Think of when you drop a pebble into a pond, you will see circular waves eminate from the point where you dropped the pebble.

Figure 9.1. 

When you drop two pebbles side by side you will see a much more complicated pattern:

Figure 9.2. 

Likewise with electromagnetic waves, you can get interesting interference phenomena when waves eminate from two point sources.

Lets take a particular example of two point sources separated by a distance d. The waves emitted by point source are spherical and thus can be written To make the problem easier we will make the k 's the same for the two sources. Also lets set the E₀'s to be the same as well.

Figure 9.3. 

The the only difference in the waves will be the r 's, that is Now there is a slightly subtle point here that is important to understand. In the denominator it is sufficient to say that r₁ ≈ r₂ and just call it r . We assume that we are far enough away that the differences between r₁ and r₂ are too small to matter. However this is not true in the argument of the harmonic function. There, very small differences between r₁ and r₂ can have a big effect. So lets define r₁ = r₂ = R Now to evaluate the final term we use cos ( θ − φ ) = cosθ cosφ + sinθ sinφ and write So we have

Clearly I will be a maximum when the cosine is = +1 k Δ r = 2 n π    n = 0 , 1 , 2 … Δ r = nλ There will be a minimum when the cosine is = -1 k Δ r = nπ    n = 1 , 3 , 5 … So you get light and dark bands which are called interference fringes.To reiterate; we have two rays of light eminating from two point sources. We have looked at the combined wave at some point, a distance r₁ from the first source and a distance from the second source. In that case we find that the intensity is proportional to To make things easier we can redefine E₀ to be the amplitude of the waves at the point under consideration, that is I = ε₀c E₀² ( 1 + cosk Δ r ) . Or we can say I₀ = ε c E₀² / 2 and write I = 2 I₀ ( 1 + cosk Δ r ) .

Figure 9.4. 

Say we place a screen a distance S away from the two sources:

Figure 9.5. 

In this case we see that Δ r = d sinθ So we have maxima at Δ r = nλ = d sinθ . The angle between two maxima is given by or for small θ Notice how when the sources are moved far apart the effect maxima become very close together so the screen appears to be uniformly illuminated. If a screen is placed a distance S away the maxima on the screen will occur such that d sinθ = nλ but in the small angle limit which implies likewise minima will occur at using we can rewrite I = 2 I₀ ( 1 + cosk Δ r ) as

Young's double slit.is an excellent example of two source interference. The equations for this are what we worked out for two sources above. Interference is an excellent way to measure fine position changes. Small changes in Δ r make big observable changes in the interference fringes.

A particularly useful example of using interference is the Michelson interferometer. This can be used to measure the speed of light in a medium, measure the fine position of something, and was used to show that the speed of light is a constant in all directions.

Figure 9.6. 

When Δ r , the path length difference in the two arms is Δ r = nλ then the rays of light in the traveling down the center of the apparatus will interfere constructively. As you move off axis the light travels slightly different lengths and so you get rings of interference patterns. If you have set up the apparatus so that Δ r = nλ and then move one of the mirrors a quarter wavelength then and you get destructive interference of the central rays. Thus you can easily position things to a fraction of a micron with such a set up.

What really matters is the change in the optical pathlength. For example you could introduce a medium in one of the paths that has a different index of refraction, or different velocity of light. This will change the optical pathlength and change the interference at the observer. Thus you can measure the velocity of the light in the introduced medium.

Michelson and Morely used this technique to try to determine if the speed of light is different in different directions. They put the whole apparatus on a rotating table and then looked for changes in the interference fringes as it rotated. They saw no changes. In fact they went so far as to wait to see what happened as the earth rotated and orbited and saw no changes. They thus concluded that the speed of light was the same in all directions (which nobody at the time believed, even though that is the conclusion you draw from Maxwell's equations.)

Another application of interference is a a gyroscope, ie. as device to measure rotations.

Figure 9.7. 

If the apparatus is rotating, then the pathlengths are different in different directions and so you can use the changes in the interference patterns to measure rotations. This is in fact how gyroscopes are implemented in modern aircraft.

Suppose there is a very thin film of dielectric and light is incident on it normally. Lets consider single reflections. (We make the small angle of incidence approximation)

Figure 9.8. 

We will assume n₃ > n₂ > n₁. The physical path length difference of the reflected light is Δ r = 2 d . We will get maxima in the interference when: Δ r = 2 d = m λ₂    m = 1 , 2 , 3 … where λ₂ is the wavelength in the film. Now λ_iν_i = c / n_i . In our example we have ν₁ = ν₂ = ν₃, that is the frequency does not change moving between the media. So we have λ₁n₁ = λ₂n₂ = λ₃n₃ . Thus constructive interference will happen when where n_film = n₂. Destructive interference will happen when ( 2 d ) n_film = m λ_air / 2   m = 1 , 3 , 5 …

When destructive interference occurs then that value of λ is not reflected. Note that this is a function of both d and λ . The next effect is that different colours of light get reflected at different thicknesses of the film. This is why soap films or oil films on water give rainbow effects.

Note I have assumed that n₃ > n_film > n_air in the above, where n₃ is the material that the film sits upon.

Consider an interface between two materials with indices of refraction n₁and n₂. If n₂ > n₁. Then lets examine what happens to the phase of an electromagnetic wave upon reflection. For a transverse electric field, there is a phase change of π . For the transverse magnetic field (or E_∥ ) there is not, if the light ray is close to the normal. However if n₁ > n₂ then and the situation is reversed and the transverse electric field does not undergo a phase change and the transverse magnetic field does. In the example above, their will be no relative phase change between the rays in either case. Either both will change by π or neither will change, depending on the orientation of the E field.