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(wt).

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

E_{2} = E_{o} sin(wt + f).

Thus the total electric field at our point is:

E = E_{1} + E_{2} = E_{o} [ sin(wt) + sin(wt + f) ]

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

where A = wt + f and B = wt

This gives E = 2 E_{o} cos(f/2) sin(wt + f/2)

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

Therefore I a 4 E_{o}^{2} cos^{2}(f/2) sin^{2}(wt + f/2)

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

I_{av} a 2 E_{o}^{2} cos^{2}(f/2) or I_{av} = I_{max} cos^{2}(f/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 d = l the phase f = 2p.

This gives f/2p = d/l, so:

f = 2pd/l = 2pd sin(q)/l

Thus our expression for the average intensity is:

I_{av} = I_{max} cos^{2} (f /2) =
I_{max} cos^{2} (pd sin(q)/l)

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

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