Question

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 _______.

Comments

How ∆u = 0.5 cm
It should by 0.25cm only
Only doubt in this how ∆U become = 2times of 0.25cm

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bro, $\Delta u$ is calculated as difference of 2 points in the reading of the scale so the 0.25 error is added twice so 0.5 is the actual error now.

Answer 1

\( u=\left(x_2-x_1\right)=75-45=30 \mathrm{~cm}\)

\( \Delta u=\Delta x_2+\Delta x_1=\frac{1}{4}+\frac{1}{4}=\frac{1}{2} \mathrm{~cm}\)

\( v=\left(x_3-x_2\right)=135-75=60 \mathrm{~cm} \)

\( \Delta v=\Delta x_3+\Delta x_2=\frac{1}{4}+\frac{1}{4}=\frac{1}{2} \mathrm{~cm}\)

\( \therefore \frac{1}{v}-\frac{1}{u}=\frac{1}{f} \Rightarrow \frac{1}{60}+\frac{1}{30}=\frac{1}{f}\)

\( \therefore \mathrm{f}=20 \mathrm{~cm} \text { Also, } \frac{-d v}{v^2}+\frac{-d u}{u^2}=\frac{-d f}{f^2} \)

\(\Rightarrow \frac{d f}{f}=f\left[\frac{d v}{v^2}+\frac{d u}{u^2}\right]=20\left[\frac{1}{60^2}+\frac{1}{30^2}\right] \frac{1}{2} \)

\( \therefore \frac{d f}{f} \times 100=10\left[\frac{1}{36}+\frac{1}{9}\right]=\frac{50}{36}=1.38 \text { and } 1.39 \text { (both) }\)

Answer 2

% error = 1.38 or 1.39