According to Huygens' principle, each wavefront is an infinite superposition of the secondary wavelets created by the previous wavefront. So if we take an infinite number of slits and light undergoes interference, I expect to see general illumination and not bands or fringes. What do you guys say? The width and distance between the slits is finite.

  • $\begingroup$ The question isn't clear. Are these slits still of finite width and separated by some nonzero distance? Why do you expect to see "general illumination" (which I guess means uniform brightness)? Do you know any equations that might express the brightness as a function of position? $\endgroup$ – Mike Feb 27 '17 at 11:49
  • $\begingroup$ @Mike Yes, slits are of finite width and separated by finite distance there are just infinite no of them, and yeah as I said that Huygens Rule says that next wavefront is an infinite superposition of secondary wavelets so when a wave passes through infinite slits all those diffracted wavefronts interfere with one another and form the next wavefront.Since when a single wavefront hits the screen it would be of uniform brightness I think. $\endgroup$ – Sarthak Sharma Feb 27 '17 at 11:58
  • $\begingroup$ Well, first off, the amplitude of a single wavefront falls off with one over the distance to the slit, so that's clearly not uniform brightness. Second, even though the waves at the screen are a superposition of waves from all of the slits, the sum of an infinite number of waves isn't just 1; it can vary from point to point. So the question is: do you know how to sum those waves together? Do you know how to do it with just two slits? $\endgroup$ – Mike Feb 27 '17 at 12:12
  • $\begingroup$ @Mike Sorry for saying uniform brightness, I wanted to convey that it would be a general illumination and with two slits we can add wavefronts using phasors, and we get an interference pattern. $\endgroup$ – Sarthak Sharma Feb 27 '17 at 12:17
  • $\begingroup$ With three slits, you can also add wavefronts using phasors. Or any number of slits. Why would that be different? $\endgroup$ – Mike Feb 27 '17 at 12:39

I think the answer is not the uniform brightness. In fact you can compute the N slit experiment with finite width for the slits. For a supposed 0-width, you'll get an interference pattern that is getting sharper and sharper :

enter image description here

Then, if you suppose a non zero width, you'll get an enveloppe on that pattern :

Case of 2 wide slits

(This is the case of 2 wide slits)

Finally, if N goes to infinity, you'll get a "Dirac comb" topped by the pattern due to the width of the slits. Still, in reality, you cannot make light go through this infinite number of slips (because of the infinite amount of energy it would require to produce such light), so in the end I guess the pattern would be as sharp as you manage to light up more slits.

  • $\begingroup$ Thanks for some details, but see we are using interference right? In this experiment slits are really narrow such that only 1 or 2 wavefronts are able to diffracted through 1 slit, but the pattern you have shown represents self interference(when slit is wide), so that is not an interference pattern but a self-interference(diffraction) $\endgroup$ – Sarthak Sharma Feb 27 '17 at 13:03
  • $\begingroup$ See according to Huygens principle you can trace the path of a wavefront by considering each point on a wavefront as source of secondary wavelets, and then you can interfere all those wavelets ( infinitely many) and get the next wavefront, now shouldn't the same thing happen if you did the infinite slit experiment, and since there is only a single wavefront obtained why the pattern gets more and more shaper? $\endgroup$ – Sarthak Sharma Feb 27 '17 at 13:16
  • $\begingroup$ Actually, I showed both case. And in the real world the width of the slits are never 0. So you always get that topping pattern. Interference and diffraction are basically the same phenomenon. Why would you get a single wavefront ? If you consider a point on a screen at say distance D from your slits, each wave (for the slits) won't be arriving with the same phase. $\endgroup$ – Naptzer Feb 27 '17 at 13:39
  • $\begingroup$ Yeah Naptzer Interference occurs when two diffracted wavefronts mix up with one another, see I was studying how a wavefront propagates , so Huygens said that if you have a wavefront and if you consider each point on the wavefront as a source of secondary wavelets, since there are infinite points on a wavefront so they will create infinite secondary wavelets and when those wavelets combine, they do so in such a way that the next wavefront is generated, shouldn't the same thing happen if you take the ideal case where width of slit approaches 0 and no of slits approaches infinite? $\endgroup$ – Sarthak Sharma Feb 27 '17 at 14:03
  • $\begingroup$ Yes for sure ! But that doesn't mean that the front will be uniform (intensity). If you mean (correct me if I'm wrong) that since there is infinity both in the number of sub-wavelets that generates the front and in the number of slits from which the first fronts start, we should have the same result (uniform intensity) then you are wrong : the slits are infinite in number but you can count them (there are as much as there are integers, since you said there is a finite non zero distance between slits), whereas there are much more sub-wavelets (as much as the number of real numbers). $\endgroup$ – Naptzer Feb 27 '17 at 14:40

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