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Natural Science Forum / Physics / Research / July 2008q. about the result of the dual-slit setup with an additional slit
Hi! Suppose you have the light-source and dual-slit and screen, setup so you get the classic interference pattern on a screen, and you make a new slit in the screen in the center just where you have the dark band. Does any light get through this new slit or not ? As detected by a wide-angle detector after the new slit, or by the experimenter peering through it... /Bjorn > Hi! > [quoted text clipped - 8 lines] > > /Bjorn IIRC you get a bright band not a dark band at the centre due to constructive interference, so a slit there will let plenty of light through. > Hi! > [quoted text clipped - 8 lines] > > /Bjorn In Young's interference experiment - which is what I think you're talking about - at the center you have a bright band (the two different paths are equal, thus you have constructive interference). If you cut a third slit, then the light wave coming thru that slit will either contructively interfere or destructively interfere with the other two depending on the path difference - n+1/2 wave lengths= destructive; n wave lengths, contructive and in between, in between. But even if it destructively interferes - it won't be completely dark as in the case of two slits. Think of the relative amplitudes of the three light waves. With two waves, in Young's experiment, the amplitudes are the same. I hope this addresses your question. > > Suppose you have the light-source and dual-slit and screen, setup so > > you get the classic interference pattern on a screen, and you make a [quoted text clipped - 8 lines] > talking about - at the center you have a bright band (the two > different paths are equal, thus you have constructive interference). Hi, Yes, sorry, I meant placing the slit at a site with destructive interference. The first node to the right or left of the center then, it does not matter where it is. Suppose the experiment is set up so you can get complete destructive interference at *some* point, and then open up that point for further spreading of the wave. Now you would say, of course the wave will just continue past the node, but there is a catch with this with regards to what exactly is defined as an event in the Feynman sum-over-histories approach, which is what I really wanted to get at. But one step at a time :) > But even if it destructively interferes - it won't be completely dark > as in the case of two slits. Think of the relative amplitudes of the > three light waves. With two waves, in Young's experiment, the > amplitudes are the same. I think you misundestood the experimental setup, maybe I wasn't clear: There is one light-source, two screens after each other, and one detector. The first screen is the usual dual-slit, or whatever other setup which creates a dark band on the second screen. Then you open the second screen at the dark band and put the detector behind it. /Bjorn > > > Suppose you have the light-source and dual-slit and screen, setup so > > > you get the classic interference pattern on a screen, and you make a [quoted text clipped - 42 lines] > > /Bjorn Ah,...I did misunderstand the experiment (or you weren't clear :-) - It doesn't matter) I would say that the fact that two waves destructively interfere at the slit on the second screen doesn't mean that they don't exist. They would continue and could be detected. I'm sure we've all seen interfering water waves on a pond cancel each other out at a particular point but continue on. > > There is one light-source, two screens after each other, and one > > detector. [quoted text clipped - 12 lines] > interfering water waves on a pond cancel each other out at a > particular point but continue on. Yes - I'm pretty sure this is the result of the experiment as well. But the second screen's dark band would signify an absence of interaction with electrons in the atoms there. When you remove it, what happens to the propagation ? Can the photon (or whatever particle used in the experiment) interact with virtual particles in this region of destructive interference ? If yes, what is the difference between the virtual interaction and the "real" interaction (which didn't occur in the dark band), and if no, how is this reflected in the calculation of the propagation of the wave. For example, Huygens principle (a dear subject of this newsgroup :) when applied to QED as recently discussed implies that the resulting wave and its direction arises due to the combined "virtual" interactions with the photons and atomic electrons in the medium. But in the second screen's slit, there are no interactions due to the interference and there should not be any "Huygens propagation". I must have overlooked something. /Bjorn > > > > Suppose you have the light-source and dual-slit and screen, setup so > > > > you get the classic interference pattern on a screen, and you make a [quoted text clipped - 51 lines] > interfering water waves on a pond cancel each other out at a > particular point but continue on. If that were true. putting slits where the dark patches are would decrease the brightness of the light patches. That would be a pretty impressive macroscopic entanglement effect, if you could demonstrate it works. |
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