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

We have,

$\begin{array}{l}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\frac{1}{v}-\frac{1}{u}=\frac{2}{f}\\ \Rightarrow \hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}d\left(\frac{1}{v}-\frac{1}{u}\right)=d\left(\frac{2}{f}\right)\\ \Rightarrow \hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}-\frac{1}{v^2}dv+\frac{1}{u^2}du=-\frac{2}{f^2}df\\ \Rightarrow \hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\left(\frac{1}{v^2}-\frac{1}{u^2}\right)\alpha =\frac{2}{f^2}df\\ \Rightarrow \hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\alpha \left(\frac{1}{v}+\frac{1}{u}\right)\left(\frac{1}{v}-\frac{1}{u}\right)=\frac{2}{f^2}df\\ \Rightarrow \hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\alpha \left(\frac{1}{v}+\frac{1}{u}\right)\times \frac{2}{f}=\frac{2}{f^2}df \left[\because \frac{1}{v}-\frac{1}{u}=\frac{2}{f}\right]\\ \Rightarrow \hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\frac{df}{f}=\alpha \left(\frac{1}{u}+\frac{1}{v}\right)\end{array}[\because du=dv=\alpha ]$

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