Question

(ii) Light waves each of amplitude “a” and frequency “ω”, emanating from two coherent light sources superpose at a point. If the displacements due to these waves is given by y1 = a cosωt and y2 = acos(ωt + ∅) where ∅ is phase difference between the two, obtain the expression for the resultant intensity at the point

Answer 1

Light wave is a transverse wave. We find the energy. power, intensity this way.

y1(t) = a Cos ωt        y2(t) = a Cos (ωt+Ф)

Superposition of the two waves results in the net displacement of a particle at that given point x as:
    y(t) = a [cos ωt  + cos (ωt+Ф) ]
          = [2 a Cos(Ф/2)]  Cos(ωt + Ф/2)
   Velocity of the particle = v(t) = dy(t)/dt = -[2aω Cos(Ф/2)] Sin(ωt+Ф/2)
   Let the medium have a density of ρ and Volume V.  So mass = Vρ = m.

   KE of this medium = 1/2 m v²
   KE /unit volume = 1/2 ρ v² = [2a²ω²ρ Cos²(Ф/2)] Sin²(ωt+Ф/2)
   This is to be averaged over t = 0 to T=2π/ω.
   The average of Sin²(ωt+Ф)  is = 1/2.  We can do this by integration.

   KE/unit volume, (averaged over in 1 time period) = a² ω² ρ Cos²(Ф/2)
   Intensity = Power / unit cross section area
               = Energy / unit volume * wave speed
              = a² ω² ρ Cos²(Φ/2) * c
              = a² ω² ρ c * Cos(Φ/2),   where c = speed of light wave in the medium

 Here unit volume
          = unit area of cross section * unit length in the direction of propagation.
          = unit area * wavelength 
Then we take time as the time period to travel one wavelength distance.