JEE Main 2024 · 4 April · morning · Single correct · +4 / −1
In an experiment to measure focal length (f) of convex lens, the least counts of the measuring scales for the position of object (u) and for the position of image (v) are Δ u and Δ v, respectively. The error in the measurement of the focal length of the convex lens will be:
- (a) 2 f[(Δ u)/u+(Δ v)/v]
- (b) f[(Δ u)/u+(Δ v)/v]
- (c) f²[(Δ u)/u²+(Δ v)/v²]
- (d) (Δ u)/u+(Δ v)/v
Answer and solution
Answer (c) f²[(Δ u)/u²+(Δ v)/v²]
Solution
Physics sir · 3 steps
Do not touch f = uv/(u+v); differentiate the lens formula, where each term carries one variable.
- Start from lens formula. Using magnitudes of the distances for a real image, 1/f = 1/u + 1/v. Each term on the right holds only one measured quantity, so the errors are easy to track.
- Differentiate each term. Since d(1/x) = -dx/x², every term picks up a square in the denominator. For the maximum error, add the magnitudes of the two contributions and never subtract them.
(Δ f)/f² = (Δ u)/u² + (Δ v)/v²
- So. Multiply through by f² to get Δ f = f²[(Δ u)/u² + (Δ v)/v²]. That is option (c).
The idea to keep. When a formula has reciprocals like 1/f = 1/u + 1/v, the error in 1/x is (Δ x)/x². Add these magnitudes, then solve for the error you want.
Answer from the NTA final answer key. Solution by PrepWiser.