# Non resonant modes in cavity

I have a simple question. Suppose we have a laser with a cavity such that the frequency of the light that the laser emits does not match with any longitudinal modes allowed by the cavity.

In practice it would mean that the line width of the emitted light would be very small (a few Hz). The cavity must be small enough and such that the allowed mode does not match with the frequency of the light.

In addition, let's consider that there is no damping mechanism to absorb the light. Theoretically the mode can not build up and the waves emitted at the frequency do not interfere constructively.

Then my question is : In that case, what happens to the energy of the waves since the wave is supposed to interfere destructively ??

P.S.: the question could also stand for acoustic waves.

## 1 Answer

The energy is reflected from the cavity.

In general, an optical cavity acts as a variable transmissivity mirror for a light source with very narrow linewidth (a laser). As you change the length, the cavity can go between highly transmissive and highly reflective. The specifics of how tranmissive and how reflective it can be depend on the reflectivity of the mirrors which make up the cavity.

In the particular case of a two mirror cavity with equal reflectivity mirrors, the transmissivity is shown below for two different values of finesse. Finesse is a measure of how narrow the transmission line of the cavity is (as can be seen below). For the case of a two mirror cavity with equal reflectivity mirrors, it is given simply by $$F=\frac{4R}{(1-R)^2}=\frac{\delta\lambda}{\Delta\lambda},$$ where $\delta\lambda$ is the FWHM of the transmission peak and $\Delta\lambda$ is the free spectral range (as shown below). Since laser line mirrors can have reflectivities as high as 0.999995, it is possible to have cavities with a Finesse in the range of 100,000.

• Great it makes perfect sense to me now! – Ronan Tarik Drevon Mar 19 '15 at 13:59
• I have the feeling there is a misprint though the finesse should be : $F=\frac{4R}{(1-R)^2}$ – Ronan Tarik Drevon Mar 19 '15 at 14:32
• @RonanTarikDrevon You are correct. I fixed it. – Chris Mueller Mar 19 '15 at 14:34