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
$-\frac{2}{f^2}df=-\frac{1}{v^2}dv-\left(-\frac{1}{u^2}\right)du$
$=-p\left(\frac{1}{v}-\frac{1}{u}\right)\left(\frac{1}{v}+\frac{1}{u}\right)\left(\because dv=du=p\right)$
$=\frac{-2p}{f}\left(\frac{1}{v}+\frac{1}{u}\right)$ [using Eq. (i)]
$\therefore \frac{df}{f}=p\left(\frac{1}{v}+\frac{1}{v}\right)$