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The focal length of a mirror is given by $\large\frac{2}{f}=\frac{1}{v}-\frac{1}{u}.$ In finding the values of $u$ and $v$, the error are equal and equal to 'p'. Then, the relative error in f is

\[(a)\;\frac{p}{2}\bigg(\frac{1}{u}+\frac{1}{v}\bigg)\quad (b)\;p\bigg(\frac{1}{u}+\frac{1}{v}\bigg) \quad (c)\;\frac{p}{2}\bigg(\frac{1}{u}-\frac{1}{v}\bigg) \quad (d)\;p\bigg(\frac{1}{u}-\frac{1}{v}\bigg)\]

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$ (b)\;p\bigg(\frac{1}{u}+\frac{1}{v}\bigg)$
answered Nov 7, 2013 by pady_1
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