An optical bench has 1.5   m long scale having four equal divisions in each c m . While measuring the…

Solution

Let, $\left(\begin{array}{c}{x}_1=distance of object from O\\ x_2=distance of lens from O \\ x_3=distance of image from O\end{array}\right.open=[$
$\therefore u=\left(x_2-x_1\right)=75-45=30 \mathrm{c}\mathrm{m}$
$\Rightarrow ∆u=∆x_2+∆x_1=\frac{1}{4}+\frac{1}{4}=\frac{1}{2}\mathrm{c}\mathrm{m}\Rightarrow$ error in measurement of $“u”$
And, $v=\left(x_3-x_2\right)=135-75=60 \mathrm{c}\mathrm{m}$
$\Rightarrow ∆v=∆x_3+∆x_2=\frac{1}{4}+\frac{1}{4}=\frac{1}{2} \mathrm{c}\mathrm{m}\Rightarrow$ error in measurement of $“v”$
Using Lens formula $\therefore \frac{1}{v}-\frac{1}{u}=\frac{1}{f}\Rightarrow \frac{1}{60}+\frac{1}{30}=\frac{1}{f}\Rightarrow \frac{1}{f}=\frac{1}{20}$
$\therefore f=20 \mathrm{c}\mathrm{m},\mathrm{ }\mathrm{A}\mathrm{l}\mathrm{s}\mathrm{o},\frac{-dv}{v^2}+\frac{-du}{u^2}=\frac{-df}{f^2}\to$ [differentiating lens formula]
$\Rightarrow \frac{df}{f}=f\left[\frac{dv}{v^2}+\frac{du}{u^2}\right]=20\left[\frac{1}{{60}^2}+\frac{1}{{30}^2}\right]\frac{1}{2}$
$\therefore \frac{df}{f}\times 100=10\left[\frac{1}{36}+\frac{1}{9}\right]=\frac{50}{36}=1.38 \mathrm{a}\mathrm{n}\mathrm{d}\mathrm{ }1.39\mathrm{ }\left(\mathrm{b}\mathrm{o}\mathrm{t}\mathrm{h}\right)$ `

Asked in: JEE Advanced 2019 (Paper 2)

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