Basic laser question

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Why is a laser monochromatic? I read somewhere that the reason is because of a Fabry-Perot cavity, and not necessarily because of stimulated emission, so that if you have an ordinary light bulb in such a cavity, it would produce monochromatic waves.

Can you just send the stimulated light through a light filter instead of such a cavity? Also, how high is the bandwidth for atomic transitions? It can't be that high, since you can still see fine lines in spectroscopy? So do you need a filter at all?

So if I'm understanding it right, the length of the Fabry-Perot cavity has to be half a multiple of the wavelength of the atomic transition? How accurate do you have to make this length, since it seems that the light should be close to monochromatic anyways just from the fine spectral lines?
 

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I'm not very expert in the field of lasers and I'm discussing from physics point of view.
The important point in the lasers are that all EM waves be in phase. stimulated emission is the main mechanism which amplifies the incoming wave by this way that when an electron is in its excited state and one EM comes, the electron emits a EM wave which is in phase with the incoming wave and so amplifies the amplitude of EM waves and this process continues.
Fabry-Perot cavity just filters out the EM waves and not amplify it so {I think} you can not make a laser just by a cavity without stimulated emission.
 
  • #3
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I'm not very expert in the field of lasers and I'm discussing from physics point of view.
The important point in the lasers are that all EM waves be in phase. stimulated emission is the main mechanism which amplifies the incoming wave by this way that when an electron is in its excited state and one EM comes, the electron emits a EM wave which is in phase with the incoming wave and so amplifies the amplitude of EM waves and this process continues.
Fabry-Perot cavity just filters out the EM waves and not amplify it so {I think} you can not make a laser just by a cavity without stimulated emission.
I too am not an expert.

I think it's okay to have a very weak laser, with the amplitude not so high.

How about if you put a very bright light bulb, or many light bulbs, in the cavity? The amplitudes would be very high, and inside a filtering cavity, it will come out monochromatic.

If the filtering cavity is not enough, you can always add an extra colored piece of glass for your filter.

I think if you make the opening where the light comes out very small, then the light will be collimated.

So if the light is monochromatic and collimated and intense, then is it a laser?

Am I missing something?

There is missing spatial coherence, which I think means a definite phase relationship along the cross-section of the beam. But really who cares about that, especially since the beam is so thin?

Actually, is it possible to even collimate a spatially incoherent beam? Shouldn't diffraction out of the laser opening ruin everything? Yes I think this is the case, but if this is true, one shouldn't be able to collimate the light coming from a light bulb, but I think you can.

So confused.
 
  • #4
Drakkith
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LASER stands for "Light Amplification By Stimulated Emission of Radiation". For something to be considered a laser it MUST be produced in this way. Since light bulbs do NOT produce their light in this way, they can never produce "Laser" light. Even if the light is the same as what the laser would put out.
 
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LASER stands for "Light Amplification By Stimulated Emission of Radiation". For something to be considered a laser it MUST be produced in this way. Since light bulbs do NOT produce their light in this way, they can never produce "Laser" light. Even if the light is the same as what the laser would put out.
I am agree with Drakkith. Also consider that when you filter the EM waves you get constructive interference just in very narrow and specific angles with very very low portion of incoming power.
 
  • #6
Redbelly98
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Just having mirrors and a light source are not enough. There are four key elements for having a laser, which are listed on page 1 of the book Lasers by Milonni and Eberly. The list is:

1. As Drakkith said, you must have a material that amplifies light by stimulated emission.
2. Mirrors, or some means of redirecting the amplified light back through the medium for repeated amplification.
3. A means of extracting the amplified light. This is usually accomplished with at least one partially transmitting mirror (for example, one that is 99% reflective, and 1% transmissive). Sometimes, but not often, it is accomplished by having a small hole in one mirror.
4. A means of replenishing the energy in the amplifying medium.

So if I'm understanding it right, the length of the Fabry-Perot cavity has to be half a multiple of the wavelength of the atomic transition?
Yes.

How accurate do you have to make this length, since it seems that the light should be close to monochromatic anyways just from the fine spectral lines?
Not accurate at all. It doesn't matter whether the cavity is 100000 times the wavelength, or 100000.5 times, or 100001, or ....

You be wondering, what if the cavity length is xxxx.25 times "the wavelength", i.e. not a multiple of 0.5·λ? The answer is that there are are actually a range of wavelengths that the amplifying medium can produce, and at least of them will "match up" with the cavity length.

Actually, is it possible to even collimate a spatially incoherent beam? Shouldn't diffraction out of the laser opening ruin everything? Yes I think this is the case, but if this is true, one shouldn't be able to collimate the light coming from a light bulb, but I think you can.
Calling a beam of light "collimated" is a bit of a misnomer. Because of diffraction, any beam can be reasonably collimated only over some limited distance. At a long enough distance away from the light source, there will be a noticeable spreading out of the beam due to diffraction, whether it is coherent or incoherent light. For coherent, monochromatic light, this distance is of the order of D2, where D is the diameter of the "collimated" beam.
 
  • #7
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You be wondering, what if the cavity length is xxxx.25 times "the wavelength", i.e. not a multiple of 0.5·λ? The answer is that there are are actually a range of wavelengths that the amplifying medium can produce, and at least of them will "match up" with the cavity length.
How can there be a range when the atomic transitions are discrete? Doesn't the hydrogen atom only have the Balmer series, which totals 4 lines? So wouldn't the cavity have to be half of one of the 4 lines?


