In an experiment for determination of the foc | vedclass

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Question

$u=10 \pm 0.1 \mathrm{~cm}, \quad v=20 \pm 0.2 \mathrm{~cm}$

$\frac{1}{v}-\frac{1}{u}=\frac{1}{f} \Rightarrow \frac{1}{v^2} d v+\frac{1}{u^2} d u=-

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$1$

Vedclass · Answer

$u=10 \pm 0.1 \mathrm{~cm}, \quad v=20 \pm 0.2 \mathrm{~cm}$

$\frac{1}{v}-\frac{1}{u}=\frac{1}{f} \Rightarrow \frac{1}{v^2} d v+\frac{1}{u^2} d u=-\frac{1}{f^2} d f$

$\frac{1}{20}+\frac{1}{10}=\frac{1}{f} \Rightarrow \frac{1}{f}=\frac{3}{20} \Rightarrow f=\frac{20}{3} \mathrm{~cm}$

$\Rightarrow \frac{1}{(20)^2}(0.2)+\frac{1}{(10)^2}(0.1)=\frac{9}{400} d f$

$	ext { df }=\frac{1}{9}\left(\frac{400}{400} 	imes 0.2+\frac{400}{100} 	imes 0.1ight)$

$d f=\frac{1}{9}(0.2+0.4) \Rightarrow \mathrm{df}=\frac{0.6}{9}$

$\frac{\mathrm{df}}{\mathrm{f}}=\frac{0.6}{9} 	imes \frac{3}{20}=\frac{1}{100}$

$\% 	ext { error }=1 \%$

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