A single-slit diffraction experiment uses bd/D = mλ to determine slit width b. D and d are measured with least counts 1 cm and 1 mm, respectively. λ = 600 nm and m = 3 are exact. For d = 5 mm and D = 1 m, find the absolute error in b in μm.
Physics · Circular Motion
JEE Advanced 2025 Paper 1

Question

A single-slit diffraction experiment uses bd/D = mλ to determine slit width b. D and d are measured with least counts 1 cm and 1 mm, respectively. λ = 600 nm and m = 3 are exact. For d = 5 mm and D = 1 m, find the absolute error in b in μm.

Physics · Circular Motion

JEE Advanced 2025 Paper 1

G1 Physics · Accepted Answer

First, b = mλD/d = 3(600 × 10⁻⁹)(1)/(5 × 10⁻³) = 3.6 × 10⁻⁴ m = 360 μm. Using maximum instrumental errors equal to the least counts, Δb/b = ΔD/D + Δd/d = 0.01/1 + 0.001/0.005 = 0.01 + 0.20 = 0.21. Therefore Δb = 0.21 × 360 = 75.6 μm. The source key accepts 75–79 μm; it also lists an alternate accepted range 94–95 μm. The direct least-count propagation gives 75.6 μm.