At the central maxima of Fraunhofer diffraction, all the secondary wavelets constructively add up (I am looking at a 2-D picture here with the plane of the slits and screen in-out). So why isn't that a point of infinite intensity?

In double-slit experiments, the amplitude at the central maxima is 2 times that of the waves so the intensity is 4 times, with 3 slits amplitude of the wave at central maxima is 3 times and intensity is 9 times. So, with infinite slits, wouldn't the amplitude and therefore the intensity also be infinite? Please give an answer within the scope of high school.

(not necessary for answering the question) I also checked by math. I found amplitude at any point on the screen in terms of the number of slits and phase difference between light from two adjacent slits (which is $\delta \phi =d\sin\theta$ where $d$ is the distance between slits and $\theta$ is the angle made by a line joining either of slits and the point with the horizontal). I got (using complex numbers for adding up the n sine waves) $$A_{r}=A \frac{\sin\left(\frac{N\delta \phi}{2}\right)}{\sin\left(\frac{\delta \phi}{2}\right)}$$

Now, taking the limit of $N$ to $\infty$ we get $$A_{r}= N A \frac{\sin\left(\frac{\pi a \sin\theta }{\lambda}\right)}{\frac{\pi a \sin\theta}{\lambda}}$$ which is infinity since $N$ is infinity. I also got the central intensity as $N A$.

Wikipedia (under slit of infinite depth) got the coefficient different but I can't really follow what they did (what is $xx'$?) and I don't know Fourier transforms and all that.

  • $\begingroup$ Have you heard of Zeno's paradoxes of motion? You might like to look them up. $\endgroup$ Commented Jan 5, 2022 at 22:58

1 Answer 1


When you start introducing infinity as though it was a number and/or something which can be realised in practice you come up with problems, one of which you have highlighted.

In the ideal system that you describe with an infinite number of perfectly align and equal width, equal separation slits then the Mathematics may well give you an infinite intensity.
Of course you would need to have a light source of infinite intensity as well.

What a diffraction grating (many slits) does is to split up light into its component colours and channel the light into very narrow beams which have a high intensity.

In practice there is a limit to what can be achieved as to the width of a grating and the number of slits it has.
One of the purposes of having lots of slits is to reduce the width of the (principal) maxima and at the same time make them brighter.
It is like squeezing the 2-slit interference fringes thus making them narrower but higher. If this is done then wavelength which are close together from a light source will produce fringes which are close together but are still distinguishable.
It was the ability to measure wavelengths very accurately which provided a lot of the experimental evidence which showed that ideas on which Quantum Mechanics is based were correct.

On a minor point, the intensity is proportional to the amplitude squared.

  • $\begingroup$ What if you consider a gap in an opaque screen through which the light passes through to fall on another screen. The huygens principle states that every point on a wave front acts as a point source emitting light. So what I did is to add up the light from the N sources and then make N infinity to include all points. In that case, won't N actually be a limit to infinity and not a large number? $\endgroup$
    – Skawang
    Commented Oct 13, 2016 at 13:29
  • $\begingroup$ So you have a gap which you split into an infinite number of infinitely thin (point) sources and that is not the same as splitting the slit into $N$ sources of width d/N where $d$ was the width of the original slit and then allowing $N$ to become very large ie tend to infinity. You see that as you increase the number of "almost point" sources you reduce the amplitude of the point sources. $\endgroup$
    – Farcher
    Commented Oct 13, 2016 at 14:06
  • $\begingroup$ Why is the amplitude reduced? $\endgroup$
    – Skawang
    Commented Oct 13, 2016 at 14:22
  • $\begingroup$ If you have a finite amount of energy per second passing through a slit then if you consider it as two slits of half the width of the original then half the energy per second passes through each of the slits. $\endgroup$
    – Farcher
    Commented Oct 13, 2016 at 19:32
  • $\begingroup$ I've accepted your answer since your answer is sufficient for the question I asked. But on a side note, What is the energy of light in classical optics ( I am talking about before Planck's theory of quantisation of energy)? Since it is a 3-d wave, shouldn't the expression be different from that of a one dimensional wave like one on a string? Do we take the energy emitted by a wavefront or something? $\endgroup$
    – Skawang
    Commented Oct 14, 2016 at 14:20

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