In a Young's double slit experiment, the path difference, at a certain point on the screen, between two interfering waves is 1/8th of the wavelength. The ratio of the intensity at this point to that at the centre of a bright fringe is close to:
0.94
0.74
0.85
0.80
Solution:
Δx = λ/8 Δφ = (2π/λ)λ/8 = π/4 I = I₀cos²(π/4) I/I₀ = cos²(π/4) = (1/√2)² = 1/2 = 0.5 The closest option is 0.74