Correlation between wavelength and beam’s width

Discussion in 'Physics & Math' started by Eagle9, Aug 28, 2011.

  1. Eagle9 Registered Senior Member

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    Imagine the laser beam with the wavelength of 0.1 μm (this is UV light, right?). I would like to know what the minimal diameter/width this beam can have. Can the beam be narrower than its wavelength-0.1 μm? Actually is there any correlation between wavelength and beam’s width?

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  3. James R Just this guy, you know? Staff Member

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    The beam can have no width. If it consists of photons in a line, then essentially each photon has no width, so the beam will have no width.

    There's no restriction on width due to wavelength.

    In practice, the width on a laser beam is determined by the characteristics of the source.
     
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  5. Trippy ALEA IACTA EST Staff Member

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    However, doesn't restricting the beamwidth increase dispersion (heisenbergs uncertainty principle and all that). It's my dim recollection from many moons ago that this is at the heart of the double slit experiment - by passing the beam through the first slit[sup]1[/sup] the uncertainty in it its location in one direction, and so increase the uncertainty in the momentum in that direction, resulting in the observed dispersion.

    I don' know if that would be applicable to a source or not though.

    1. First slit refers to the 'classic' setup using a non-coherent lightsource, but the same principle applies passing a LASER beam through a two slit setup.
     
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  7. James R Just this guy, you know? Staff Member

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    Collimating light into a beam usually involves an aperture of some kind and often at least one lens. Any time that light passes through an aperture of any kind it diffracts. All lenses are also apertures, because they have finite size.

    The two-slit experiment depends on diffraction through two apertures - namely the two slits.
     
  8. Trippy ALEA IACTA EST Staff Member

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    Beam divergence on Wiki
    Sounds like precisely what I was referring to. If we ignore focusing mechanisms, and concentrate on the LASing chamber, then the minimum minimum diameter of the beam is at the point that it leaves the LASing chamber.

    According to that, the smaller the device, the greater the divergence - which sounds exactly like the heisenberg uncertainty principle at work to me.
     
  9. AlexG Like nailing Jello to a tree Valued Senior Member

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    Wouldn't the minimum width of the beam be the amplitude of the wave? In a LASER, all the photons are collated, so the minimum width would be of a single photon/wave, and the minimum width would be from positive wave crest to negative wave crest.
     
  10. Trippy ALEA IACTA EST Staff Member

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    Maybe while they're in the LASing chamber, sure, but remember, they have to pass through an appeture (the end of the tube) in order to leave the LASing chamber, and so uncertainty applies (or it should anyway).
     
  11. AlexG Like nailing Jello to a tree Valued Senior Member

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    But we haven't established the width of the beam, just it's minimums. Uncertainty could make it much wider.
     
  12. James R Just this guy, you know? Staff Member

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    It leaves the cavity by passing through an aperture (usually a part-silvered mirror). The divergence of the beam after that is due to diffraction by the exit aperture.

    Yes, as in single-slit diffraction, the smaller the slit the greater the diffraction.

    Those crests and troughs in a light wave are peaks in the electric/magetic field magnitude of the wave. There isn't literally a wobbly wave that extends from the beamline out into space perpendicular to the beam direction.
     
  13. AlexG Like nailing Jello to a tree Valued Senior Member

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    Yes, but they have a physical dimension. That should establish the minimums.
     
  14. Trippy ALEA IACTA EST Staff Member

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    Implicit in all of this is my assumption that we were talking about minimum beamwidths.
     
  15. AlexG Like nailing Jello to a tree Valued Senior Member

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    It was in the op.
     
  16. Trippy ALEA IACTA EST Staff Member

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    Yes, and even after passing through focussing mechanisms, the infomation I have been able to track down, as limited as it may be (at least, for an ideal, gaussian beam), that there is a minimum beam width, the size of which is dependent on the quality of the beam, the angle of divergence (which presumably depends on the focal length of the lens assembly) and the wavelength of the light.
     
  17. Trippy ALEA IACTA EST Staff Member

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    According to the Wiki article I linked to, for a gaussian beam:

    \(\theta = \frac{\lambda}{\pi\omega}\)

    For a beam that has been collimated with a lens:
    \(\Theta=\frac{D_m}{f}\)

    and:

    \(\theta=\frac{\Theta}{2}\)

    Where
    \(\Theta\) = Angle of Dispersion.
    \(\omega\) = The diameter of the narrowest point of the beam.
    \(D_m\) = The diamater of the narrowest point of the beam before the Lense assembly
    \(f\) = The focal length of the lense assembly.

    In theory, at least, you should be able to rearrange those to give the diameter of smallest spot possible for any given LASER and lense assembly.

    I'll be honest and say I'm not even going to try and wrap my head around the N-slit interferometric equation.

    As far as how small the LASER itself can be, there are other practical constraints at play as well - it's constrained by the pumping requirements, and the physical properties of the LASing medium, among other things.
     
  18. DRZion Theoretical Experimentalist Valued Senior Member

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  19. Eagle9 Registered Senior Member

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    So, if I want to make some very little/narrow hole in cell/petal or elsewhere I can use an ordinary laser that will emit very narrow beam (red light, for example), right?

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  20. DRZion Theoretical Experimentalist Valued Senior Member

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    There is a limit to how small the beam can be, and it will always diverge, unless you consider some very radical technologies, for instance self-contained airy beams which coincidentally go around corners.

    wiki:
    However, if you want to make a very little/narrow hole you can use a micro-pipette, as is frequently done to measure voltage differences between inner cell and exterior.

    AFAIK lasers aren't great for drilling because of their divergence and also because they don't have the ability to pull a substance out of the hole that is being drilled, hence it has to be vaporized, making it somewhat energy inefficient.

    At the same time you should take into consideration that small holes can be made chemically, as is done with nano-filters. A nano filter is constructed using powerful, randomly emitted radiation which basically blows holes in a UV coat on a membrane. After this UV light eats neat little tubes in the material, but only in the exposed regions.

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  21. Eagle9 Registered Senior Member

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    DRZion
    Thanks for comprehensive and interesting answer

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  22. Neddy Bate Valued Senior Member

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    There have been some really good explanations given already, but I thought I'd offer something simple. Microwave ovens have transparent doors covered with a grid of small holes. Visible light can pass through the holes, because its wavelength is short enough that the "width" of the rays can fit through the holes. Microwaves cannot pass through the holes because their longer wavelength makes them too "wide" to fit through. In this sense there is a correlation between wavelength and a beam's width.
     

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