How sensitive is double-slit result to a fixed wavelength?

Question

I was watching a video on the double-slit experiment, and the lecturer (*) mentioned, in passing, that the frequency of the waves must be kept the same. (Still in the context of water waves at this point.) Which made me wonder how sensitive is the result (the striped interference pattern) to a fixed frequency?

Concretely, I am creating a 50 Hz wave, so one every 20 ms, and I see the expected interference pattern. Now if I change it so the time between waves is uniform random between say 19 and 21 ms would I still see basically the same pattern, a bit ragged round the edges? Or would that small change be enough to completely break it up? If I increased the range to 18 to 22 ms, would it be twice as ragged, and 17 to 23 ms is three times as ragged, and so on? Or would it be fairly recognizable as I gradually increase the variance but then suddenly fall apart at a critical point?

I also wondered about regular patterns. Say, alternating between 15 ms and 25 ms.

My guess on this one would be that waves 1, 3, 5, ... would continue to interfere, and waves 2, 4, 6, ... would also interfere, and I'd see a pattern that is the sum of a 15 ms pattern and a 25 ms pattern. Which would be both logical and quite surprising (i.e. that wave 2 didn't disrupt the pattern waves 1 and 3 were making together).

*: A certain Prof. Feynman, in the 6th Messenger Lecture. Specifically, from about 18:45 in https://www.feynmanlectures.caltech.edu/fml.html#6

I wiggle my finger up and down here, and I have a little piece of wood here, and ripples start out here, and then I’ve arranged in a tank to put boards in the way here so that I have these two holes, and then I have this so—called detector.

Then what I do with the detector—what the detector detects is how much the water is jiggling: for instance, I put a cork in the water and measure how it moves up and down. What I’m going to measure, in fact, is the energy of the agitation of the cork,which is exactly proportional to the energy carried by the waves.

Also, I forgot to say that this jiggling is made very regular and perfect, so that the waves are all of the same spacing from one another, and then I’ll describe what we get under those circumstances.

Answer 1

Short answer

Since you don't have a steady-state input wave anymore your pattern, probably, will look similar but change a little bit over time.

Long answer

The patterns of diffraction seen for the double-slit experiment are due to a incident wave that is monochromatic an plane. Mathematically, this is represented as

$$u_0(x, t) = A e^{i(\omega t - kx)}\, .$$

Because of the form of this wave you end up with a steady diffraction pattern.

What you are proposing is to replace this by a transient wave that looks something like this

$$u_0(x, t) = A e^{i(\omega(t) t - kx)} = A e^{i([\omega_0 + \omega_1(t)] t - kx)}\, ,$$

where $\omega_1(t)$ is the deviation you propose and has a magnitude that is small in comparison with $\omega_0$. So, I guess, that you could rewrite the problem as

$$\frac{u_0(\zeta, \tau)}{A} = e^{i([1 + \Omega(\tau)] \tau - \zeta)}\, ,$$

with $\Omega = \omega_1/\omega_0$, $\zeta = kx$, and $\tau = \omega_0 t$. Thus,

$$\frac{u_0(\zeta, \tau)}{A} = e^{i(\tau - \zeta)}e^{i\Omega(\tau)\tau} \, ,$$

and, since $\Omega \ll 1$, I think that we could use an asymptotic expansion and use Kirchoff's integral theorem to obtain the solution as the usual solution plus a small correction term.

Answer 2

Your water experiment with 2 sources instead on one (like in the DSE) can have a variety of interference patterns, the math is fairly straightforward as you predict. But the 2 source water experiment you propose would not be a good model for the traditional light based DSE. It would be like having a blue laser at one slit and a red laser at the other slit. You would just get 2 mixed "single slit diffraction" patterns on the screen.

Your experiment does however show an important difference between light and water waves. The water waves work in the water medium, you add some energy and get the standing waves going, all energy is in the kinetic up and down motion of the water. You can observe it without destroying it.

For light the EM field is the medium, and we cannot observe any wave motion at all, we only see a pattern emerge, and this occurs after the energy is absorbed in a camera or your eye.

More importantly while we assume the blue and red light can cross each other (or superimpose) in space we can never directly observe it. The blue and red photons as far as we are concerned are only absorbed or visible where they are seen .... photons truly never interfere, that's quantum optics.