A \(2\,meter\) long scale with least count of \(0.2\,cm\) is used to measure the locations of objects on an optical bench. While measuring the focal length of a convex lens, the object pin and the convex lens are placed at \(80\,cm\) mark and \(1\,m\) mark, respectively. The image of the object pin on the other side of lens coincides with image pin that is kept at \(180\,cm\) mark. The \(\%\) error in the estimation of focal length is

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Answer & Solution

Correct Answer

(C) \(1.70\)

Detailed explanation

Least count \(=0.2\,cm\)

\(u =(100 \pm 0.2)-(80 \pm 0.2)=(20 \pm 0.4)\,cm\)

\(v =(180 \pm 0.2)-(100 \pm 0.2)=(80 \pm 0.4)\,cm\)

From lens formula,

\(\frac{1}{v}-\frac{1}{u}=\frac{1}{f} \Rightarrow \frac{1}{f}=\frac{1}{80}-\frac{1}{-20}\)

\(f=16\,cm\)

Also \(\frac{\Delta v}{v^2}+\frac{\Delta u}{u^2}=\frac{\Delta f}{f^2}\)

\(\Rightarrow \frac{\Delta f}{f} \times 100=\left(\frac{\Delta v}{v^2}+\frac{\Delta u}{u^2}\right) \times f \times 100\)

\(\Rightarrow \% f=\left(\frac{0.4}{400}+\frac{0.4}{6400}\right) \times 16 \times 100\)

\(=1.70\)

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