Q111 / 160 · AP EAMCET 2013 · Mathematics
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 to ' $p$ '. Then, the relative error in $f$ is
- Correct answerA$\frac{p}{2}\left(\frac{1}{u}+\frac{1}{v}\right)$
- B$p\left(\frac{1}{u}+\frac{1}{v}\right)$
- C$\frac{p}{2}\left(\frac{1}{u}-\frac{1}{v}\right)$
- D$p\left(\frac{1}{u}-\frac{1}{v}\right)$
Solution
Given, equation is $\frac{2}{f}=\frac{1}{v}-\frac{1}{u}$
Differentiating the given equation, we have
$$
\begin{aligned}
-\frac{2}{f^2} d f & =-\frac{1}{v^2} d v+\left(-\frac{1}{u^2}\right) d u \\
& =-p\left(\frac{1}{v}-\frac{1}{u}\right)\left(\frac{1}{v}+\frac{1}{u}\right) \quad\left(\because \frac{d v}{v}=\frac{d u}{u}=p\right) \\
& =\frac{-2 p}{f}\left(\frac{1}{v}+\frac{1}{u}\right) \quad \text { [using Eq. (i)] } \\
\therefore \quad \frac{d f}{f} & =p\left(\frac{1}{v}+\frac{1}{u}\right)
\end{aligned}
$$
Hence, the correct answer is option: 'A'.