Question

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

• A. $\frac{2}{\alpha }$
• B. $\alpha (\frac{1}{u}+\frac{1}{v})$
• C. $\alpha (\frac{1}{u}-\frac{1}{v})$
• D. none of these

Answer 1

Answer: (B)

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

Given, $\frac{1}{v}-\frac{1}{u}=\frac{2}{f}$
$\Rightarrow -\frac{\Delta v}{v^2}+\frac{\Delta u}{u^2}=-\frac{2\Delta f}{f^2}$
Since, given equal errors in measuring u and v i.e.$\Delta u=\Delta v=\alpha$
$\Rightarrow \alpha (\frac{1}{u}-\frac{1}{v})(\frac{1}{u}+\frac{1}{v})=-\frac{2\Delta f}{f^2}$
But, $\frac{1}{v}-\frac{1}{u}=\frac{2}{f}$
$\Rightarrow \frac{\Delta f}{f}=\alpha (\frac{1}{u}+\frac{1}{v})$
Hence, correct option is B