Color of light:
Distance to Screen:

Intensity of a Two-Source Interference Pattern

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

E1 = Eo sin(wt).

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

E2 = Eo sin(wt + f).

Thus the total electric field at our point is:

E = E1 + E2 = Eo [ 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 Eo cos(f/2) sin(wt + f/2)

The intensity of the wave is proportional to E2.

Therefore I a 4 Eo2 cos2(f/2) sin2(wt + f/2)

Averaging over time using the fact that the average value of sin2(q) = 1/2 gives:

Iav a 2 Eo2 cos2(f/2) or Iav = Imax cos2(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:

Iav = Imax cos2 (f /2) = Imax cos2 (pd sin(q)/l)

For small angles Iav = Imax cos2 (pd y/lL)

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