JEE Main · 2023 · 06 Apr 2023 · Shift 2 · Units & Measurement

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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 lm 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:

(A)1.02
(B)0.85
(C)1.70✓ Correct
(D)0.51

Answer: (C)

Solution

The lens is at the

100

cm mark, the object pin at

80

cm and the image pin at

180

cm, so the object distance is

u = 100 - 80 = 20

cm and the image distance is

v = 180 - 100 = 80

cm. From the lens equation

\frac{1}{f} = \frac{1}{v} + \frac{1}{u} = \frac{1}{80} + \frac{1}{20} = \frac{1}{16}

, so

f = 16

cm. Each mark is read with an uncertainty equal to the least count

0.2

cm, and since

u

and

v

are each the difference of two readings,

Δ u = Δ v = 0.4

cm. Differentiating

\frac{1}{f} = \frac{1}{v} + \frac{1}{u}

gives

\frac{Δ f}{f ^{2}} = \frac{Δ v}{v ^{2}} + \frac{Δ u}{u ^{2}}

, so

\frac{Δ f}{f} \times 100 = f (\frac{Δ v}{v ^{2}} + \frac{Δ u}{u ^{2}}) \times 100 = 16 (\frac{0.4}{6400} + \frac{0.4}{400}) \times 100

. The percentage error in the focal length is

1.70%

JEE Main · 2023 · 06 Apr 2023 · Shift 2 · Single correct option

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JEE Main · 2023 · 06 Apr 2023 · Shift 2 · Units & Measurement

Solution

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