This question is inspired by the recent "double-slit experiment in time" experiment that was popularized. See here and here. I have not looked into the original paper in detail, so my question may not be strictly related.

In any case, let me explain my understanding of the role of Fourier transforms in the Fraunhofer single- and double-slit setups. This part is not too relevant to the main question, so the reader can skip the next two bullet points. Note that the functions involved will not be precise (hence why I will use $\sim$ instead of $=$). Lastly to clarify, by "box function" I mean the function of the form $\text{rect}(x) = 1$ if $x\in [-\tfrac{1}{2}, \tfrac{1}{2}]$ and $0$ otherwise.

  • When we send light through a single-slit, the light comes out with a box-function amplitude profile $A(x) \sim \text{rect}(x)$. The Fourier transform of the box-function is the sinc-function, so in the far-field when the light reaches the screen, it will roughly have a sinc-function amplitude profile $A(x)\sim \text{sinc}(x)$ up to some rescaling of the input and output.
  • When we send light through a double-slit, the light comes out with two box distributions side-by-side: $A(x)\sim \text{rect}(x-D)+\text{rect}(x+D)$. The Fourier transform of this is of the form $A(x)\sim \text{sinc}(x)\cos(xD)$ up to some rescaling of the input and output. The additional cos-term is interference.

So far this makes sense. Now I am thinking about the following. Note that in the illustration, the refractive prism can be substituted by a diffraction grating. The point is to have anything that can split different wavelengths of light.

  • Suppose I create a light pulse with amplitude in the form $A(z, t) \sim \text{rect}(z-ct)$. Now when I send it through a refractive prism, I expect to the wavelengths of the light to split and get a wavelength distribution similar to the sinc-distribution as above. An illustration is given below (I'm only illustrating the middle fringe of the sinc function; also remember that when we take the intensity we must take the modulus-square of the amplitude).

enter image description here

  • Suppose now I create two light pulses that together have amplitude in the form $A(z, t) \sim \text{rect}(z-ct) + \text{rect}(z-ct-D)$. Now when I send it through a refractive prism, since the splitting of wavelengths that get projected onto the screen is a form of Fourier transform, by analogy of the double-slit setup, I expect the distribution of wavelengths to be roughly sinc*cos-distribution. An illustration is given below (contrast this to the previous image; again remember that when we take the intensity we must take the modulus-square of the amplitude). The cos-term is interference in this case.

enter image description here


My question is, if my description of the refraction experiments is correct, then how can the two distinct, separate light pulses possibly interact with one another? Given that there is a gap of length $D-1$ between them and assuming $D-1 > 0$, there is no point at which there is any interference of any kind possible.

Despite this, the reasoning for why there should be interference is that the refractive prism acts as a kind of Fourier decomposer of wavelengths, and two separate box light pulses clearly have a different spectrum than a single box light pulse. So which line of reasoning here is fallacious? Can someone clarify this?

More to the point, I am wondering how to gain the right intuition for the specific setup I sketched. What exactly would happen visually as I imagine the two pulses enter and exit the refractive prism?


1 Answer 1


From a purely mathematical point of view, if I take a function consisting of two distinct pulses, compute its spectrum and then filter out a portion of this spectrum, the inverse Fourier transform of this portion would not in general reproduce the original two pulses. Depending on the portion that is selected, the inverse Fourier transform of the portion would produce two blurry pulses that overlap. So, even though the original to pulses are distinct as a function of time, the individual single frequency signals that make up the function of time, are not. Therefore, when we consider the function in terms of its spectrum, we'll see interference.

  • $\begingroup$ I think this is a good answer, and I certainly agree with it. I suppose what I am asking for is, what would we see if we simulated the two light pulses going through the prism? I don't have the intuition wrt that part. $\endgroup$ Commented May 2, 2023 at 18:22
  • $\begingroup$ How do you want to simulate it? $\endgroup$ Commented May 3, 2023 at 3:10

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