Interference and How The Path Difference Causes It

Interference is a fundamental phenomenon which occurs everywhere in nature and technology where waves are involved. Many things around us consist of waves.

• Water waves on the sea.

• Sound waves that allow us to communicate verbally with other people.

• Light waves emitted by our sun.

• Matter waves of smallest particles, like electrons, which also show interference phenomena.

Other applications include anti-reflective coatings, such as anti-reflection coating for eyeglasses. But also interferometers, with which the smallest changes in length can be measured. These include, for example, Michelson interferometers, Mach-Zehnder interferometers and Sagnac interferometers which you may learn about later in life as you follow the path of physics.

In physics, you can find interference in the double-slit experiment, in acoustic beat or Bragg reflection.

To understand what happens when waves are overlap, you first have to create waves.

• For example, you can throw two stones next to each other into the water. The stones create water waves, the waves collide and interfere.

• You can turn on two speakers. These generate sound waves that can interfere with each other.

To keep things from getting too complicated, let's look at interference using a very simple but very basic example:

One-dimensional sine wave propagating along the \(x\) axis.

The sine wave is a sequence of wave crests and wave troughs and has two properties important for interference:

1. Wavelength \( \lambda \) ("Lamb-da!") - this is the distance between two wave crests (or troughs).

2. Amplitude \( A \) - in the case of a sound wave it indicates how loud the sound is.

If you turn on a loudspeaker sine tone, the sound is emitted in different directions so that everyone in the room can hear it. Let's set the amplitude to the value \( A ~=~ 1\) so that we can easily overlap two waves with it.

One wave alone is not enough to study interference. We need another wave for this purpose. This can be generated, for example, by another loudspeaker standing next to it. Let's switch on exactly the same sine wave at the second loudspeaker. This has the advantage that the wavelengths of the two sine waves are the same, making the calculation a bit easier. Of course, you can also consider any other waves.

As soon as the two sine waves meet, they overlap. We sent them off in such a way that a wave crest of one sine wave meets a wave crest of the other sine wave. Also each wave trough meets corresponding wave trough at the same point.

To perform the interference mathematically, you proceed as follows:

1. Select any position \(x\) where you want to superimpose the two waves.

2. Add the amplitude \( A_1 \) of one wave to the amplitude \(A_2\) of the other wave at the same position \(x\). The sum of the two amplitudes gives the amplitude \(A(x)\) of the resulting wave at the considered location: \( A(x) = A_1(x) + A_2(x) \).

3. Plot the value of the new amplitude in an amplitude-position graph at the considered position \(x\).

For example, take two maxima (crests) at the same location \(x\). Since both crests have amplitude 1, the resulting amplitude is:

Then take for example two minima (wave troughs) at the same location \(x\). Since both wave troughs have the amplitude -1, the resulting amplitude is:

Proceed analogously with all other locations \(x\). If you plot all the resulting points in a diagram and connect them, you will get a resulting wave.

In this example, it was a so-called constructive interference because the amplitude of the new sine wave has become twice as large as the amplitude of the original waves. The resulting sound described by this sound wave is twice as loud!

So let's summarize: for constructive interference of two waves to occur, the path difference must be a multiple of the wavelength:

Here \(m\) is a number which can take negative and positive integers: \( m ~=~ ...-2,-1,0,1,2... \).

But what if one sine wave precedes the other wave not by a multiple \( m \, \lambda \) , but by exactly half the wavelength, i.e. by \( \lambda/2 \) ? Then each wave crest meets a wave trough. The deflections at the wave trough and wave crest always add up to zero at each location. At the zero points the deflections also add up to zero: 0 + 0 = 0. And also at every other place the deflections add up to zero. The two waves cancel out completely. In this case, we have destructive interference. It always occurs when one wave precedes the other by \( \lambda/2 \). Or also by \( \lambda/2 +1\lambda \). Or \( \lambda/2 +2\lambda \), or \( \lambda/2 +3\lambda \) and so on.

Here \( m \) is again an integer with negative and positive values: \( m ~=~ ...-2,-1,0,1,2... \).

So far we have considered two special cases: where one wave crest has perfectly met the other wave crest, i.e. constructive interference, and where one wave crest has perfectly met a wave trough, i.e. destructive interference. But this does not always have to be the case! The waves could just as well interfere only partially with each other. This would mean that the wave crest does NOT exactly meet the other wave crest... But these cases are rather unimportant, because in applications mostly complete amplification or complete cancellation are useful.

Sometimes the path difference \( \Delta s \) is given as phase difference \( \Delta \varphi \). What is a phase, you may ask?

We can express our sine waves mathematically as follows:

Here \(A_1\) and \(A_2\) are their amplitudes, \(\omega\) is the angular frequency. In our case it is the same for both sine waves and it contains the information about the wavelength. The angle \(\varphi_1\) of one wave and the angle \(\varphi_2 \) of the other wave gives the displacement of the wave to the left or to the right - with respect to a previously defined starting position. This angle is called the phase of a wave.

The phase difference, as the name implies, is the difference of the two angles (phases):

Each path difference \( \Delta s \) corresponds to a certain phase difference \( \Delta \phi \). You can specify the phase difference either in radians (from \( 0 \) to \( 2\pi \)) or in degrees (from 0 to 360°).

You can calculate the phase difference in degrees using the path difference as follows:

And you can convert the path difference into phase difference in radians as follows:

We have already derived an expression for the path difference for constructive (eq. 3) and destructive interference (eq. 5). You can now simply substitute the condiitions into the equations for phase difference in degrees or radians.

Phase difference for destructive interference in degrees, looks like this, for example:

Here the wavelength \(\lambda\) cancels.