# Question Video: Identifying the Correct Expression for the Phase Difference between Two Light Waves Physics

The diagram shows some apparatus used in holography, including a cylindrical object. The laser shown emits light waves that have a wavelength π. Which of the following is the phase difference between the light waves that travel the path π΄π΅πΆ and the light waves that travel the path π΄πππΆ? [A] 2ππ/(π΄πππΆ β π΄π΅πΆ) [B] 2π(π΄πππΆ β π΄π΅πΆ)/π [C] 2π(π΄πππΆ β π΄π΅πΆ)π [D] 2π(π΄πππΆ + π΄π΅πΆ)π [E] 2π(π΄πππΆ + π΄π΅πΆ)π. Which of the following is the phase difference between the light waves that travel the path π΄π΅π· and the light waves that travel the path π΄πππ·? [A] 2π(π΄πππ· β π΄π΅π·)/π [B] 2π(π΄πππ· β π΄π΅π·)π [C] 2ππ/(π΄πππ· β π΄π΅π·) [D] 2π(π΄πππ· + π΄π΅π·)/π [E] 2π(π΄πππ· + π΄π΅π·)π

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### Video Transcript

The diagram shows some apparatus used in holography, including a cylindrical object. The laser shown emits light waves that have a wavelength π. Which of the following is the phase difference between the light waves that travel the path π΄ capital π΅ capital πΆ and the light waves that travel the path π΄ lowercase π lowercase π capital πΆ? (A) Two ππ divided by π΄ lowercase π lowercase π capital πΆ minus π΄ capital π΅ capital πΆ. (B) Two π times the quantity π΄ lowercase π lowercase π uppercase πΆ minus π΄ capital π΅ capital πΆ all divided by π. (C) Two π times the quantity π΄ lowercase π lowercase π uppercase πΆ minus π΄ times capital π΅ capital πΆ all times π. (D) Two π times the quantity π΄ lowercase π lowercase π uppercase πΆ plus π΄ times capital π΅ capital πΆ all times π. (E) Two π times the quantity π΄ lowercase π lowercase π uppercase πΆ plus π΄ capital π΅ capital πΆ all divided by π.

Letβs begin by identifying the two paths, π΄ capital π΅ capital πΆ and π΄ lowercase π lowercase π capital πΆ, on our diagram. The first path from π΄ to capital π΅ to capital πΆ looks like this. A beam of light from our laser light source follows this path. We notice that at point πΆ, the light in this beam combines with the light in the other beam in our diagram. That beam follows this pathway from capital π΄ to lowercase π to lowercase π to uppercase πΆ. Evaluating by eye, it appears that this pathway in orange is of a different length than this pathway in blue. In fact, clearing some space at the top of our screen, we can write out that path length difference as the longer of these two paths, that appears to be π΄ lowercase π lowercase π uppercase πΆ, minus the shorter of the two paths, π΄ capital π΅ capital πΆ.

Like we said earlier, this is the path length difference between these two interfering beams of light. Our question though asks us to identify the phase difference between these two light waves. That phase difference is related to the path length difference, and this is because the source of all light in this apparatus is a laser, which emits coherent radiation. That tells us that any phase difference between our beams of light at point uppercase πΆ is due to a path length difference, a difference in distances that these beams traveled.

Letβs say that light from our two beams interferes at point πΆ like shown here. If we were to line up these interfering waves so that they were parallel, we would see that they have a phase difference, in this case, of π radians. That corresponds to a path length difference of one-half of one wavelength. That is, the wave that followed this path, π΄ lowercase π lowercase π uppercase πΆ, traveled an overall distance of one-half of one wavelength farther than the wave that followed this path.

Recall that this quantity here is a distance perhaps in units of nanometers or millimeters. It is the path length difference between the two waves, and we want to convert it to express the phase difference between these waves. To begin doing that, weβll want to express this distance in terms of the wavelength π of our radiation. Specifically, if we divide the distance by that wavelength π, then we now have a number of wavelengths that our path length difference is equal to. What weβve done then is weβve taken a distance, and by dividing by π, we now have that path length difference expressed in terms of a number of wavelengths of our wave.

