an optical cavity is "an arrangement of mirrors that forms a standing wave cavity resonator for light waves" (wikipedia).

The possible standing wave patterns for such structure are like these:

enter image description here

As you can see, the vertical black lines (which are the mirrors) are the nodes of the standing waves, since they force the wave to be 0 at those points.

Well, I have studied a similar situation for electromagnetic resonant cavities. In such devices, the mirrors were replaced by walls made of perfect electric conductors, and these faces were the nodes of the standing waves because they forced the tangential electric field to be 0 along them (which is the interface condition for perfect electric conductors). But in this case, the walls are generic mirrors, so I do not understand why they force the wave to be 0 along them.

So my question is: why do the mirrors force the wave to assume always 0 amplitude, i.e. mirrors are the nodes of the standing waves?


The reflected wave obtains a phase shift of $\pi$. If 100% of the light is reflected, the amplitude at the mirror vanishes, because the phase shift flips the sign, $e^{i \pi}=-1$.

Refering to the comments: The following graph shows the situation, where the mirror reflects only 50% of the incident light: enter image description here

The blue points are the superposition. Personally I would not describe this as a superposition of a standing wave and a propagating wave. Although this formulation is mathematically fine, it does not describe what I see in the graphic.

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  • $\begingroup$ Is this true also if the mirrors are partially reflecting surfaces (for instance this is what happens in a Fabry Perot Cavity)? $\endgroup$ – Kinka-Byo Jun 30 at 19:57
  • $\begingroup$ @Kinka-Byo Yes, that's also how you build a laser. The partially reflecting mirrors (with 99% reflectivity) allow only certain wavelenghts to constructively interfere and resonate inside the cavity $\endgroup$ – ErickShock Jun 30 at 20:10
  • $\begingroup$ @ErickShock but does it work only because 99% is almost a totally reflecting surface? What I do not understand is if the standing wave is due to the presence of a generic wall, or of a totally reflecting wall $\endgroup$ – Kinka-Byo Jun 30 at 20:16
  • $\begingroup$ @Kinka-Byo: The phase shift is true whenever light is reflected on a medium, which is optically denser than the medium is it initially propagating in. However, your images suggest that the electric field is equal to zero at the knots. To check whether or not this is the case for a wave which is reflected only partially, I invite you to do the calculation yourself. Simply add two counter propagating waves with different amplitudes and see what you obtain. $\endgroup$ – Semoi Jun 30 at 20:34
  • $\begingroup$ @Kinka-Byo if you just calculate for a single pass on a partially reflecting mirror the amplitude will not cancel exactly at the mirror. However, after many roundtrips in the cavity only the wavelengths that satisfy $\lambda_n = 2L/n$ will reinforce each other, while all the others will destructively interfere and vanish over many roundtrips $\endgroup$ – ErickShock Jun 30 at 20:46

Assuming that the wave patterns in your diagram are meant to represent electric field strength patterns, whoever drew the diagrams assumed that the mirror was a perfectly conducting one. Therefore they showed the nodes of the E field at the mirrors.

If the mirrors were dielectric boundaries, the E field nodes would be slightly beyond the mirror locations.

If you formed the mirrors from a multilayer dielectric coating, you could arrange to have essentially whatever phase shift you want from the reflection, and therefore put the E field nodes wherever you like relative to the first surface of the coating.

But in this case, the walls are generic mirrors, so I do not understand why they force the wave to be 0 along them.

Mirrors formed by a thin layer of metal (silver or gold, for example) on a glass plate are fairly common, and these will act very much like a perfectly conductive coating, producing nodes very close to the metal surface.

Even if the mirror is not a metallic one, if the cavity length is much longer than a wavelength (i.e. if you have a few cm long cavity, even) then assuming that field nodes occur at the mirror surfaces won't in many cases result in any errors that obscure the physics of the cavity behavior.

If you're manufacturing a glass etalon, for example, you don't, for example, calculate you need a cavity length of 4.2190000 mm to fit 10,000 wavelenths of 632.8 nm (vacuum wavelength), and then manufacture to that dimension. You make the cavity about 4.2 mm and then polish it while testing with a reference laser and stop polishing when you get high transmission. If it turns out you need the physical dimension to be 4.2188 mm to allow for phase shift at the mirrors, you'll never know the difference.

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  • $\begingroup$ So whatever screen forces the wave to be a standing wave, and the choice of its material determines the position of the nodes. Correct? $\endgroup$ – Kinka-Byo Jul 1 at 7:27
  • $\begingroup$ @Kinkya-Byo if the mirrors are multilayer coatings then there's a lot more free parameters than that. But if they're just a single interface, then yes. $\endgroup$ – The Photon Jul 1 at 13:44

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