Wave Optics

Physics · Class 12

Simulation · Physics · Class 12

Coherent and incoherent addition

From the lesson Coherent and incoherent addition in Wave Optics. Change the values and watch what happens.

Coherent and incoherent additionPhysics · Class 12

The idea behind it

NCERT §10.4

  • 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.