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Diffraction limited laser

  1. Jan 2, 2016 #1
    Why can a laser only be focussed to within its wavelength? It says due to it being diffraction limited, what about focussing lenses and mirrors they don't involve diffraction. Please explain
     
  2. jcsd
  3. Jan 2, 2016 #2

    blue_leaf77

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    First of all, only plane waves focused using a lens of infinite aperture can produce exactly a point of zero dimension in the focal plane, clearly such arrangement is not possible. Second, laser light is hardly exactly a plane wave. It always has certain curve in its phase front and moreover the field distribution is obviously not uniform across any perpendicular plane to the propagation direction. In most practical situations, a laser light is in fact slightly diverging. These two reasons limit why laser light cannot be focused lower than its wavelength. The mathematical reasoning can be understood from the fact that the far-field field distribution is related by Fourier transform with the field at the input plane.
     
  4. Jan 2, 2016 #3

    sophiecentaur

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    Ah well - they do! Diffraction is not only limited to tiny holes and slots. It occurs whenever a wave is constrained to go through any hole or around any object - the diffraction pattern of large apertures is very near the conventional 'ray' diagram but lenses (telescopes and microscopes have a resolution that's fundamentally limited by diffraction effects). However 'sharp' you try to make the beam of a laser, at the place where the beam is said to be focussed, the actual location of where the spot occurs is still governed by diffraction. If the spot were much less than the wavelength then the image would have to consist of wavelengths that are shorter than the wavelength of the light. Also, how would you measure where the spot is?
    It is possible to produce an array of RF transmitting antennas (EM is the same at all frequencies) which can have a main beam which is narrower than the width that's limited by diffraction (supergain) BUT the pattern has 'sidelobes' that are higher than the equivalent sidelobes of a non-supergain array. I expect there is a similar trick that can be done optically but I don't know how far the process can hbe taken. So you don't get anything for nothing.
     
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