- #106
collinsmark
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Don't forget about phase shifts. A reflected wave gets a phase shift of [itex] \pi [/itex]*. Are there any differences in the total phase shifts between the two beams in question (one on the laser side of the interferometer, and the other on the wall side of the interferometer), even if their path lengths are the same?Devin-M said:Yes at the very end he end he asks the viewer if the path difference of the 2 mirrors is 1/2 wavelength, why does the light going to one screen cancel out and the other add constructively. He doesn’t answer this question and leaves it for the viewer to answer. It seems like before he adds the second detection screen, half the light from the mirrors is going to the screen and the other half is going back to the laser source. When he adds the second detection screen now 1/2 is going to the 1st screen, 1/4th is going to the 2nd screen, and 1/4th is going back to the laser. As far as why on one screen there’s a half wavelength path difference and the other there isn’t, I believe it’s due to the thickness of the beam splitter. It seems there must be an extra half wavelength path length (or odd multiple of 1/2 wavelength) for the light from the left mirror that goes back through the beam splitter. I’m not yet fully convinced that when the light cancels out its really going back to the source because once its adding destructively the path length back to the source should be identical all the way back to the source so it should add destructively all the way back to the source.
For example with a different thickness beam splitter it may be possible to have the light on both detection screens add constructively at the same time or both destructively at the same time.
*(For the purposes here, ignore the nuances about reflections caused by the glass itself -- the reflections going from one index of refraction to another. That makes things things overly complicated, and can be ignored for this discussion. A good beam splitter will have coatings to minimize these types of reflections anyway. Instead, treat all reflections as being caused solely by the very thin, half mirrored surface -- the light either gets reflected with a phase shift of [itex] \pi [/itex], or it gets transmitted with no phase shift at all.)