The focal length of a mirror is given by $$1v$$−$$1u=2f$$. If equal errors are made in measuring u and v, then the relative error in f, is

We have, $$1v$$−$$1u=2f$$⇒ $$d1v$$−$$1u=d2f$$⇒ −$$1v2dv+1u2du=$$−$$2f2df$$⇒ $$1v2$$−$$1u2$$α$$=2f2df$$⇒ α$$1v+1u1v$$−$$1u=2f2df$$⇒ α$$1v+1u$$×$$2f=2f2df$$ ∵$$1v$$−$$1u=2f$$⇒ $$dff=$$α$$1u+1v$$[∵ $$du=dv=$$α]⇒ Relative error in $$f=$$α$$1u+1v$$

Questions

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

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By Expert Faculty of Sri Chaitanya

\frac{2}{α}

α (\frac{1}{u} + \frac{1}{v})

α (\frac{1}{u} - \frac{1}{v})

none of these

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detailed solution

Correct option is B

We have,

$\begin{array}{l} \frac{1}{v} - \frac{1}{u} = \frac{2}{f} \\ \Rightarrow d (\frac{1}{v} - \frac{1}{u}) = d (\frac{2}{f}) \\ \Rightarrow - \frac{1}{v^{2}} d v + \frac{1}{u^{2}} d u = - \frac{2}{f^{2}} d f \\ \Rightarrow (\frac{1}{v^{2}} - \frac{1}{u^{2}}) α = \frac{2}{f^{2}} d f \\ \Rightarrow α (\frac{1}{v} + \frac{1}{u}) (\frac{1}{v} - \frac{1}{u}) = \frac{2}{f^{2}} d f \\ \Rightarrow α (\frac{1}{v} + \frac{1}{u}) \times \frac{2}{f} = \frac{2}{f^{2}} d f [∵ \frac{1}{v} - \frac{1}{u} = \frac{2}{f}] \\ \Rightarrow \frac{d f}{f} = α (\frac{1}{u} + \frac{1}{v}) \end{array} [∵ d u = d v = α]$