# Why is the double slit interference intensity modulated by single slit intensity?

I am a high school student. I understand that the pattern of the intensity for double-slit interference is packaged within the intensity pattern for single slit interference because the slit width, in reality, it is not negligible. But I don't have an intuitive understanding of why that is the case?

Any help much appreciated :), just trying to grasp the concept.

• Each slit projects a single-slit pattern. When the two patterns overlap, they form additional cancellation spots within the single-slit pattern. Hence the double-slit pattern looks the way it does. Commented Apr 18, 2020 at 11:28
• Commented Apr 18, 2020 at 12:20
• The answer should also take into account that individual edges also produce fringes. So, if the double slit result is the overlay not only of the single slits, but finally the overlay from all edges. Commented Apr 19, 2020 at 8:08

This comes about from concepts relating to Fourier transforms. The interference intensity pattern is the Fourier transform of the aperture function. A single slit can be modelled as a rectangular ("top hat") function, and a double slit configuration is the convolution of a rectangular function and two delta functions. The convolution theorem tells us that in the Fourier domain, a convolution is turned into a multiplication, so the original pattern (FT of rectangular function) modulates the FT of the two deltas.

Sadly, I don't think any of this is intuitive. It is mostly due to the weird way that waves behave.

As Fehértói-Nagy Lili pointed out, the observed diffraction pattern is a convolution between two functions:

1. the diffraction pattern of a single aperture (e.g. rectangular or circular) and
2. a function which accounts for the fact that we have two of these apertures.

Let's try to understand and visualise this effect. However, since the diffraction patterns of these apertures are complicates and in 2D, we will use simpler functions for the illustration: So let's suppose the single aperture is described by a "triangular" function (blue), and the fact that we have to apertures is describes by the "step" function (red):

(Just to make sure: There is no connection between these functions and the actual function. These functions are merely used for illustration.)

The effect of the convolution is to slide the blue triangle across the red step function. For each point we multiply these functions and calculate the area underneath this product. So let's do that:

In green I plot the product. The red dot indicates the area underneath the product curve (I use the average of all the green curves in the pictures). I keep the red for previous positions of the triangle in the image, so that we can see a trend.

These images show that the triangular function acts as a weighted average of the "step" function. The triangular shape smoothes the steps and generates a fuzzy / blurred result. However, the important point (regarding your question) is that if we do not take a triangular but a different function for "smoothing" we obtain a different result. This is exactly why the double slit experiment for a rectangular and circular aperture generate different diffraction patterns.

Here is a very simple-minded answer. As you know, for a single slit the central maximum fades off to a (local) zero within an angle, $$\theta_1$$ say, of the straight-through direction, and there are other zeros at larger angles, $$\theta_m$$. With two such slits, side by side, there will be no light emerging at angles $$\theta_m$$ to interfere, even if there is a path difference of $$n \lambda$$ between light from the centres of the slits travelling at these angles. In this way the single slit pattern modulates the two slit pattern.