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An optical bench has $1.5 m$ long scale having four equal divisions in each $cm$. While measuring the focal length of a convex lens, the lens is kept at $75 cm$ mark of the scale and the object pin is kept at $45 cm$ mark. The image of the object pin on the other side of the lens overlaps with image pin that is kept at $135 cm$ mark. In this experiment, the percentage error in the measurement of the focal length of the lens is. . . . .
$0.69$
$0.75$
$0.80$
$0.85$
Solution
For the given lens
$u =-30 cm$
$v =60 cm$
$\& \frac{1}{ f }=\frac{1}{ v }-\frac{1}{ u }$ on solving $: f =20 cm$
also $\frac{1}{ f }=\frac{1}{ v }-\frac{1}{ u }$
on differentiation
$\frac{d f}{f^2}=\frac{d v}{v^2}+\frac{d u}{u^2}$
$\frac{d f}{f}=f\left[\frac{d v}{v^2}+\frac{d u}{u^2}\right]$
$\& \frac{d f}{f} \times 100=f\left[\frac{d v}{v^2}+\frac{d u}{u^2}\right] \times 100 \%$
$f=20 cm , d u=d v=\frac{1}{4} cm$
Since there are $4$ divisions in $1 cm$ on scale
$\therefore \frac{ df }{ f } \times 100=20\left[\frac{1 / 4}{(60)^2}+\frac{1 / 4}{(30)^2}\right] \times 100 \%$
$=5\left[\frac{1}{3600}+\frac{1}{900}\right] \times 100 \%$
$=5\left[\frac{5}{36}\right] \%=\frac{25}{36} \% \approx 0.69 \%$
