Guide 28-1. Interference (Calculating phase changes)

The key concept to use in determining how two waves interfere is the difference in phase between the two waves. If they're in phase, the interference is constructive. If they're 180° out of phase, the interference is destructive. Of course, the interference can also be partially destructive if the waves are out of phase by an amount between 0° and 180°.

In many interference problems, we have a situation were two waves from two closely-spaced sources travel different distances to an observer. This was, for example, the subject of P209 and P223. Whether the observer sees constructive or destructive interference depends on three things:

1. the difference in the distances traveled by each wave from the sources to the observer
2. the wavelength
3. any difference in phase created by reflection

Consider, for example, a Young's double-slit experiment. The waves passing through the two slits are initially in phase and there are no reflections along the way. Therefore, we don't have any phase differences resulting from item 3 above. So we just need to find the difference in the distances traveled by the waves from the slits to the screen and compare that to the wavelength. If the difference in the distances--called the path difference--is an integral number of wavelengths, then the interference is constructive. If the path difference is an odd-integral number of half wavelengths, the interference is destructive.

Beyond this, the only thing that may be different for different interference situations is how any phase difference is calculated. The situation in which one typically encounters phase differences is in thin-film interference, since that involves interference of waves reflected from different surfaces. Phase differences may result from the following:

1. one reflected wave travels a greater distance than the other
2. one or both reflected waves undergo phase inversion on reflection

Thin-film Interference

Figure 1

We now examine an example of determining the phase difference in thin-film interference. See Figure 1 to the right. Light waves of a particular frequency are produced by a light source and travel along path 1 to a thin film of thickness t and index of refraction n2. We'll take n1 = n3 = 1.00. What the observer sees depends on how the two reflected rays 2 and 6 interfere. Let's trace the paths of the light rays. Some of the light traveling along path 1 refracts into the thin film, bending closer to the normal and taking path 3. The rest of the light traveling along path 1 reflects from the upper surface of the film and takes path 2. Some of the light traveling along path 3 refracts back out into the air along path 4. The observer doesn't see this light, so we won't be concerned about it. The rest of the light traveling along path 3 reflects from the lower surface of the film and takes path 5. This light refracts back out into the air along path 6. Light waves traveling paths 2 and 6 interfere with each other. If the interference is destructive, the observer will see bright light. If the interference is constructive, the observer will see dim light or no light all. In order to determine how the waves interfere, one calculates the difference in phase changes between the light traveling path 1-2 and the light traveling path 1-3-5-6. This process will be described next.

In order to simplify the math, we'll move the light source so that the angle of incidence is 0°. See Figure 2 below. The incident wave corresponding to path 1 of Figure 1 is shown in green. The reflected wave corresponding to path 2 is shown in blue in Figure 3. Note that this wave is flipped 180° about the normal. This phase change of one-half wavelength occurs because the medium from which the light reflects has a greater index of refraction than the incident medium.

Now consider Figure 4. The wave that continued into the film has reflected from the lower surface. This reflected wave, which corresponds to path 5 in Figure 1, is shown in red. Note that the incident and reflected waves at the lower surface coincide. There is no phase inversion on reflection this time, because the index of refraction of the air is less than that of the incident medium, which is the film in this case.

Finally, consider Figure 5. The wave reflected from the lower surface refracts back into the air at the upper surface. There is no change in phase on refraction. The exiting wave, shown in magenta, corresponds to path 6 in Figure 1. Note that the two waves (blue and magenta) moving upward in air from the upper surface of the film are out of phase by half a wavelength. This means that the interference of these two waves is completely destructive. So the observer would see no light.

Now continue below the figures to see how the difference of phase changes is calculated.

Figure 2	Figure 3
Figure 4	Figure 5

The difference of phase changes consists of 3 parts:

1. The change in phase of the wave upon reflection from the upper surface of the film
2. The change in phase of the wave upon reflection from the lower surface of the film
3. The change in phase of the wave as it traverses the film, both down and back up

We've already determined that the phase changes for i) and ii) are 1/2 wavelength and 0 wavelengths, respectively. Now we consider the phase change for iii). For this, we need to calculate the number of wavelengths of the light that fit between the lower and upper surfaces of the film, keeping in mind that the wavelength in the film is different from the wavelength in the air. The wavelength in medium 1 is λ₁ = v₁/f, where v₁ is the speed of light in medium 1 and f is the frequency of the light. Likewise, the wavelength in medium 2 is λ₂ = v₂/f, where v₂ is the speed of light in medium 2. Note that the frequency is the same in both media. Forming the ratio of the wavelengths and dividing out the frequency, we have:

λ₂/λ₁ = v₂/v₁.

Since n = c/v, we can rewrite the above equation in terms of indices of refraction.

λ₂/λ₁ = n₁/n₂

Finally, we solve for λ₂ and substitute n₁ = 1.00 for the index of refraction of air.

λ₂= λ₁/n₂

The number of wavelengths between the upper and lower surfaces of the film is simply t/λ₂ = n₂t/λ₁.

We can now calculate what is called the effective path length, l_eff for each of the reflected rays. For the ray reflected from the upper surface, this is simply the half wavelength introduced on reflection.

l_eff,1 = λ₁/2

For the ray reflected from the lower surface, the effective path length is:

l_eff,2 = (2n₂t/λ₁)λ₂ + 0.

Note that the 0 is included as a reminder to include the phase change on reflection. Finally, we can find the difference in phase changes of the two reflected rays.

Difference in phase changes = l_eff,2/λ₂ 1 l_eff,1/λ₁

Substituting the expressions for effective path length and simplifying, we have:

Difference in phase changes = (2n₂tλ₂/λ₁)/λ₂ - (λ₁/2)/λ₁= (2n₂t/λ₁) - 1/2.

The interference will be constructive if the difference in phase changes is an integer and will be destructive if the difference in phase changes is an odd multiple of 1/2.

See Thin Films in Section 28-3 of the text for example problems. Note that the textbook uses the notation n = n₂, λ_n= λ₂, and λ_vacuum = λ₁.

Also see the section Air Wedge for a related application.

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