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Diffraction through circular apertures

  1. Jan 7, 2014 #1
    Hi all,

    This is probably an error resulting from my qualitative understanding of diffraction, but in most basic descriptions of the phenomena they talk about diffraction occurring when light encounters an aperture that is equal to or smaller than the wavelength of the incident light with the relationship at its most basic being minimum resolvable separation distance of two objects = lambda/aperture diameter. Thats a very simple relationship to understand, however lets say the apertures i'm talking about are lenses and the light is emitted from the object plane not transmitted. The physical diameter of a lens is much greater than the wavelength of the light so why does diffraction still occur?

    Many thanks
     
  2. jcsd
  3. Jan 8, 2014 #2

    Simon Bridge

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    Diffraction always happens - everywhere you have an edge, or more than one source close together.

    The simple relationship for diffraction at a slit is a special case where the effect is particularly clear.

    If you are thinking of the relation for the resolving power of a telescope (say) then it is the small angular size of the distant object/separations that makes the effect apparent.
    http://en.wikipedia.org/wiki/Angular_resolution
     
  4. Jan 8, 2014 #3
    Thanks simon, diffraction at a slit is a good example. I was actually thinking about microscope objective lenses. In that case the aperture is given by nsinΘ. The important thing is that any lens has a finite aperture which dictates the achievable resolution.

    Thanks again.
     
  5. Jan 8, 2014 #4

    Simon Bridge

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    With the microscope it's the same thing ...

    If you uniformly illuminated a spherical aperture, without the lens in it, you get a bright spot with rings around it.
    How distinct the rings are depends on how small the aperture.

    Put a lens in there and things get trickier.
    The effect, when you crunch the numbers, is that point (or very small) sources get rings around their images.
    Each image is made up of light that passes through every part of the aperture - so the extent of the diffraction effect depends on the size of the aperture. It's just not as direct-a calculation as with the diffraction at slits case...

    You need a big aperture to lower the diffraction effect, but a small aperture to lower the aberration effect.

    There is also a limitation involving the wavelength of the stuff used to illuminate the object.
     
  6. Jan 12, 2014 #5
    Thanks simon, that makes things clearer.
     
  7. Jan 12, 2014 #6

    Simon Bridge

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    No worries.

    There's usually quite a lot going on in any real-life setup and when you get into limiting cases, like "how small an object can you see?" is when small influences start to get important.
    I tried to figure a way of showing you without the math - but it didn't work so the best I can do is give you an idea of things interrelating.
     
  8. Jan 13, 2014 #7

    sophiecentaur

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    When you see an object with 'nothing in the way' the light that enters your eye is the result of all possible propagation paths between the object and you. Now the contribution from paths that are way off the straight line is vanishingly small but, once you restrict the range of angles (with a small(ish) aperture), the missing bits start to be noticeable. (That's diffraction and can best be expressed with Maths, of course - involving Integral Calculus) The impairment to what you see, due to the light waves that are missing will take the form of 'fringes' round the object and will blur two nearby objects into one non-resolvable image. When the aperture gets really small, the (minuscular amount of) light that gets through it spreads out in all directions from that tiny spot and all objects appear to be very big blobs, all in the same place..

    It's interesting to note that the diffraction pattern for an aperture is the complete inverse (white for black fringes and black for white fringes) of the pattern for an object the same size and shape as the aperture. Not surprisingly, when you add the two, you will see a perfect image. (i.e. no obstruction)
     
  9. Jan 17, 2014 #8

    Claude Bile

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    Lenses transform plane waves into spherical waves.

    However, since real lenses have a finite aperture, the plane wave (and the resulting spherical wave) becomes truncated.

    These "truncated" spherical waves focus into an Airy disc rather than a single point.

    Claude.
     
  10. Jan 17, 2014 #9

    sophiecentaur

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    And spherical waves into other spherical waves (more commonly).
     
  11. Jan 17, 2014 #10
    As an aside from the above question but related to spherical waves and circular apertures. A wave from a point in space propagates as a sphere expanding in all directions. When this spherical wave encounters a lens, the lens only captures the "cap" of the wave, with the light at the periphery of the aperture encoding the highest resolution information. So does that mean that the further a lens is from an object the more the spherical wave has expanded meaning the lens captures less of the light. Does that also explain the resolution improvement of near-field optics over far-field?
     
  12. Jan 17, 2014 #11

    sophiecentaur

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    The further away the lens is, the nearer the wave front approaches a plane. (Curvature is 1/x)
    The energy admitted into the aperture will follow the Inverse Square law, as soon as the source is far enough away to appear point like.
    In the near field, calculation of the pattern needs to include more terms than in the far field. Google Fraunhoffer and Fresnel diffraction.
     
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