# What if the optical length of a cavity is not a half integer scalar of the standing wave wavelength?

Pardon the confusing title, but I'll try and rephrase what I'm asking. Longitudinal modes in a wavelength of light require the optical length of the cavity to be such that the wavelengths need to form two nodes at either end of the cavity - essentially starting and ending at the same effective 0 point.

The equation is as follows:

$$L_o = \frac{m\lambda_m}{2}$$

Where $L_o$ is the optical length of the cavity, $m$ is an integer, and $\lambda_m$ is the standing wave wavelength.

My question has two parts:

• Do we measure optical length because the light travels slower through the gain medium as thus it would effectively travel a longer distance if we keep its speed constant, so we need to make that kind of adjustment by using the optical length?
• If this longitudinal mode condition is not met, where the lasing wavelength (I assume that's what the standing wave wavelength is supposed to be) does not form two nodes at either end of the cavity? Why is this bad? What happens? Does the light leak out somehow?

Wavelengths that do not fit the cavity are not amplified, and are not present in the laser output.

A laser cavity typically has a partially reflecting mirror on one end and a fully reflecting mirror on the other.

Suppose you feed a long train of waves into a cavity. The waves are reflected back and forth. If the wavelength does not match the cavity length, the phase will be different after a round trip. Suppose the phase change is $180^\circ$. Incoming waves will cancel the reflected wave. For other phase changes, cancellation is partial. But it gets more complete after multiple round trips.

Interference does not make energy disappear. It rearranges where energy is found. Cancellation in one place is always matched by reinforcement in another. For wavelengths that do not match the cavity length, very little energy is found in the cavity. Therefore, you would expect to find the net effect of all the reflections is to reflect the beam away from the cavity.

A laser cavity is filled with an active medium that amplifies light inside. If a photon in the cavity strikes an excited atom, the atom drops to a lower energy state and two photons with the same phase continue downstream. Light in the cavity is amplified. Light outside is not amplified.

This highlights the counter intuitive quantum mechanical nature of interference. A photon that travels down a cavity and destructively interferes with a photon coming back (which can be itself) is seldom found in the cavity. It is never found there for perfectly destructive interference.

To answer the first part of your question, you are right. Light travels slower in a medium, and so has a longer wavelength than in vacuum. You must use the wavelength in the medium.

You can read more about laser cavities and related topics in the RP-Photonics Encyclopedia. For example, check out the article on Resonator Modes.

• But different wavelengths that are allowed modes will destructively interfere as well. Yet they are useful to us through mode-locking. How are these two types of interferences different, one being not very light amplifying and one preferential? Does the wave interfere with itself if it is not nodelocked at either end of the cavity? – sangstar May 6 '17 at 18:13
• See rp-photonics.com/mode_locking.html?s=ak. It describes how a mode locked pulse train can be built out of a sum of frequencies that match the cavity length. Note that the index of refraction can be different for each frequency. – mmesser314 May 6 '17 at 19:05