An optical bench has a 1.5 m long scale havin | vedclass

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Question

(A) Given,the lens is at $75 cm$,object pin at $45 cm$,and image pin at $135 cm$.Object distance $u = 45 - 75 = -30 cm$.Image distance $v = 135 - 75 =

Vedclass · Accepted Answer

$0.69$

Vedclass · Answer

(A) Given,the lens is at $75 cm$,object pin at $45 cm$,and image pin at $135 cm$.Object distance $u = 45 - 75 = -30 cm$.Image distance $v = 135 - 75 = 60 cm$.Using the lens formula $\frac{1}{f} = \frac{1}{v} - \frac{1}{u}$,we get $\frac{1}{f} = \frac{1}{60} - \frac{1}{-30} = \frac{1+2}{60} = \frac{3}{60} = \frac{1}{20}$. Thus,$f = 20 cm$.The scale has $4$ divisions per $cm$,so the least count $\Delta u = \Delta v = \frac{1}{4} cm = 0.25 cm$.Differentiating the lens formula $\frac{1}{f} = \frac{1}{v} - \frac{1}{u}$,we get $-\frac{df}{f^2} = -\frac{dv}{v^2} + \frac{du}{u^2}$.Taking magnitudes for maximum error: $\frac{df}{f^2} = \frac{dv}{v^2} + \frac{du}{u^2}$.Percentage error $\frac{df}{f} 	imes 100 = f \left[ \frac{dv}{v^2} + \frac{du}{u^2} ight] 	imes 100$.Substituting values: $\frac{df}{f} 	imes 100 = 20 \left[ \frac{0.25}{60^2} + \frac{0.25}{30^2} ight] 	imes 100$.$= 20 	imes 0.25 \left[ \frac{1}{3600} + \frac{1}{900} ight] 	imes 100 = 5 \left[ \frac{1+4}{3600} ight] 	imes 100 = 5 	imes \frac{5}{36} = \frac{25}{36} \% \approx 0.69 \%$.

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