What happens when light of two different colours interfere destructively?

Question

In one of my lab courses, we are using an interferometer. I can understand how the interference works when it comes to monochromatic light, but for white light, I am unable to picture how the interference should look and why.

My main problem is with different phases of the wavelengths. In the case of monochromatic light, they interfere to cause dark spots. Though, with white light, this interference seems to also cause different colours to become more visible than others which I don't fully understand the cause of.

For my problem, I will be restricting the wavelengths to that of red, green and blue. Suppose two of the colours are in phase when they reach your eye, but the other in out of phase. What do you see in this case? There should be destructive interference in some way, but I don't know what colours would cancel.

And for the case where there is not restriction on the wavelengths, how do you go from different amplitudes of different wavelengths at a certain point to which colour you will see?

Answer 1

As a general rule, light sources of two different frequencies cannot interfere, either constructively and destructively.

Destructive interference happens when a monochromatic wave $E_1(t)=E_0\cos(\omega t)$ is superposed with another wave at the same frequency, $\pi$ radians out of phase, $E_2(t)=E_0\cos(\omega t+\pi)$, in which case the total wave completely cancels: \begin{align} E(t) & = E_1(t)+E_2(t) \\ & = E_0(\cos(\omega t)-\cos(\omega t)) \\ & \equiv 0. \end{align}

On the other hand, if the other wave is at a different frequency, say, $E_2(t)=E_0\cos(2\omega t+\pi)$, then the two waves will not cancel out, even if they start off giving $E(0)=0$:

In other words, waves of different frequencies cannot remain in phase for any significant length of time, and on average they go in and out of phase constantly, so that the intensity is just the sum of the intensities of the two waves.

Answer 2

The title of your question is not quite the same as the physics described in the question itself, and this why existing answers (at the time of writing) have not explained what is going on.

I will briefly describe observations with two different types of interferometer: Youngs slits and the Michelson interferometer.

Young's slits does not have to be called an interferometer, but it is a useful starting-point. If you have monochromatic light for Young's slits then the interference pattern is a sequence of bright and dark stripes. If the observation plane is at distance $L$ from the slits and the angles are small then adjacent maxima are separated on the observation plane by $$ s = L \frac{\lambda}{d} $$ where $d$ is the separation of the slits.

If you have two different wavelengths then you get two overlapping interference patterns, with different $s$. For example, for blue light at $\lambda = 460\,$nm with $L=1\,$m and $d=1\,$mm we get $s = 0.46\,$mm and for red light at $\lambda = 690\,$nm we get $s = 0.69\,$mm. In this case, at a distance of $0.69\,$mm from the central maximum there is a maximum of red light and a minimum of blue light (because $0.69 = 1.5 \times 0.46$ so we are at a minimum of the blue interference pattern). Also, at a distance of $0.345\,$mm from the centre there is a minimum of red light and the blue light is quite strong (not maximum, but not far off its peak at $0.46\,$mm). Hence the overall appearance in this case is a central bright fringe which looks magenta (blue plus red), then a blue stripe, then a red stripe, and further out these colours repeat.

If you have a continuous spread of wavelengths then similar things happen. For white light the central fringe is white, and then you get some coloured fringes ranging from blue to red-ish, and then all the patterns overlap and it looks near-white further out.

In all the above notice that each wavelength interferes with itself, but not with the other wavelengths. When the maximum brightness of one wavelength falls at the same location as the minimum brightness of another wavelength this is not called interference; it is just adding up the intensities of the various colours at the given place. In fact on a very fast timescale each wavelength can interfere with another, but this interference effect oscillates at the difference frequency so it is much too rapid for our eyes to detect.

Finally, a similar description to the above also applies to the Michelson interferometer, only now instead of locations on an observation screen we are talking about the intensity on a detector at a fixed position at the output, as a function of the position of one of the mirrors.

Answer 3

Coherence length

The key concept here is the coherence time (or coherence length) of the incoming light. This is the time over which the signal "looks the same" (the maximal shift for which the signal has a strong correlation with its shifted self).

Monochromatic light has infinitely long coherence length : you can shift it by as many periods as you wish, it's still the same sine signal, and therefore as you noted, it can interfere with itself. (Note that the exact position of constructive and destructive interferences is determined by where the path difference is an integer/half-integer number of wavelengths : it depends on the wavelength itself.)

Conversely, white light has a coherence length not much larger than its typical wavelength : you can do interference with it, as long as the path difference is less than or around 1 wavelength.

The coherence length is inversely proportional to the bandwidth of the emitted spectrum. This is easily understood if you remember that the Fourier transform of a gaussian of width $α$ is a gaussian of width $1/α$. Therefore, a light source with a wavelength $λ±δλ$ will have a coherence length of $λ²/δλ$.

Some lasers have coherence lengths of tens of kilometers. Typical cheap laser diodes will be of the order of a few millimeter. White light sources filtered with narrow band-pass filters will be coherent over a few tens of micrometers at best.

So it is entirely possible to do interference with white light, with a few caveats :

• you have to adjust the path difference to within ~ 1 wavelength (i.e. 1µm). That's usually extremely inconvenient (read: nervous breakdown…), unless you use a common-path interferometer, where the path difference is ~0 by construction.
• you will see a very limited number of fringes (typically, one…),
• it's better if you filter the light, to increase coherence length. Note that taking a picture of the interferogram with a color camera is equivalent to taking pictures of 3 interferograms filtered with red, blue and green filters ; each of which has about 100nm bandwidth which allows a few fringes. Wikipedia : white light interferometry has a nice example. Note that the fringes will be at different distances for the different colors ; summing them is equivalent to removing the filters, decreasing coherence length.

This has been done many times, most notably by Michelson at mount Wilson to measure the diameter of stars.

Beatings

All the above refers to a static (time-average) observation of interference pattern, of a time-delayed version of a signal with itself.

However, when you consider superposition of 2 different signals, even if each of them is monochromatic, the story is fundamentally different. What you get is a time-varying signal whose time-average (over long enough times) is zero.

Consider superposition of 2 monochromatic waves. The addition will be a wave with the average frequency, modulated by the half-difference of the frequencies: $$sin(ω₁t)+sin(ω₂t)=2 sin((ω₁+ω₂)t/2) cos((ω₁-ω₂)t/2)$$ Considering the huge frequencies of visible light, the beating will be too fast to be noticed unless, for a measurement bandwidth Δω :

• the frequencies are extremely close ($ω₁-ω₂<<Δω$),

• both sources are indeed monochromatic (bandwidth $<< Δω$, which pulls in the above discussion of coherence length)

As some lasers have bandwidth in the 10s of kHz and photodetectors can go well into GHz bandwidth, that's actually possible (and called optical heterodyne detection). Note that it's the exact equivalent of some radio recievers, but in the optical frequency range…