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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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:

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G1 Physics · Accepted Answer

Object position = 80 cm Lens position = 100 cm Image position = 180 cm Therefore, u = 20 cm v = 80 cm Focal length: f = uv/(u+v) = (20×80)/(100) = 16 cm The least count of the scale is 0.2 cm. Since u and v are each obtained from the difference of two position readings, the maximum error in each is Δu = Δv = 0.2 + 0.2 = 0.4 cm For f = uv/(u+v), Δf/f = [v/(u+v)](Δu/u) + [u/(u+v)](Δv/v) = (80/100)(0.4/20) + (20/100)(0.4/80) = 0.016 + 0.001 = 0.017 Percentage error = 1.7%. Hence option D follows.