Calling a beam of light "collimated" is a bit of a misnomer. Because of diffraction, any beam can be reasonably collimated only over some limited distance. At a long enough distance away from the light source, there will be a noticeable spreading out of the beam due to diffraction, whether it is coherent or incoherent light. For coherent, monochromatic light, this distance is of the order of D2, where D is the diameter of the "collimated" beam.
I don't understand how noncoherent light can spread out. I understand the D2 figure for coherent light, since a line of sources of length D that are in phase can cause interference. But a line of incoherent sources of length D cannot interfere by definition. Diffraction is caused by interference, but how can incoherent light interfere?

Then again, if you try to collimate incoherent light by having it pass through a slit in the screen, then if there is another screen behind the slit, that whole screen doesn't get lit up. Shouldn't the slit act as a source, and if all the phases are random at the slit, then the slit should be incoherent, and hence there can be no interference on the screen, so that the whole screen is lit up?
 
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There is a difference that you should consider between N_excited_atoms laser and N_emitting_atoms light bulb that first generates an in_phase EM wave with power of N^2 order of magnitude and second with power of N order of magnitude so there is a huge huge difference.
You can put a filter on your eye glass and look to outside but it doesn't mean that you make a laser since you filter out all of incoming waves other than a very very small piece of incoming waves' spectrum so you suppress the light instead of amplifying it.
 
  • #9
Drakkith
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I don't understand how noncoherent light can spread out. I understand the D2 figure for coherent light, since a line of sources of length D that are in phase can cause interference. But a line of incoherent sources of length D cannot interfere by definition. Diffraction is caused by interference, but how can incoherent light interfere?

Maybe Redbelly was talking about diffraction of the light with the generating/collimating medium?
 
  • #10
Claude Bile
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Why is a laser monochromatic? I read somewhere that the reason is because of a Fabry-Perot cavity, and not necessarily because of stimulated emission, so that if you have an ordinary light bulb in such a cavity, it would produce monochromatic waves.
No laser is truly monochromatic, all lasers experience spectral broadening to some degree. The basic parameter that determines the spectral width of a laser is the lifetime of the laser transition; longer lifetimes = smaller spectral widths as a direct consequence of Heisenberg's Uncertainty Principle. Other effects such as Doppler broadening can further increase the spectral width.

Can you just send the stimulated light through a light filter instead of such a cavity?
Absolutely, although you get much more loss using a filter.

Also, how high is the bandwidth for atomic transitions? It can't be that high, since you can still see fine lines in spectroscopy? So do you need a filter at all?
The bandwidth is determined by the lifetime of the transition. The lines you see in spectroscopy though are probably determined by the resolution of the instrument, unless the gratings are particularly good (i.e. > US$20,000 worth).

So if I'm understanding it right, the length of the Fabry-Perot cavity has to be half a multiple of the wavelength of the atomic transition?
Yes. In other words, the number of longitudinal modes is related to the transmission bandwidth (Finesse) of the FP cavity and the bandwidth of the laser transition.

How accurate do you have to make this length, since it seems that the light should be close to monochromatic anyways just from the fine spectral lines?
To get a laser output, you need sufficient overlap between the transmission spectra of the FP cavity and the output spectrum of the laser transition. It is impossible to prevent the laser output completely, what will generally happen is the highest transmission of the FP cavity over the transition bandwidth will dominate resulting in laser emission on that line.

Claude.
 
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  • #11
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Maybe Redbelly was talking about diffraction of the light with the generating/collimating medium?
Actually, I had a misconception. If you have an incoherent source such as a lightbulb, and a single slit that serves as a collimator, then the beam will be collimated. What happens is that you treat each point in your lightbulb as a single source, and this source causes the entire slit to be filled with sources a la Huygen. The sources in the slit created by this single point do have a phase relation, and because the wavelength of light is so small, you'll get a beam only in one direction. Then you take another point on the light bulb, and you'll get the same thing - that it creates sources lined up on the slit and you'll get a beam in only one direction. Since the two different points on the light bulb are not related in phase, the intensities would be additive.

So there's a huge difference between imaginary sources at an aperture caused by a real incoherent source, and putting a real incoherent source at the aperture. In the latter case you will get spreading of the light in all directions, and the former is really collimated.
 
  • #12
Redbelly98
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You be wondering, what if the cavity length is xxxx.25 times "the wavelength", i.e. not a multiple of 0.5·λ? The answer is that there are are actually a range of wavelengths that the amplifying medium can produce, and at least of them will "match up" with the cavity length.
How can there be a range when the atomic transitions are discrete? Doesn't the hydrogen atom only have the Balmer series, which totals 4 lines? So wouldn't the cavity have to be half of one of the 4 lines?
Atomic transitions still have a finite spread, generally between a MHz to several x 10 MHz. But the biggest cause of a spread in wavelengths, in a gas laser, is due to Doppler broadening. Gas atoms typically have speeds of several hundred m/s. Owing to the Doppler shift, that means the neon atoms in a HeNe laser can emit in a range of +/- 0.001 nm around the main wavelength. This is somewhat more than the wavelength spacing in a 30 cm long laser cavity, so it is guaranteed than some of the atoms can emit light that matches a cavity wavelength.

Calling a beam of light "collimated" is a bit of a misnomer. Because of diffraction, any beam can be reasonably collimated only over some limited distance. At a long enough distance away from the light source, there will be a noticeable spreading out of the beam due to diffraction, whether it is coherent or incoherent light. For coherent, monochromatic light, this distance is of the order of D2, where D is the diameter of the "collimated" beam.
RedX said:
I don't understand how noncoherent light can spread out. I understand the D2 figure for coherent light, since a line of sources of length D that are in phase can cause interference. But a line of incoherent sources of length D cannot interfere by definition. Diffraction is caused by interference, but how can incoherent light interfere?
For incoherent light there is a still a width, less than D, in which the light may be considered coherent. The light is coherent, or reasonably so, within that smaller width. Since this width is less than D, the diffraction is even more than what you would get for a coherent light.
 

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