Question

The focal length of a mirror is given by $\frac{2}{f}=\frac{1}{v}-\frac{1}{u}$ . If in the measurement of $u$ and $v$ , the errors are equal to $p$ each, then 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)

Given, equation is $\frac{2}{f}=\frac{1}{v}-\frac{1}{u}$ ....(i)
Differentiating the given equation, we have
$-\frac{2}{f^{2}} \, df=-\frac{1}{v^{2}}dv-\left(- \frac{1}{u^{2}}\right) \, du$
$=-p \, \left(\frac{1}{v} - \frac{1}{u}\right) \, \left(\frac{1}{v} + \frac{1}{u}\right)\left(\because dv = du = p\right)$
$=\frac{- 2 p}{f}\left(\frac{1}{v} + \frac{1}{u}\right)$ [using Eq. (i)]
$\therefore \frac{df}{f}=p \, \left(\frac{1}{v} + \frac{1}{v}\right)$