Question

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

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

Answer 1

Answer: (B)

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

Answer 2

We have,

$\frac{1}{v} - \frac{1}{u}$ = $\frac{2}{f}$

Ž d$\left( \frac{1}{v} - \frac{1}{u} \right)$ = d $\left( \frac{2}{f} \right)$

Ž –$\frac{1}{v^{2}}$dv + $\frac{1}{u^{2}}$du = –$\frac{2}{f^{2}}$df

Ž $\left( \frac{1}{v^{2}} - \frac{1}{u^{2}} \right)$ a = $\frac{2}{f^{2}}$ df

[Q du = dv = a]

Ž a$\left( \frac{1}{v} + \frac{1}{u} \right)$ $\left( \frac{1}{v} - \frac{1}{u} \right)$ = $\frac{2}{f^{2}}$ df

Ž a$\left( \frac{1}{v} + \frac{1}{u} \right)$ × $\frac{2}{f}$ = $\frac{2}{f^{2}}$ df$\left\lbrack \because\frac{1}{v} - \frac{1}{u} = \frac{2}{f} \right\rbrack$

Ž $\frac{df}{f}$ = a$\left( \frac{1}{u} + \frac{1}{v} \right)$

Ž Relative error in f = a $\left( \frac{1}{u} + \frac{1}{v} \right)$.

Hence (2) is the correct answer.