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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0.85

1.70

1.02

0.51

answer is B.

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Detailed Solution

LC= 0.2 cm
$\begin{array}{l} u = - (20 \pm 0.4), v = (80 \pm 0.4) \\ \frac{1}{f} = \frac{1}{v} - \frac{1}{u} \\ f^{- 1} = v^{- 1} - u^{- 1} \\ - \frac{df}{f^{2}} = - \frac{dv}{v^{2}} + \frac{du}{u^{2}} \\ \frac{df}{f} \times 100 = 16 [\frac{0.4}{80^{2}} + \frac{0.4}{20^{2}}] 10^{- 4} \\ = 16 [\frac{0.4}{64} + \frac{0.4}{4}] \\ = 16 \times 0.4 [\frac{1}{64} + \frac{1}{4}] = 16 \times 0.4 [\frac{1 + 16}{64}] \\ = 1.70 % \end{array}$