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: (a) 1.70 (b) 0.51 (c) 1.02 (d) 0.85

Answer: (a). u = 100 - 80 = 20 cm, v = 180 - 100 = 80 cm, f = uv/u+v = 16 cm. Each distance is a difference of two readings: Δ u = Δ v = 0.2 + 0.2 = 0.4 cm. From 1/f = 1/u + 1/v: (Δ f)/(f) = f((Δ u)/(u²) + (Δ v)/(v²)) = 16((0.4)/(400) + (0.4)/(6400)) (Δ f)/(f) = 0.0170 ⇒ 1.70%. So the answer is (a).

A 2 meter long scale with least count of 0.2 cm is used to measure… | JEE Main 2023 (6 Apr, Shift 2)

Q61JEE Main 2023 · 6 APR · SHIFT 2Medium

Spherical Surfaces & Thin Lenses

A 2 meter long scale with least count of $0.2\text{ cm}$ 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 $80\text{ cm}$ mark and $1\text{ m}$ mark, respectively. The image of the object pin on the other side of lens coincides with image pin that is kept at $180\text{ cm}$ mark. The $\%$ error in the estimation of focal length is:

Show answer and solution

Answer: (a)

$u = 100 - 80 = 20$ cm, $v = 180 - 100 = 80$ cm, $f = \frac{uv}{u+v} = 16$ cm.
Each distance is a difference of two readings: $\Delta u = \Delta v = 0.2 + 0.2 = 0.4$ cm.
From $\frac{1}{f} = \frac{1}{u} + \frac{1}{v}$ : $\frac{\Delta f}{f} = f\left(\frac{\Delta u}{u^2} + \frac{\Delta v}{v^2}\right) = 16\left(\frac{0.4}{400} + \frac{0.4}{6400}\right)$
$\frac{\Delta f}{f} = 0.0170 \Rightarrow 1.70\%$ . So the answer is (a).

Report a mistake

Spherical Surfaces & Thin Lenses question from JEE Main 2023 (6 Apr, Shift 2)

More questions on Spherical Surfaces & Thin Lenses