# Destructive and interference waves with different waves

Yesterday I posted a question and, between the comments, I noted this:

I wrote: ok, but for example can a blue laser and a violet laser interfere? I think that there isn't much different between their wavelenghts

The answer: Yes, in theory, there can be an interference pattern between them, but this is highly theoretical. Keep in mind that the coherence length of a common laser pointer isn't usually much more than 10cm and the wavelengths in that vary by only about 10 to 20nm. If you mix blue and violet, you probably wont see any interference patterns more than a millimeter or so from where the waves mixed. As I said, it's possible but extremely difficult

I wrote: Blue laser and violet laser encounter only in one point for cycle, so we don't notice the interference. But if I take two blue laser, I'll see the interference because the waves concide. Is it right?

The answer: If I interpreted you correctly, then yes, that is correct

So I image that the bold text is something like this: The waves, which have different wavelenghts because they are different, encounter only in two points in the image, so there is interference but it isn't very visible.

But now I'm thinking that also the destructive waves, which have the same wavelenght because they are the same waves,encounter in only two points: And this waves interfere a lot between them, in fact they annul. Why don't the first two waves make a visible interference? Why don't they annul or interfere? Does this mean that the bold text isn't right? Is it due to the different wavelenghts?

• Short answer: add the y-values at every x-position to get the resulting wave. In the second picture, it always adds to zero; you get total destructive interference. In the first picture (let's make it easier and make amplitudes the same), you get constructive and destructive interference at different points. For short wavelengths, it'd be hard to tell that there is any interference at all without extremely precise (and possibly non-existent) equipment.
– Jim
Mar 29, 2017 at 12:46

A key feature of waves is that they move in both space and time. The reason destructive interference is so interesting is that, at certain points in space, the equation for amplitude is 0, independent of time. For those points of destructive interference, it doesn't matter what phase one beam is in, the other beam is always exactly 180 degrees out of phase with it, perfectly canceling it out.

If you like math, remember that the 1-d wave equation is $A(x,t)=A_{max}\sin(\frac{\lambda}{2\pi} x + \omega t)$ Destructive interference occurs when you have two waves with amplitudes $A_1$ and $A_2$ and different path lengths $x_1$ and $x_2$ such that the sum of their amplitudes is $$A_{total}(x_1,x_2,t)=A_1\sin(\frac{\lambda}{2\pi} x_1 + \omega t) + A_2\sin(\frac{\lambda}{2\pi} x_2 + \omega t)$$

Permit me some rearranging which makes the math more straightforward, I'll express the second wave as the sum of $A_1$ and $A_2-A_1$. This is just a mathematical rearranging which makes the destructive interference more visible in the equations $$A_{total}(x_1,x_2,t)=A_1[\sin(\frac{\lambda}{2\pi} x_1 + \omega t) + \sin(\frac{\lambda}{2\pi} x_2 + \omega t)] + (A_2-A_1)\sin(\frac{\lambda}{2\pi} x_2 + \omega t)$$

By notating it this way, we can see some common factors. when $x1-x2$ is a multiple of $\frac{\lambda}{2}$ (i.e. there's a half-wavelength difference in their path lengths), then we see the two sine terms are always exactly $\frac{\pi}{2}$ out of phase or exactly in phase. This is true regardless of what $t$ is. This means that at these points, you have either maximally destructive or maximally constructive interference.

However, consider if the frequencies are different. If the frequencies are different, then you no longer have that super-easy cancelation. The sine terms will not always be perfectly out of phase with each other at all times.

There is a highly related concept called "beats," which occur when two signals which are close in frequency interfere in time. They interfere in a way that's similar to the spatial interference you are looking at, but they do it in time. It's most noticeable in sound waves. If you have two frequencies which are close together, say 5000Hz and 5005Hz, you will hear them pulsing at the difference between those frequencies (in this case, 5 Hertz).

In your example, blue and violet do interfere this way, creating beats. However, those beats are in the 10-20THz region. This is way too fast to see pulsing directly, but far too slow for our human eyes to detect it as a photon... That happens to be in the microwave region!

• I didn't understand your explanation about the 1-d wave equation. Can you tell me it with other words? Mar 25, 2017 at 22:23
• If you have one frequency of coherent light, you get interference patterns because you get nulls where one light wave cancels the other out at all points in time. If you have two frequencies, there is no point in space which always has 0 energy on it because the two wave are not in phase. Mar 25, 2017 at 22:45
• One thing that might help is to remember that what you've drawn is just the wave at one moment in time. The wave moves as time progresses. Mar 25, 2017 at 22:46
• Well, so if the frequencies are very close, isn't there a noticeable interference patterns? Mar 27, 2017 at 20:57
• In theory, yes you could see beats appear if you got the frequencies close. Of course, given that the light frequencies you are looking at are roughly 600THz, and the human eye can just barely see flashing patterns of 24 Hertz (if you have very good eyes), if you had two lasers tuned within 0.000000000004% of each other's frequencies, you'd just barely make out the beats. Any less perfectly tuned than that and the beats would be too fast to notice. Mar 27, 2017 at 20:59

I think the point you're missing here at the moment is that these waves are both sinusoidal waves, and their interference pattern will simply be the algebraic sum of both waves. On the second sketch you posted, it's not that the two waves only meet up at two points during any given period and cancel the whole wave out. Instead, notice that at any point along the pink wave, there is a negative "copy" of it which resides on the yellow wave. As such, when we add both contributions up, the net result is zero. This happens for every point along the wave on your second drawing, and would be called destructive interference.

Your first sketch is obviously different. Here, it's not as obvious what the resulting curve will look like, but you can figure out what the sum of the two waves will be at any given point by adding each of their $y$ values. I took a moment and graphed two waves that are similar to yours on Desmos, and then in orange you can see what happens when you add both waves together. The resulting orange wave is the interference pattern. You can check it out here.

• all right, but if I try to interfere blue laser with violet laser, I don't see a visible interference, so the waves don't add like the first sketch, which represents the two lasers Mar 23, 2017 at 20:26
• When you say that you try it, do you mean you have two physical lasers and are combining them? What exactly are you doing?
– Germ
Mar 23, 2017 at 20:44
• I updated the question Mar 23, 2017 at 20:50
• Alright, well I'd imagine it might have something to do with the waves being so close together in frequency that you don't get much of an interference pattern. Looking on Wikipedia, it says that violet light is between 668-789 THz, and blue light is between 606-668 THz. As such, perhaps you aren't seeing a real interference pattern because they are close to each other.
– Germ
Mar 23, 2017 at 21:02
• Ok, but then is there an interference between radio waves and blue waves? From what you said, I only have to sum the waves and in this case their frequencies are very different Mar 23, 2017 at 21:07