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
Answer & explanation
Correct answer: option 2
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 $f=\alpha\left(\frac{1}{u}+\frac{1}{v}\right)$