $\alpha$ are made in measuring $\mathrm{u}$ and $\mathrm{v}$ then relative error in $\text{⨍ }$ is
 

$$\begin{gathered} \begin{array}{l}\text{ Given, }\frac{1}{\mathrm{v}}-\frac{1}{\mathrm{u}}=\frac{2}{\mathrm{f}}\\ \Rightarrow -\frac{\mathrm{\Delta v}}{{\mathrm{v}}^2}+\frac{\mathrm{\Delta u}}{{\mathrm{u}}^2}=-\frac{2\mathrm{\Delta f}}{{\mathrm{f}}^2} \\ \text{ Since, given equal errors in measuring }u\text{ and v i.e. }\mathrm{\Delta u}=\mathrm{\Delta v}=\alpha \\ \begin{array}{l}\Rightarrow \alpha \left(\frac{1}{\mathrm{u}}-\frac{1}{\mathrm{v}}\right)\left(\frac{1}{\mathrm{u}}+\frac{1}{\mathrm{v}}\right)=-\frac{2\mathrm{\Delta f}}{{\mathrm{f}}^2}\\ \text{ But, }\frac{1}{\mathrm{v}}-\frac{1}{\mathrm{u}}=\frac{2}{\mathrm{f}}\\ \Rightarrow \frac{\mathrm{\Delta f}}{\mathrm{f}}=\alpha \left(\frac{1}{\mathrm{u}}+\frac{1}{\mathrm{v}}\right)\end{array}\end{array} \end{gathered}$$