Why is the white light of the interferogram produced by using Michelson Interferometer necessarily symmetric? This is really hard to think. white http://upload.wikimedia.org/wikipedia/en/d/dc/Colored_and_monochrome_fringes.png

  • $\begingroup$ It isn't, the wave-fronts that produced it are not symmetric about any particular axis, unless it was produced by a white light laser... $\endgroup$ Nov 28, 2013 at 4:04
  • $\begingroup$ How can white light is a laser? Laser is a monochromatic source. $\endgroup$ Nov 28, 2013 at 10:33
  • $\begingroup$ Questions about how to make a white light laser, should have an optical or electromagnetic engineering tag. $\endgroup$ Nov 28, 2013 at 14:35
  • $\begingroup$ @elpsyCongroo All laser light begins as spontaneous emission, which is amplified by stimulated emission and recirculated in a cavity. If you engineer your cavity and the spontaneous emission to be broad band and if you have enough energy around, the amplified output can be very broad band. $\endgroup$ Nov 29, 2013 at 1:15

1 Answer 1


If we assume the wave-fronts that initially enter the system to produce this image are symmetric, than the result is intuitively symmetric as all the components used to produce it are too. And anyone who looks at the images you have shown, will clearly see a symmetry about the Y-axis. Assuming they do not look for minuscule errors due to wave-front discontinuity.

Let's make sure we have all the right ideas by verifying the definition of "symmetry". "Symmetry across a Y-axis" is a reflectional symmetry. Meaning that both sides, if split down the center and one side flipped, are equal to each-other.

I suppose the other question is why this symmetry is visible for such a system? It was made using a "Michelson Interferometer", if we look at such a device it is clearly also reflectively symmetric (Neat). Now, keep in mind that all wave fronts curve. Given the reflective path, the wave-fronts on the edge would start to fall further behind the center wave-front, giving it a different position. The shorter path will have edge wave-fronts in different positions than the longer path. This will create the interference we see. Both paths are almost equal, we are talking about a difference in distance on the order of fractions of wavelengths.

Hope that helps.


Due to the nature of any light source, the vibrating particles making the light can be out of sink with each-other. This is not true for a laser, once the laser is started the selected waveform for that laser will reinforce the motion of the particles, making them move in synchronization with each-other. If the particles producing the light are moving together the discontinuities will diminish.

  • $\begingroup$ I still don't understand what you mean by wave-front discontinunity sorry Luke. $\endgroup$ Nov 28, 2013 at 10:32
  • $\begingroup$ Hi, Luke. So I understand that you mean the laser ( which is a coherent light source ) will be continuous, whereas a white light comprised of different wavelengths will result in cancelling. However, is the pattern necessarily symmetry after all? If yes, what's the reason behind this? Thanks ! $\endgroup$ Nov 28, 2013 at 18:40

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