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A How a laser resonator generates the standing wave?

  1. Mar 23, 2017 #1
    Hi, guys. I'm interrest in the question about how optical (laser) resonator can generates the standing wave? As i know, there is a condition that the resonator base (lenght between two mirrors) should be multiple to the lenght of the wave. (L=k*(lamda/2)). But in practice no one never measures the distance between the mirrors of the resonator when creating the laser system. So i would like to ask you, how actually laser resonator can generates the standing wave without the measuring the distance between two mirrors?
     
  2. jcsd
  3. Mar 23, 2017 #2

    tech99

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    The frequency will adjust itself slightly so that a whole number of wavelengths fit in the resonator.
     
  4. Mar 24, 2017 #3
    How can "frequency adjust itself"? How this process called?
     
  5. Mar 24, 2017 #4
    Ok, i specify the question. Can you write the condition in which the optical resonator generates the standing wave? According to the wikipedia "an optical cavity is an arrangement of mirrors that forms a standing wave". Is it mean that any pair of 2 or more mirrors (which forms different types of the optical resonators) gives the standing wave in the output beam?
     
  6. Mar 24, 2017 #5
    I almost realize the idea of generating the standing wave in optical resonator. The main concept is that the reflected and falling light beams must propagates towards each other. So the fabry-perot interferometer makes the standing wave, but ring resonator is don't make the SW. Am I right??
     
  7. Mar 24, 2017 #6
    I understand it this way. You don't need standing waves, but you need a certain wavelength for interaction at each end. This allows for multiple journeys back and forth which increases the chances of photons escaping at that wavelength.
    (Just noticed... my 1000th post!)
     
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