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The focal length $(f)$ of a curved mirror is related to the object distance $u$ and the image distance $v$ in accordance with the mathematical relation, $\frac{1}{f} = \frac{1}{u} + \frac{1}{v}$ .
The maximum relative error in calculating the focal length $f$ of the mirror is

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$\frac{Δ f}{f^{2}} = \frac{Δ u}{u^{2}} + \frac{Δ v}{v^{2}}$

$\frac{Δ f}{f} = \frac{Δ u}{u} + \frac{Δ v}{v} + \frac{Δ u}{u + v} + \frac{Δ v}{u + v}$

$\frac{Δ f}{f} = \frac{1}{Δ u / u} + \frac{1}{Δ v / v}$

$\frac{Δ f}{f} = \frac{u v}{u + v} (\frac{Δ v}{v^{2}} + \frac{Δ v}{v^{2}})$

answer is A.

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Detailed Solution

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

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