In an experiment to measure the focal length | vedclass

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Question

(B) The lens formula is given by $\frac{1}{f} = \frac{1}{v} - \frac{1}{u}$,which can be written as $f^{-1} = v^{-1} - u^{-1}$.Differentiating both sid

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$f^2 \left[ \frac{\Delta u}{u^2} + \frac{\Delta v}{v^2} \right]$

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(B) The lens formula is given by $\frac{1}{f} = \frac{1}{v} - \frac{1}{u}$,which can be written as $f^{-1} = v^{-1} - u^{-1}$.Differentiating both sides with respect to the variables,we get:$-f^{-2} df = -v^{-2} dv - u^{-2} du$.Multiplying by $-1$,we have:$\frac{df}{f^2} = \frac{dv}{v^2} + \frac{du}{u^2}$.Substituting the least counts $\Delta u$ and $\Delta v$ for the errors $du$ and $dv$,the error in focal length $\Delta f$ is:$\Delta f = f^2 \left[ \frac{\Delta u}{u^2} + \frac{\Delta v}{v^2} \right]$.

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