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

$\alpha\left(\frac{1}{u}+\frac{1}{v}\right)$

We have,

$\frac{1}{v}-\frac{1}{u}=\frac{2}{f}$

$\Rightarrow d\left(\frac{1}{v}-\frac{1}{u}\right)=d\left(\frac{2}{f}\right)$

$\Rightarrow -\frac{1}{v^2} d v+\frac{1}{u^2} d u=-\frac{2}{f^2} d f$

$\Rightarrow \left(\frac{1}{v^2}-\frac{1}{u^2}\right) \alpha=\frac{2}{f^2} d f$           $[∵ d u=d v=\alpha]$'

$\Rightarrow \alpha\left(\frac{1}{v}+\frac{1}{u}\right)\left(\frac{1}{v}-\frac{1}{u}\right)=\frac{2}{f^2} d f$

$\Rightarrow \alpha\left(\frac{1}{v}+\frac{1}{u}\right) \times \frac{2}{f}=\frac{2}{f^2} d f$       $\left[∵ \frac{1}{v}-\frac{1}{u}=\frac{2}{f}\right]$  

$\Rightarrow \frac{d f}{f}=\alpha\left(\frac{1}{u}+\frac{1}{v}\right)$

⇒ Relative error in $f=\alpha\left(\frac{1}{u}+\frac{1}{v}\right)$