What Is the Maximum Length of a Plasma Tube to Excite Only One Frequency?

In summary, the problem involves determining the largest value for the length of a plasma tube, which is used to create a laser, in order to produce a unique frequency of light. This frequency must not be a harmonic of any other frequencies in order to propagate in the tube. The wavelength of this unique frequency is determined by the resonant cavity formed by two mirrors at either end of the tube. The length of the tube can be calculated using the equation L = n*lambda/2, where n is an integer. The spectral width of the plasma tube also plays a role in determining the maximum value for L.
  • #1

Homework Statement

Alright this is from the MIT intro series book by AP French on Waves and Optics, Problem 6-10 (If you happen to have the book)

"A laser can be made by placing a plasma tube in an optical presonany caity formed by two highly reflecting faltmirrors, which act like rigid walls for light waves. The purpose of the plasma tube is produce light by exciting normal modes of the cavity."

"Supose that the plasma tube emits light centered at frequency 5E14 Hz and that it has a spectral width of +- 1E9 Hz. What is the largest value of the Length of the Tube (L) where only one frequency in the spectrum will be excitied in the plasma tube? Assume the speed of light to be c=3E8 m/s"

Any help on this would be great, I have no idea how to do it.

Homework Equations



The Attempt at a Solution

So, we know that this wavelength has to be completely unique, and so tiny that no other wave can propigate. This means that we need to find the largest wavelength whose fundamental isn't not a harmonic of any other frequencies, which would allow those frequencies to propigate in the tube.

I ahve no clue how to set this up mathematically
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  • #2
Here's my guess at a starting point:
Lasers work because of the resonant cavity. For a cavity with mirrors at either end, a resonant wavelength is one such that after making a round trip in the cavity, there's some multiple of 2pi phase shift. (That's where your [tex]\lambda = 2L/n [/tex] comes from, right? The n there is an integer, not refractive index) So you can also say that the cavity length L = n*lambda/2, and then consider how big n can be before another lambda is also satisfied. This is probably where the spectral width comes in.

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