A 2 meter long scale with least count of 0.2 cm is used to measure the locations of objects on an op

Question

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\

Professor PC Thomas Classes · Answer

(a) : As, $u=(100 \pm 0.2) \mathrm{cm}-(80 \pm 0.2) \mathrm{cm}$ $=(20 \pm 0.4) \mathrm{cm}$\ $v=(180 \pm 0.2)-(100 \pm 0.2)=(80 \pm 0.4) \mathrm{cm}$.\ $\frac{1}{f}=\frac{1}{v}-\frac{1}{u}=\frac{1}{80}+\frac{1}{20}=\frac{5}{80}$\ $f=16 \mathrm{~cm}$\ $\frac{1}{v^{2}} \Delta v+\frac{1}{u^{2}} \Delta u=\frac{1}{f^{2}} \Delta f \quad \frac{\Delta f}{f^{2}}=\frac{0.4}{6400}+\frac{0.4}{400}$\ $\frac{\Delta f}{f}=\left[\frac{2720}{6400 	imes 400}ight] f$\ $\frac{\Delta f}{f} 	imes 100=0.017 	imes 100=1.70 \%$\