My professor told us that the fringes formed by interference will always be countable or finite. But I am just wondering, if, it is possible to have infinite fringes? Is there such a case? I am thinking that if the slit is placed too far from the screen, an infinite number of bright fringes will form. Or, in two-slit interference, suppose the screen is placed very far from the slits, will there be an infinite number of bright fringes formed on the screen? Your take on this?

  • 2
    $\begingroup$ Just a heads-up: countable does not mean finite, per se. $\endgroup$
    – Danu
    Dec 5, 2013 at 9:01
  • $\begingroup$ @Danu I guess he means finite or infinite countable. $\endgroup$
    – jinawee
    Dec 5, 2013 at 9:33
  • $\begingroup$ I thought the two are interchangeable. Anyhow, thanks for telling that there is a distinction between the two. I'll keep that in mind. Thanks, @Danu. $\endgroup$
    – ellekaie
    Dec 6, 2013 at 1:38

1 Answer 1


Strictly according to Double Slit experiment calculations, yes infinite fringes are possible if (1) the slits are infinitesimal and placed infinitely apart, but actually, the fringes will be very close to each other so your eyes won't be able to differentiate and secondly each of the fringe will have almost $0$ intensity.So for obtaining infinite fringes you also need to have an infinite screen.

The above results come from the fact that for a fringe: $\frac{dsin\theta}{\lambda}=n$ Now for $n$ to be grow without bound, $d$ should be grow without bound.

Also replying on your suggestion that a screen placed infinitely far from the sources, you seem to assume that the distance between the fringes will get small, but this isn't the case as the distance between two consecutive fringes is $\frac{\lambda D}{d}$ so if you get $D \rightarrow \infty$ the fringes also get infinitely apart, thus even if you will have an infinite screen length, you will only be able to say a finite number of fringes on the screen as $\frac{\infty}{\infty}$ will be finite in this case as both are dependent on exponent $1$ of length hence are of same order, here the ratio of infinities is taken to denote number of fringes on the infinite screen.

Intuitively, you may think that the $sin\theta$ of the maxima gets smaller, as the screen is getting farther away but what also is happening is that $n$ and $\lambda$ are finite so to get to second maxima, you have to travel an infinite distance on the screen to change$\theta$ by an appreciable amount.

EDIT : It actually if you see carefully is a case of similar triangles. If the distance from the source gets infinity, then the fringe width gets infinity, if you draw out the case of two screens such that one is behind other, you'll see because for fixed $n,d,\lambda$ $\rightarrow$ $\sin\theta$ is constant.


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