The focal length of a mirror is given by frac | vedclass

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(B) Given the equation: $\frac{2}{f} = \frac{1}{v} - \frac{1}{u}$ $(i)$Taking the differential of both sides:$-\frac{2}{f^2} df = -\frac{1}{v^2} dv +

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

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(B) Given the equation: $\frac{2}{f} = \frac{1}{v} - \frac{1}{u}$ $(i)$Taking the differential of both sides:$-\frac{2}{f^2} df = -\frac{1}{v^2} dv + \frac{1}{u^2} du$Given that the absolute errors in $u$ and $v$ are $p$, we have $du = p$ and $dv = p$.Substituting these into the differential equation:$-\frac{2}{f^2} df = -\frac{1}{v^2} p + \frac{1}{u^2} p$$-\frac{2}{f^2} df = -p \left( \frac{1}{v^2} - \frac{1}{u^2} \right)$Using the identity $a^2 - b^2 = (a - b)(a + b)$:$-\frac{2}{f^2} df = -p \left( \frac{1}{v} - \frac{1}{u} \right) \left( \frac{1}{v} + \frac{1}{u} \right)$Since $\frac{1}{v} - \frac{1}{u} = \frac{2}{f}$ from equation $(i)$:$-\frac{2}{f^2} df = -p \left( \frac{2}{f} \right) \left( \frac{1}{v} + \frac{1}{u} \right)$Dividing both sides by $-\frac{2}{f}$:$\frac{df}{f} = p \left( \frac{1}{v} + \frac{1}{u} \right)$Thus, the relative error in $f$ is $p \left( \frac{1}{u} + \frac{1}{v} \right)$.

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

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