In an experiment for determination of the foc | vedclass

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(A) Given: $u = 10 \pm 0.1 	ext{ cm}$,$v = 20 \pm 0.2 	ext{ cm}$.Using the lens formula: $\frac{1}{f} = \frac{1}{v} - \frac{1}{u}$.For a convex lens

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$1$

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(A) Given: $u = 10 \pm 0.1 	ext{ cm}$,$v = 20 \pm 0.2 	ext{ cm}$.Using the lens formula: $\frac{1}{f} = \frac{1}{v} - \frac{1}{u}$.For a convex lens,$u$ is negative,so $\frac{1}{f} = \frac{1}{v} - \frac{1}{-u} = \frac{1}{v} + \frac{1}{u}$.Calculating $f$: $\frac{1}{f} = \frac{1}{20} + \frac{1}{10} = \frac{1+2}{20} = \frac{3}{20} \implies f = \frac{20}{3} 	ext{ cm}$.Differentiating the lens formula: $-\frac{df}{f^2} = -\frac{dv}{v^2} - \frac{du}{u^2} \implies \frac{df}{f^2} = \frac{dv}{v^2} + \frac{du}{u^2}$.Substituting the values: $\frac{df}{f^2} = \frac{0.2}{(20)^2} + \frac{0.1}{(10)^2} = \frac{0.2}{400} + \frac{0.1}{100} = \frac{0.2 + 0.4}{400} = \frac{0.6}{400}$.Relative error $\frac{df}{f} = f^2 	imes \frac{0.6}{400} = \left(\frac{20}{3}ight)^2 	imes \frac{0.6}{400} = \frac{400}{9} 	imes \frac{0.6}{400} = \frac{0.6}{9} = \frac{6}{90} = \frac{1}{15}$.Wait,re-evaluating the derivative: $\frac{df}{f^2} = \frac{dv}{v^2} + \frac{du}{u^2} \implies df = f^2 \left( \frac{dv}{v^2} + \frac{du}{u^2} ight)$.Percentage error = $\frac{df}{f} 	imes 100 = f \left( \frac{dv}{v^2} + \frac{du}{u^2} ight) 	imes 100 = \frac{20}{3} \left( \frac{0.2}{400} + \frac{0.1}{100} ight) 	imes 100 = \frac{20}{3} \left( \frac{0.2 + 0.4}{400} ight) 	imes 100 = \frac{20}{3} 	imes \frac{0.6}{400} 	imes 100 = \frac{20 	imes 0.6}{3 	imes 4} = \frac{12}{12} = 1 \%$.Therefore,$n = 1$.

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