In an experiment for estimating the value of focal length of converging mirror, image of an object placed at 40 cm from the pole of the mirror is formed at distance 120 cm from the pole of the mirror. These distances are measured with a modified scale in which there are 20 small divisions in 1 cm. The value of error in measurement of focal length of the mirror is 1/K cm. The value of K is __________.

Solution

Physics sir · 4 steps

Get f from the mirror formula, then differentiate it: each distance's error adds, weighted by f squared.

1. Find f. Object and image are both real, in front of the mirror. With magnitudes, 1/f = 1/40 + 1/120 = 4/120, so f = 30 cm. The scale has 20 divisions in 1 cm, so its least count is Δ u = Δ v = 1/20 = 0.05 cm.
2. Differentiate. Differentiate 1/f = 1/u + 1/v. For the maximum error, the errors in u and v always add.

   (Δ f)/f² = (Δ u)/u² + (Δ v)/v²

3. Put in numbers. Δ f = 900 × 0.05 (1/1600 + 1/14400) = 45 × 10/14400 = 450/14400 = 1/32 cm.
4. So. Δ f = 1/K cm with K = 32.

The idea to keep. For a reciprocal formula like the mirror or lens equation, differentiate it directly. Each error gets divided by its own distance squared, and the errors always add.

Answer from the NTA final answer key. Solution by PrepWiser.