Just by way of example, say that our wavelength π was 700 nanometers, the color of red light. And say further that our path length difference between these two waves turned out to be 1400 nanometers. In that case, this fraction would equal two, and this is the number of wavelengths that our path length difference of 1400 nanometers is equal to, keeping in mind that these are just example values to give us a sense for what this fraction means. Notice that our answer, two, is a difference expressed in terms of wavelengths, but it is not a phase difference like we want. A phase difference is expressed as an angle, say angles measured in radians.

Now, letβs think about this. If our path length difference in terms of wavelengths is two full wavelengths, then looking at our waves, that would involve a path length difference of this plus this, two complete wave cycles. Each one of those wave cycles corresponds to a change in phase of two π radians. For example, if we said that this point on our wave had a phase of zero radians, then this point would have a phase of two π radians, this point would have a phase of four π radians, and so on. For every one wavelength of our wave, we have two π radians of phase. Therefore, to take this number, our path length difference expressed in wavelengths, and turn it into a phase difference, we would need to multiply it by two times π. This is the phase difference, we could say, across one wavelength of our wave.

Letβs now remove our example values of 1400 nanometers for a path length difference and 700 nanometers for our wavelength. These example values have served their purpose because theyβve helped us understand a general equation for phase difference between these light waves. If we take the path length difference between our two beams, divide that by the wavelength of the light, and multiply that by two π, we have an expression for the phase difference between these two waves.

Reviewing our answer options, we see that the solution we found corresponds to option (B). The phase difference between the light waves that travel the path π΄ capital π΅ capital πΆ and the light waves that travel the path π΄ lowercase π lowercase π uppercase πΆ is given by this expression, two π times the quantity π΄ lowercase π lowercase π uppercase πΆ minus π΄ uppercase π΅ uppercase πΆ all divided by π.

Letβs look now at part two of our question.

Which of the following is the phase difference between the light waves that travel the path π΄ capital π΅ capital π· and the light waves that travel the path π΄ lowercase π lowercase π capital π·? (A) Two π times the quantity π΄ lowercase π lowercase π capital π· minus π΄ capital π΅ capital π· all divided by π. (B) Two π times the quantity π΄ lowercase π lowercase π uppercase π· minus π΄ capital π΅ capital π· all multiplied by π. (C) Two π times π divided by the quantity π΄ lowercase π lowercase π uppercase π· minus π΄ capital π΅ capital π·. (D) Two π times the quantity π΄ lowercase π lowercase π capital π· plus π΄ capital π΅ capital π· all divided by π. And (E) two π times the quantity π΄ lowercase π lowercase π uppercase π· plus π΄ capital π΅ capital π· all multiplied by π.

Like before, letβs start out by identifying these two paths on our diagram. Here, we show the path π΄ capital π΅ capital π· in blue, while here we have the path π΄ lowercase π lowercase π uppercase π· in orange. Just as the waves in part one met at point uppercase πΆ on our screen, now they meet at point uppercase π·. These waves interfere, and we want to identify the correct expression for their phase difference. In part one, we saw that we could start by calculating the path length difference of the two waves. That involves taking the longer of the two paths, in this case the path from π΄ lowercase π lowercase π to uppercase π·, and subtracting from that the shorter of the two paths, π΄ capital π΅ capital π·.

This difference, if we were to calculate it, is a difference in distance perhaps in units of nanometers or millimeters. If we were to divide this difference though by the wavelength of our radiation, then this fraction would equal the path length difference between our two waves in units of wavelengths. This gets us closer to expressing the phase difference between these light waves. We can recall that one wavelength of a wave corresponds to a phase difference across that wave of two π radians. So to turn our path length difference in wavelengths into a path length difference in terms of phase, we multiply this fraction by two π. This expression then will give us the phase difference between the light waves that travel these two paths, π΄ capital π΅ capital π· and π΄ lowercase π lowercase π capital π·.

The expression weβve identified corresponds to answer option (A). This is the phase difference between the light waves that travel these two paths identified in our diagram.