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 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. 0.51
• B. 0.85
• C. 1.02
• D. 1.70

Answer 1

The Correct Option is D

Based on the data provided: \[ U = 100 - 80 = 20 \, \text{cm}, \quad V = 180 - 100 = 80 \, \text{cm} \] Using the formula: \[ \frac{1}{f} = \frac{1}{u} + \frac{1}{v} \quad \text{or} \quad f = \frac{uv}{u + v} \quad \Rightarrow f = \frac{20 \times 80}{20 + 80} = 16 \, \text{cm} \] For error analysis: \[ \frac{1}{f} = \frac{1}{u} + \frac{1}{v} \] Differentiating: \[ \frac{Df}{f^2} = \frac{Du}{u^2} + \frac{Dv}{v^2} \] Now: \[ \Delta u = 0.4 \, \text{cm}, \, \Delta v = 0.4 \, \text{cm} \] Now, \[ \frac{\Delta f}{f} = \left[ \frac{16 \times 0.4}{(80)^2} + \frac{16 \times 0.4}{(20)^2} \right] \] \[ \Rightarrow \frac{\Delta f}{f} = 16 \times 0.4 \left( \frac{17}{400} \right) \quad \Rightarrow \% \, \text{Error} = \frac{17 \times 0.4}{400} \times 1000 = 1.7 \]