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Newton's rings and FTL

  1. Apr 30, 2013 #1
    Probably this is an easy question, given that Newton's rings are probably one of the most common and famous light phenomena. In any case, I was wondering that given that say you have the second interface at a position in which the reflectivity on the first is 0, but then you are allowed to rotate this interface so that the reflectivity would jump up to 4%. Well, you could put a detector very close to the interface so that just when it reflects you detect it, but the probability of detecting it will depend on something that may be spatially separated. Does this have an easy resoltion?

    Thanks beforehand,

    Guille
     
  2. jcsd
  3. May 1, 2013 #2
    this is not clear enough. Can you draw a picture?
     
  4. May 3, 2013 #3
    Yeah, sorry, it's actually quite simple, but should have made it clearer. I decided instead of rotating the bottom surface to displace it a bit, which is maybe more practical?. Anyway, in the picture, the space between the first surface and either of the bottom ones is glass say. The yellow beam is light of wavelength lambda inside the glass, and the little black box is a photon detector. My prediction is that moving the bottom surface (by some idealized means) from 0 to 1 would immediately determine the probability of the photon reflecting on the first surface. As the moment of the detection can occur arbitrarily close to the moment of reflection, the probability of the photon being detected will depend nearly instantaneously on the position of the bottom surface, thus allowing FTL signaling if a bunch of several photons is sent every time (to calculate averages). Of course, I'm pretty sure there must be something wrong in here..
     

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