So, we know where the intensity is maximum and where it's zero. How does the intensity change in between these points?

Pick a random point on the screen. Let's say the light arriving from the first source has an electric field which varies as

E_{1} = E_{o} sin(ωt).

The electric field from the second source is almost the same, just with a phase shift:

E_{2} = E_{o} sin(ωt + φ).

Thus the total electric field at our point is:

E = E_{1} + E_{2} = E_{o} [ sin(ωt) + sin(ωt + φ) ]

Use the trig. identity sin A + sin B = 2 sin [(A+B)/2] cos [(A-B)/2]

where A = ωt + φ and B = ωt

This gives E = 2 E_{o} cos(φ/2) sin(ωt + φ/2)

The intensity of the wave is proportional to E^{2}.

Therefore I α 4 E_{o}^{2} cos^{2}(φ/2) sin^{2}(ωt + φ/2)

Averaging over time using the fact that the average value of sin^{2}(θ) = 1/2 gives:

I_{av} α 2 E_{o}^{2} cos^{2}(φ/2) or I_{av} = I_{max} cos^{2}(φ/2)

The phase difference depends on the path length difference. When the path length difference is one wavelength, for instance, what is the phase difference between the waves?

When δ = λ the phase φ = 2π.

This gives φ/2π = δ/λ, so:

φ = 2pd/λ = 2πd sin(θ)/λ

Thus our expression for the average intensity is:

I_{av} = I_{max} cos^{2} (φ /2) =
I_{max} cos^{2} (πd sin(θ)/λ)

For small angles I_{av} = I_{max} cos^{2} (πd y/λL)

where y is the distance along the screen measured from the center.