A 2,m long scale with a least count of 0.2,cm | vedclass

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(C) Least count $\Delta x = 0.2\,cm$.Object distance $u = (100 \pm 0.2) - (80 \pm 0.2) = (20 \pm 0.4)\,cm$.Image distance $v = (180 \pm 0.2) - (100 \p

Vedclass · Accepted Answer

$1.70$

Vedclass · Answer

(C) Least count $\Delta x = 0.2\,cm$.Object distance $u = (100 \pm 0.2) - (80 \pm 0.2) = (20 \pm 0.4)\,cm$.Image distance $v = (180 \pm 0.2) - (100 \pm 0.2) = (80 \pm 0.4)\,cm$.Using the lens formula $\frac{1}{f} = \frac{1}{v} - \frac{1}{u}$, we have $\frac{1}{f} = \frac{1}{80} - \frac{1}{-20} = \frac{1+4}{80} = \frac{5}{80} = \frac{1}{16}$.Thus, $f = 16\,cm$.Differentiating the lens formula $\frac{1}{f} = \frac{1}{v} - \frac{1}{u}$, we get $\frac{\Delta f}{f^2} = \frac{\Delta v}{v^2} + \frac{\Delta u}{u^2}$.Percentage error in focal length is $\frac{\Delta f}{f} 	imes 100 = \left( \frac{\Delta v}{v^2} + \frac{\Delta u}{u^2} ight) 	imes f 	imes 100$.Substituting the values: $\% 	ext{error} = \left( \frac{0.4}{80^2} + \frac{0.4}{20^2} ight) 	imes 16 	imes 100$.$\% 	ext{error} = \left( \frac{0.4}{6400} + \frac{0.4}{400} ight) 	imes 1600 = \left( \frac{0.4}{4} + \frac{0.4}{64} ight) 	imes 16 = (0.1 + 0.00625) 	imes 16 = 1.6 + 0.1 = 1.70\%$.

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