Question

The focal length of a mirror is given by $\frac{2}{f}=\frac{1}{v}-\frac{1}{u}$. In finding the values of u and v, the errors are equal and equal to 'p'. The the relative error in f is

• A. $\frac{p}{2}\left(\frac{1}{u}+\frac{1}{v}\right)$
• B. $p\left(\frac{1}{u}+\frac{1}{v}\right)$
• C. $\frac{p}{2}\left(\frac{1}{u}-\frac{1}{v}\right)$
• D. $p\left(\frac{1}{u}-\frac{1}{v}\right)$

Answer 1

Answer: (B)

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

$\frac{2}{f}=\frac{1}{v}-\frac{1}{u}$
$-2\frac{\Delta f}{f^2}=\frac{-\Delta V}{v^2}+\frac{\Delta u}{u^2}$
$\Rightarrow \frac{2\Delta f}{f^2}=\frac{\Delta v}{v^2}-\frac{\Delta v}{u^2}$
$\Rightarrow \frac{\Delta f}{f}=\frac{fP}{2}[\frac{1}{v}-\frac{1}{u}][\frac{1}{v}+\frac{1}{u}]$
$\frac{\Delta f}{f}=\frac{fp}{2}\times \frac{2}{f}[\frac{1}{v}+\frac{1}{u}]$
$\frac{\Delta f}{f}=p[\frac{1}{v}+\frac{1}{u}]$