I'm trying to understand the diffraction pattern for a single slit compared to a double slit. I understand that the wavelets passing through the slit diffract and interfere with each other producing a pattern. Why is the central maximum double the width, and why do the intensity of the maximum's diminish compared to that of the double slit pattern?

  • $\begingroup$ are you familiar with the concept of Fourier transforms (FT)? If yes, just do the math. The FT of the box function defining a single slit is the sinc² function and the double slit consists of two such functions that are shifted (Fourier shift theorem). Google found e.g. this site where they show the math $\endgroup$ – riddleculous Feb 16 '17 at 16:35
  • $\begingroup$ Hi John. Have you read the derivation of the diffraction pattern for a single slit? If not there must be a hundred easily googlable articles on the web. $\endgroup$ – John Rennie Feb 16 '17 at 16:37
  • $\begingroup$ This won't answer your question, but you may find it helpful. youtube.com/watch?v=9D8cPrEAGyc $\endgroup$ – Lambda Feb 16 '17 at 16:43

I boil everything down to Fourier transforms and I'll refer to the useful Wikipedia article several times. I won't care about pre-factors in the Fourier transforms for the sake of simplicity. You can look them up in the referred transforms.

First, consider the single slit with width a, it has the function rect(ax) and the Fourier transform sinc(x/a) (No. 201). Square it (Intensity) and you get the diffraction pattern.

Second, the double slit. It is basically a convolution (I denote it by ** and multiplication by *) of two delta functions δ(x-d/2)+δ(x+d/2) for the positions of the slits spaced by d and the box function rect(ax) giving the single slit(s): f(x)=(δ(x-d/2)+δ(x+d/2))**rect(ax). Now the FT of the two delta functions is either directly the reverse of No. 304 or you can compute it as the sum of two shifted delta functions (shift theorem No. 102 and delta function No. 302). So FT[(δ(x-d/2)+δ(x+d/2))]=cos(dx).

By the convolution theorem (No 108) we get that FT[g(x)**h(x)]=FT[g(x)]*FT[h(x)] and thus the resulting FT[f(x)]=cos(dx)*sinc(x/a). Since the pattern displays the intensity which is the square of the field we have to square everything and get the resulting interference pattern.

Now, computing the minima is just maths and property of the sinc² function resp. the cos². It turns out that the sinc²function has minima wheresin² has minima (i.e. where sin is zero, with π periodicity) except for zero because of the definition of sinc that is 1 there. This explains the double width of the central maximum. The `cos², however, has equally spaced minima (π periodicity again)

If you have infinitely small slits, you just get a uniform fourier transform and thus the double slit would be just the cos².


A single slit diffraction pattern results from diffraction at the edges of the slit, which creates a diffraction envelope at the target and generates destructive interference at the edges. Waves passing through the single slit are most likely to be in phase, and to augment one another, at the center of the slit. So the diffraction pattern is most intense at the center.

Huygens' Principle provides one way to understand why the intensity of diffraction maximums diminish on the target the farther they are from the edge of the geometric shadow of the slit.

Huygens postulated that each point on a wave front of light may be thought of as a point source of further wavelets that spread semi-circularly at the speed of the prior wave. So, if you consider the wave front passing through the slit as many point sources spanning the width of the slit, each interfering with its neighbors, augmentation and diminution patterns grow weaker away from the center of the slit. You can see a diagram of this and how to quantify it at page O.3 of this lab manual.

Double slit interference generates constructive interference. The two slits effectively become two coherent sources of light waves, which generate wave maxima and minima that can be in synch and can equally augment each other at multiple positions on the target, although the interference pattern will be modulated by a single-slit diffraction pattern.


Maybe you wish to think about the problem qualitatively. In that case, consider two scenarios: a single slit, of total length 2D, and double slit, with two very fine slits a distance D apart.

Now slowly broaden the very fine openings of the double slit, without moving the center of either. Generally speaking, broadening the source will ''smear out'' the regions of constructive (and destructive) interference, since the phase of adjacent ''rays'' will not be slightly mismatched in the observation plane.

You can continue opening both slits until they are a width D, in which case you have now an identical single slit case. Your intensity maxima and minima haven't changed, but they are more dilute.

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    $\begingroup$ ConfusinglyCuriousTheThird Nice answer. Thinking this to the end, one will get a diffraction pattern from the left edge of a single slit of the width "2D" and a pattern from the right edge. Indeed behind every sharp edge a intensity distribution takes place. So whenever we have to explain the intensity distribution of photons or electrons behind slits we have to be able it from diffraction behind single edges. Even for individuals ejected particles. Any explanation about interferences between waves from opposite edges is irrelevant $\endgroup$ – HolgerFiedler Feb 18 '17 at 16:45

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