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Why is resolving limited by wavelength?

  1. Jan 27, 2012 #1
    Hi!

    This might very well be a silly question. In many courses I've been presented as an obvious fact that it won't work to use (e.g.) light with a wavelength larger (at least, not much larger) than the thing you want to resolve. Why is this exactly? Thinking of photons I could find no obvious explanation, but I couldn't really get my head around it thinking of classical waves either. Wouldnt you be able to, say, work out the position of a pole planted in water by examining the refraction pattern of water waves with wavelengths longer than the pole diameter? Might not be a useful analogy but that's what came to mind.

    Hope somebody can enlighten me!
     
  2. jcsd
  3. Jan 27, 2012 #2

    DaveC426913

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    Your pole analogy is apt. A small pole in large waves will not show up. It is only once the wavelength gets down to on the order of the object that the diffraction will become apparent.
     
  4. Jan 27, 2012 #3
    Thanks for your answer, good to know that my analogy works at least:) However, I obviously still don't understand the why wavelength limits resolution, so I hope you can elaborate a bit.

    When sketching up a diagram of the pole situation, I can't for the life of me understand why the wavelength/diameter ratio matters. Some fraction of the incoming wave would surely be reflected off the pole. How does this not create a (detectable, but possibly minute) perturbation of the wave pattern?
     
  5. Jan 27, 2012 #4

    DaveC426913

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    Yes. Detectable but minute. So: not much use for resolving.
     
  6. Jan 28, 2012 #5
    Thinking of photons the explanation may look like the following. According Heisenberg uncertainty principle dx*dp if of order of Plank constant h, dp – uncertainty of photon impulse, and dx – is uncertainty of location of its origin. Since any microscope collects photons only from limited angle, dp is limited by its value. dp = p*NA, where p – impulse of a photon (p=h/λ), NA – so called numerical aperture of the microscope objective (it is equal to sine of half of the collecting angle). So dx is approximately equal to λ/NA. Typical value for NA of micro-objective is within 0.2-0.45. There are immerse objectives with higher NA, but the value can’t be more than 1 in principle.

    Concerning classical wave consideration, I could propose the following consideration. Lets consider an optical system that produces an image of a micro-object. If the system is high-performance, we can replace it by an equivalent thin lens. Then we can write an obvious ratio: H/h = L/l, where h and H – size of the object and its image, l – distance from the object to the lens, L – distance from the lens to the image. The image size H can’t be less than the value limited by diffraction on the lens aperture D, that is approx. equal 2.44*lambda/D*L. So resolved size of the object can’t be less than about 2.44*lambda/D*l. l is always more than focal length of the lens f, though under microscopic observation it is very close to f. D/f for high-performance micro-objective is about 1, maybe a bit higher.


    I should note that the consideration above deal with transversal resolution only. With respect to longitudinal resolution, it can be much better than the wavelength in some cases.
     
  7. Jan 28, 2012 #6
    Also Fourier Transform very well explain resolution limit.
     
  8. Jan 29, 2012 #7
    Much clearer now, thanks a lot guys!
     
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