Quantum Tunneling and Heat Distribution in Laser Cavities?

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SUMMARY

Replacing the half-transparent mirror in a laser cavity with a 100% reflective mirror is not feasible, as no such mirrors exist; all components incur losses. The intensity of the laser would increase until losses equal the gain, leading to heat generation rather than light output. In a typical laser setup, the laser medium operates through a three-state system, where excess energy converts to heat rather than producing additional laser light. The wavelength of the laser is primarily determined by atomic or molecular transitions, which are minimally affected by temperature changes.

PREREQUISITES
  • Understanding of laser cavity design and components
  • Knowledge of quantum mechanics, specifically quantum tunneling
  • Familiarity with laser medium and population inversion concepts
  • Basic principles of thermodynamics related to heat dissipation
NEXT STEPS
  • Research laser cavity designs and the role of mirrors in laser performance
  • Study quantum tunneling effects in optical systems
  • Explore the principles of population inversion in laser mediums
  • Investigate heat management techniques in high-power laser systems
USEFUL FOR

Physicists, optical engineers, and anyone involved in laser technology and design, particularly those interested in efficiency and thermal management in laser systems.

Daniel Petka
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What if I replaced the half transparent mirror in a laser cavity with a 100% reflective mirror?
-would all photons escape due to the quantum tunneling effect?
-would the mirrors melt?
Thanks :biggrin:
 
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There are no 100% reflective mirrors. Every component of the laser has some losses. The intensity would increase until the losses match the gain.
 
Ty
 
Usually, lasers are not very efficient at converting input power to laser light. Most of it gets converted to heat, which needs to be dissipated somehow. If you made the reflectivity very high, then the laser would just stop outputting light, and instead would generate a little more heat. It probably wouldn't destroy the mirrors, since the laser wasn't very efficient to begin with, and increasing the heat a little probably won't overtax the heat sinks.

In a laser, you have a laser medium with multiple excited states. For simplicity, let's consider a three state system. G is the ground state. A is the upper state, and B is the middle state. The power supply is connected to some arc source which excites the laser medium to the A and B states. A decays to B faster than B decays to G, so you get a population inversion between B and the G. So, you get lasing for the B to G transition. But the A to B transition photons are basically lost somewhere, and I guess they get converted to heat eventually. If you turn up the reflectivity, you'll reach some steady state field energy in the cavity, and all the extra power gets converted to heat. Since there is no laser power being carried away, the laser medium becomes more excited and you have more A to B transitions creating heat.
 
So cooling the laser should make the wavelength shorter, right?
 
Nope. The wavelength is determined by what atomic or molecular transitions are amplified.
 
Khashishi said:
Nope. The wavelength is determined by what atomic or molecular transitions are amplified.

Well then I suppose you should see some leds or lasers in liquid nitrogen
 
LEDs are not lasers.

The transition energies depend weakly on the temperature, the relative intensity of different transitions (if applicable) depends more on the temperature. That is relevant for LEDs and still notable for laser diodes, for other types of lasers it is a tiny effect. Here is an estimate of 0.3 nm/K for laser diodes.
 

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