Why is resolving limited by wavelength?

[Zoroaster]

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!

 

[DaveC426913]

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!

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.

 

[Zoroaster]

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?

 

[DaveC426913]

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?

Yes. Detectable but minute. So: not much use for resolving.

 

[AlexLAV]

Hi!

Thinking of photons I could find no obvious explanation, but I couldn't really get my head around it thinking of classical waves either...

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. Let's 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.

 

[ProTerran]

Also Fourier Transform very well explain resolution limit.

 

[Zoroaster]

Much clearer now, thanks a lot guys!