JEE Advanced 2023 — Paper 1 — PHY-III solution

JEE Advanced2023Paper 1PHY-III

In an experiment for determination of the focal length of a thin convex lens, the distance of the object from the lens is $10 \pm 0.1$ cm and the distance of its real image from the lens is $20 \pm 0.2$ cm. The error in the determination of focal length of the lens is $n %$ . The value of $n$ is _______.

Your answer

Solution

Focal length. Using $\frac{1}{v} - \frac{1}{u} = \frac{1}{f}$ with $u = - 10$ cm and $v = + 20$ cm: $\frac{1}{f} = \frac{1}{20} + \frac{1}{10} = \frac{3}{20}$ , so $f = \frac{20}{3}$ cm.

Error propagation. Differentiating $\frac{1}{f} = \frac{1}{v} - \frac{1}{u}$ : $\frac{Δ f}{f ^{2}} = \frac{Δ v}{v ^{2}} + \frac{Δ u}{u ^{2}}$ . With $Δ u = 0.1$ , $Δ v = 0.2$ : $\frac{Δ f}{f ^{2}} = \frac{0.2}{400} + \frac{0.1}{100} = \frac{0.2 + 0.4}{400} = \frac{0.6}{400} .$

Thus $Δ f = \frac{0.6}{400} \cdot f^{2} = \frac{0.6}{400} \cdot \frac{400}{9} = \frac{0.6}{9}$ cm.

Percentage error. $\frac{Δ f}{f} \times 100 = \frac{0.6/9}{20/3} \times 100 = \frac{0.6}{9} \cdot \frac{3}{20} \cdot 100 = 1%$ .

Answer: $n = 1$ .

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