An optical bench has 1.5 m long scale having | vedclass

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Question

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 }=\

Vedclass · Accepted Answer

$0.69$

Vedclass · Answer

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

$\frac{d f}{f^2}=\frac{d v}{v^2}+\frac{d u}{u^2}$

$\frac{d f}{f}=f\left[\frac{d v}{v^2}+\frac{d u}{u^2}ight]$

$\& \frac{d f}{f} 	imes 100=f\left[\frac{d v}{v^2}+\frac{d u}{u^2}ight] 	imes 100 \%$

$f=20 cm , d u=d v=\frac{1}{4} cm$

Since there are $4$ divisions in $1 cm$ on scale

$	herefore \frac{ df }{ f } 	imes 100=20\left[\frac{1 / 4}{(60)^2}+\frac{1 / 4}{(30)^2}ight] 	imes 100 \%$

$=5\left[\frac{1}{3600}+\frac{1}{900}ight] 	imes 100 \%$

$=5\left[\frac{5}{36}ight] \%=\frac{25}{36} \% \approx 0.69 \%$

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