# Intuition behind the “rect” function in Fraunhofer double-slit analysis

This question is about the intuition on one end of the Fraunhofer double-slit diffraction pattern. It is not about the Fourier transform that connects a pair of rectangles to their sinc function transform, which is clear.

The square of a carefully chosen sinc function appears to model the intensity that appears when the light passes through two slits and reaches a wall or plate. The pair of square waves at the slits are what bother me.

I guess that as the wave front hits the double slit all but two "rectangular" bits are stopped. That is, the light passing through the grating is (?) instantaneously rectangular. But this raises two related questions.

(1) Without knowing in advance that the Fourier transform is an obvious candidate (because of the diffraction pattern), could we have guessed this somehow by assessing the situation at the slits? As in, "Oh, we have two rectangles here, and they are going to form a sinc pattern on yonder wall." Why? "Because that's the FT." But you can't just apply the FT to any pair of rectangles to solve any problem involving rectangles. There has to be some underlying justification.

(2) As the light enters/leaves the slits, at what point do we declare "rectangles?" The light begins to spread immediately.

So the question is about the intuition behind the rectangle end of the situation. I suppose there is a careful derivation of that end, but mostly what I find is like the linked page, a finished analysis with no insight into formulation of the problem at the slits, as in, "this is a good case for application of the FT and here is why."

A bit of intuition would help, or perhaps a reference.

The sum-of-rects function is the transmission of the aperture as a function of position. In the region where the slits are cut, the transmission is 100%. In the region where the slits aren't, the transmission is 0%. It doesn't vary from the center of the slit to the edge of the slit, or from the edge of the blocking material to a location far away from the slits. So you have a transmission function that looks like

$$T(x)= \begin{cases} 0, & x < x_1 \\ 1, & x_1 \le x < x_2 \\ 0, & x_2 \le x < x_3 \\ 1, & x_3 \le x < x_4 \\ 0, & x > x_4 \end{cases}$$

which is more compactly written as a sum of rect functions. We often consider the case of a plane wave passing through this aperture, so that the fields at the aperture are just the same function as the aperture transmission function.

The interference pattern observed on a screen in the far field is related to the Fourier transform of the field pattern at the source regardless of what that field pattern is. This is because, by Huygens' principle, we can find the field at the far field screen by summing the signals propagated to the screen from each point in the source plane, and this works out to be a kind of Fourier transform (convolution of an input function with a complex exponential kernel function).

Edit

To explain how the Fraunhofer diffreaction produces a pattern related to the Fourier transform, consider this geometry:

(Soure: Wikimedia user Epczaw)

Huygens' principle says that for propagation through the aperture, we can analyze it as if there are tiny point sources radiating omnidirectionally at each point on the aperture surface (actually, you could do this for any surface that the beam passes through, but by using the aperture surface, we get a useful result).

Since the aperture is illuminated by a plane wave from the left, the E field on the surface is given by

$$\vec{E}_a(x) = \vec{E}_0 T(x),$$

where $T(x)$ is the aperture function. $T$ doesn't have to have just 1 and 0 values, it could have fractional values to indicate partial transmission, or even complex values to indicate a phase variation on transmission, for example when the aperture has a variable-thickness sheet of glass (aka a lens) in it.

Now, if there's a screen at distance $z$, and we take the far-field approximation (so we don't have to be worried about $\theta$ being different for different parts of the aperture) and the small-angle approximation, using Huygens' principle the field on the part of the screen pointed to by angle $\theta$ will be

$$\vec{E}_s(\theta) = \int \vec{E}_a(x)e^{-i\beta(z+x\tan \theta)} {\rm d}x$$

(taking x = 0 near the top of the diagram and increasing downward). We can break this formula up and use the small-angle approximation to write it as

$$\vec{E}_s(\theta) = \vec{E}_0 e^{-i\beta z} \int T(x)e^{-i \beta x \theta} {\rm d}x$$

in which you should be able to see the Fourier transform: An input function $T(x)$ convolved with a complex exponential kernel function $e^{-i\beta x \theta}$.

• I have thought about this answer and it seems that Huygens' principle is what I lack intuition on. If you expand that portion of your answer I would accept it. In particular: "...and this works out to be a kind of fourier transform..." That is a bit vague and probably key. – daniel Dec 25 '17 at 21:04
• Sorry for delay, nice edit. – daniel Dec 30 '17 at 4:50
• @daniel, just corrected a sign error in the exponents. – The Photon Dec 30 '17 at 5:13