Coherent and incoherent addition.

Two equal sources adding at a point: phasor sum, I = 4I₀cos²(φ/2) for a locked phase difference, and a flat 2I₀ when the phase jumps at random.

Find the path difference where intensity is a quarter of the maximum (λ/3), then switch to incoherent and read a steady 2I₀.

Writing the maximum from two coherent sources as 2I₀; expecting fringes from two separate lamps.

• Interference rests on the superposition principle: where several waves meet, the resulting displacement is the vector sum of the displacements each wave would produce alone.
• Two needles dipping up and down in step in a water trough make two sets of ripples. If the phase difference between the two waves at any point stays constant in time, the sources are coherent.
• At a point P equally far from S₁ and S₂, the waves arrive in step: y₁ = y₂ = a cos ωt, so y = 2a cos ωt. Intensity goes as amplitude squared, so I = 4I₀, where I₀ is the intensity from one source alone. Every point on the perpendicular bisector of S₁S₂ gets 4I₀.
• A path difference of λ is a phase difference of 2π. A path difference of 2λ gives 4π, so the waves again arrive in step (constructive, 4I₀). A path difference of 2.5λ gives 5π, the displacements are opposite, and they cancel (destructive, zero intensity).
• Rule: path difference S₁P ~ S₂P = nλ gives constructive interference; (n + ½)λ gives destructive interference, with n = 0, 1, 2, 3, ...
• For a general phase difference φ, y = a cos ωt + a cos(ωt + φ) = 2a cos(φ/2) cos(ωt + φ/2). The amplitude is 2a cos(φ/2), so I = 4I₀ cos²(φ/2): maxima at φ = 0, ±2π, ±4π, ... and zeros at φ = ±π, ±3π, ...
• With coherent sources φ at each point is fixed, so the bright and dark places stay put: a steady interference pattern. If the phase difference changes rapidly and at random, the pattern shifts faster than it can be seen, and the time-averaged intensity is I = 2I₀ everywhere. Such sources are incoherent, and their intensities simply add, as when two lamps light a wall.
• Exercise 10.5: the intensity is K where the path difference is λ, so φ = 2π and K = 4I₀. Where the path difference is λ/3, φ = 2π/3 and cos²(π/3) = ¼, so I = 4I₀ × ¼ = K/4.

Teaching this idea? Put the simulation on your own page or LMS with this code.