Does Destructive Interference Affect Starlight in a Double Slit Experiment?

In summary: There is nowhere near twice as much red value when adding the second slit. So where did the missing energy go?
  • #106
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.
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?

*(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.)
 
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  • #108
Devin-M said:
On wikipedia it says “Also, this is referring to near-normal incidence—for p-polarized light reflecting off glass at glancing angle, beyond the Brewster angle, the phase change is 0°.”

https://en.m.wikipedia.org/wiki/Reflection_phase_change

Right. As I mentioned, let's ignore the air-glass or glass-air interfaces.

What's important here are the air-silver and glass-silver interfaces. (Well, on second though, that's just an example of one type of beam splitter. See below for clarification.)

Here's an image from the Beam Splitter entry in Wikipedia:

Wavesplitter1.gif

Phase shift through a beam splitter with a dielectric coating.

Source: https://en.wikipedia.org/wiki/Beam_splitter#Phase_shift

And as mentioned in the article, the details are dependent upon the type and geometry of the beam splitter. So the details of the experiment might change a little depending on your choice of beam-splitter type. So there isn't necessarily a one-size-fits-all simple answer to this topic.

But in any case, it will always work out, one way or the other, that energy is conserved. A decrease of energy via destructive interference somewhere will always lead to a corresponding increase of energy somewhere else. That much we can count on.
 
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  • #109
collinsmark said:
wavesplitter1-gif.gif

Phase shift through a beam splitter with a dielectric coating.

Source: https://en.wikipedia.org/wiki/Beam_splitter#Phase_shift

That diagram predicts what will be observed in the experiment but it tends to refute his claim that the destructively interfering beams on the 1st detection screen go back to the source.
 
  • #110
What I mean is in this video at 4:24 he uses quarter wave plate on the right arm to rotate the polarization of that one arm, which then prevents destructive interference even if there is a half wavelength path length difference. So suppose he used a quarter wave plate on one arm of the original setup, would the combined output on both detection screens have higher intensity from lack of destructive interference on either screen?

see 4:24:
 
  • #111
collinsmark said:
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?

*(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.)
provided the reflection is on the side of the medium with the larger index of refraction.
 
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