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(D) Using the mirror formula: $\frac{1}{v} + \frac{1}{u} = \frac{1}{f}$.Given $u = -40\,cm$ and $v = -120\,cm$ (for a real image formed by a convergin

Vedclass · Accepted Answer

$32$

Vedclass · Answer

(D) Using the mirror formula: $\frac{1}{v} + \frac{1}{u} = \frac{1}{f}$.Given $u = -40\,cm$ and $v = -120\,cm$ (for a real image formed by a converging mirror).$\frac{1}{-120} + \frac{1}{-40} = \frac{1}{f} \implies \frac{-1-3}{120} = \frac{1}{f} \implies f = -30\,cm$.The least count of the scale is $LC = \frac{1}{20}\,cm = 0.05\,cm$. Thus,$du = dv = 0.05\,cm$.Differentiating the mirror formula: $-\frac{dv}{v^2} - \frac{du}{u^2} = -\frac{df}{f^2}$.Taking magnitudes for error calculation: $|df| = f^2 \left( \frac{|dv|}{v^2} + \frac{|du|}{u^2} ight)$.$|df| = (30)^2 \left( \frac{1/20}{120^2} + \frac{1/20}{40^2} ight) = 900 	imes \frac{1}{20} \left( \frac{1}{14400} + \frac{1}{1600} ight)$.$|df| = 45 \left( \frac{1 + 9}{14400} ight) = 45 	imes \frac{10}{14400} = \frac{450}{14400} = \frac{45}{1440} = \frac{1}{32}\,cm$.Comparing with $1/K$,we get $K = 32$.

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