Graduate How a laser resonator generates the standing wave?

Click For Summary
A laser resonator generates standing waves by ensuring that the distance between mirrors allows for a whole number of wavelengths to fit within the cavity, even if this distance isn't explicitly measured. The frequency of the light adjusts to meet this condition, allowing for constructive interference of the light waves. The arrangement of mirrors, such as in a Fabry-Pérot interferometer, facilitates the formation of standing waves, while a ring resonator does not. The key concept is that the light beams must propagate towards each other to create the standing wave effect. Understanding these principles is essential for effective laser system design.
Basov
Messages
4
Reaction score
0
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 length 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?
 
Physics news on Phys.org
Basov said:
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 length 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?
The frequency will adjust itself slightly so that a whole number of wavelengths fit in the resonator.
 
How can "frequency adjust itself"? How this process called?
 
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?
 
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??
 
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!)
 
  • Like
Likes berkeman
Time reversal invariant Hamiltonians must satisfy ##[H,\Theta]=0## where ##\Theta## is time reversal operator. However, in some texts (for example see Many-body Quantum Theory in Condensed Matter Physics an introduction, HENRIK BRUUS and KARSTEN FLENSBERG, Corrected version: 14 January 2016, section 7.1.4) the time reversal invariant condition is introduced as ##H=H^*##. How these two conditions are identical?

Similar threads

  • · Replies 1 ·
Replies
1
Views
1K
  • · Replies 18 ·
Replies
18
Views
2K
Replies
4
Views
2K
  • · Replies 6 ·
Replies
6
Views
3K
  • · Replies 36 ·
2
Replies
36
Views
3K
  • · Replies 16 ·
Replies
16
Views
3K
  • · Replies 8 ·
Replies
8
Views
2K
  • · Replies 14 ·
Replies
14
Views
2K
  • · Replies 12 ·
Replies
12
Views
2K
Replies
4
Views
3K