Practice portal › Wave Optics › Interference and Young's Double Slit Experiment
Asked in GUJCET 2024 · Fringe width and fringe position
Given: d = 0.28 mm = 2.8×10⁻⁴ m, D = 1.4 m, and the fourth bright fringe at x₄ = 1.2 cm = 1.2×10⁻² m.
Idea: bright fringes sit at xₙ = (nλ D)/d, so the measured position of a fringe of known order fixes the wavelength.
Rearranged with n = 4, λ = (x₄ d)/(4D).
λ = ((1.2×10⁻²)(2.8×10⁻⁴))/(4 × 1.4) = (3.36×10⁻⁶)/(5.6).
λ = 6×10⁻⁷ m.
The wavelength used is 600 nm.
Reading this solution needs no sign-in. You only sign in to practise against the clock and keep your progress.
Practise this with a timer