Video: Displacement of a Michelson-Morley Interferometer

What is the distance moved by the traveling mirror of a Michelson interferometer that corresponds to 1500 fringes passing by a point of the observation screen? Assume that the interferometer is illumined with a 606 nm spectral line of Krypton-86.

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Video Transcript

What is the distance moved by the traveling mirror of a Michelson interferometer that corresponds to 1500 fringes passing by a point of the observation screen? Assume that the interferometer is illumined with a 606-nanometre spectral line of Krypton-86.

We’re told in this statement that the number of fringes that pass the observation screen when the mirror moves is 1500. We’ll call this 𝑁𝐹. For number of fringes, we’re also told that the wavelength of the light used in the interferometer is 606 nanometres which we’ll call πœ†. We want to solve for the distance moved by the movable mirror, which we’ll call 𝑑.

To begin our solution, let’s draw a diagram of this scenario. In a Michelson interferometer, there is a light source that sends that light to a half reflecting plate, which lets 50 percent of the light through and reflects the other half. Each of these two beams encounters a mirror, bounces back to the beam splitter, and recombines before they run into a screen. Depending on how the recombined beams interact, the screen displays a series of fringes with the number of fringes, depending on the phase relationship between the beams that combine on the screen.

Everything in the setup is fixed, except for the movable mirror, which can slide up and back and so lengthen or shorten the path length of the reflected ray of light. For a Michelson interferometer, the number of fringes that pass by the screen, π‘š, is equal to two times the distance the movable mirror is moved, 𝑑, divided by the wavelength of the light in the interferometer, πœ†.

Applying this relationship to our scenario, we write that 𝑁𝐹 the number of fringes equal two times the movable mirror’s distance moved 𝑑 over πœ†. Rearranging, we see that 𝑑 equals πœ† times 𝑁𝐹 over two. And when we plug in for πœ† and 𝑁𝐹 and enter these values on our calculator, we find that 𝑑 is equal to 4.55 times 10 to the negative fourth metres. This is how far are the movable mirror would have to translate in order for 1500 fringes to pass by on the screen.

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