A 2 meter long scale with least count of 0.2 cm is used to measure trhe 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. Th

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Solution

Q.
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 [2023]

1 1.02

2 0.85

3 1.70

4 0.51

Ans.
(3)

Least count of the scale is 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,

$\frac{1}{v} - \frac{1}{u} = \frac{1}{f} \Rightarrow \frac{1}{f} = \frac{1}{80} - \frac{1}{- 20} \Rightarrow f = 16 cm$

Also, $\frac{△ v}{v^{2}} + \frac{△ u}{u^{2}} = \frac{△ f}{f^{2}}$

$\Rightarrow \frac{△ f}{f} \times 100 = (\frac{△ v}{v^{2}} + \frac{△ u}{u^{2}}) \times f \times 100$

$\Rightarrow % f = (\frac{0.4}{400} + \frac{0.4}{6400}) \times 16 \times 100 = 1.70$

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