Physics · Units, Dimensions & Error Analysis

JEE Advanced 2019 — Paper 2 — Question 14

An optical bench has 1.5 m long scale having four equal divisions in each cm . While measuring the focal length of a convex lens, the lens is kept at 75 cm mark of the scale and the object pin is kept at 45 cm mark. The image of the object pin on the other side of the lens overlaps with image pin that is kept at 135 cm mark. In this experiment, the percentage error in the measurement of the focal length of the lens is $\_\_\_\_$ —.

Answer: 1.39

Numerical answer — enter this value.

Step-by-step solution

$\frac{1}{v}-\frac{1}{-\mathrm{u}}=\frac{1}{\mathrm{f}}$

$\Rightarrow \frac{\Delta \mathrm{v}}{\mathrm{v}^{2}}+\frac{\Delta \mathrm{u}}{\mathrm{u}^{2}}=\frac{\Delta \mathrm{f}}{\mathrm{f}^{2}}$

% error in the measurement of focal length

$==\frac{\Delta \mathrm{f}}{\mathrm{f}} \times 100=\frac{\Delta \mathrm{v}}{\mathrm{v}^{2}}+\frac{\Delta \mathrm{u}}{\mathrm{u}^{2}} \mathrm{Gf} \times 100$

$\left.=[\frac{0.5}{(60)^{2}}+\frac{0.5}{(30)^{2}}\right] \times 20 \times 100=1.39$

Answer key and solution verified before publishing.

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Exam: JEE Advanced 2019
Paper: Paper 2
Subject: Physics
Chapter: Units, Dimensions & Error Analysis
Topic: Significant Figures and Error Analysis