Resolving Power of a microscope

  • #1
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I don't know where to put this one. The resolving power of a microscope(simple) is defined on the basis of diffraction effects.I have read that and as far as my understanding goes it is 2μsinβ/(1.22λ). This was assuming circular aperture for the lens.
[Here 2β is the angle subtended by the diameter of the lens at the object which is approximately at the focus]
However in some texts I have found the 1.22 factor missing. Isn't the factor a consequence of the circular aperture of the lens? I don't know much as I am not familiar with the kind of math used to calculate it. Can someone resolve this contradiction? Which is the more appropriate expression?
 

Answers and Replies

  • #2
fluidistic
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I don't know where to put this one. The resolving power of a microscope(simple) is defined on the basis of diffraction effects.I have read that and as far as my understanding goes it is 2μsinβ/(1.22λ). This was assuming circular aperture for the lens.
[Here 2β is the angle subtended by the diameter of the lens at the object which is approximately at the focus]
However in some texts I have found the 1.22 factor missing. Isn't the factor a consequence of the circular aperture of the lens? I don't know much as I am not familiar with the kind of math used to calculate it. Can someone resolve this contradiction? Which is the more appropriate expression?

I think the 1.22 factor has to see with a zero of a Bessel function (I never went into the math yet though). It should appear with circular aperture as you said and if you consider Rayleigh's criterion instead of others.
 
  • #3
Andy Resnick
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I don't know where to put this one. The resolving power of a microscope(simple) is defined on the basis of diffraction effects.I have read that and as far as my understanding goes it is 2μsinβ/(1.22λ). This was assuming circular aperture for the lens.
[Here 2β is the angle subtended by the diameter of the lens at the object which is approximately at the focus]
However in some texts I have found the 1.22 factor missing. Isn't the factor a consequence of the circular aperture of the lens? I don't know much as I am not familiar with the kind of math used to calculate it. Can someone resolve this contradiction? Which is the more appropriate expression?

As fluidistic mentioned, the 1.22 factor arises from use of circular lenses and relates to the 'size' of a Bessel function (J_0(kr)^2/(kr)^2, specifically). Square apertures (which are sometimes used in confocal systems) give a different numerical factor.

http://www.microscopyu.com/tutorials/java/imageformation/airyna/

What you were given is called the Rayleigh criterion, and must be modified under certain circumstances: if neighboring sources are partially coherent, for example. There are also tricks to exceed the Rayleigh criterion by synthetic-aperture methods similar to radar technology or other methods such as 'structured illumination'.

Something else to keep in mind, the Rayleigh criterion holds for aberration-free lenses. This never holds in practice, and so only using the Rayleigh criterion as a performance specification is considered suspect.
 
  • #4
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Thanks everyone.
 
  • #5
sophiecentaur
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I didn't think the Rayleigh limit was based on diffraction, in particular (very unlikely really because telescope optics would almost certainly not have been diffraction limited in those days). I am sure I was taught that it says that two point images (equal brightnesses stars, I think) are said to be resolvable if there is an identifiable 'dip' to half power between them. It was just a rule of thumb, based on human perception which, nowadays, can easily be exceeded with image processing.
 
  • #6
f95toli
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Indeed, it is just a rule of thumb. The formula is based on the assumption that two point sources as easily distinguishable when the centre of the Airy disc of one falls on the first minimium of the Airy pattern of the other.
We can definitely do better than this.

I had a quick look in my copy of Hecht, and according to a footnote Rayleigh even pointed out that this was meant to be a simple -but useful- formula in his original paper.
 
  • #7
Andy Resnick
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I didn't think the Rayleigh limit was based on diffraction, in particular (very unlikely really because telescope optics would almost certainly not have been diffraction limited in those days). I am sure I was taught that it says that two point images (equal brightnesses stars, I think) are said to be resolvable if there is an identifiable 'dip' to half power between them. It was just a rule of thumb, based on human perception which, nowadays, can easily be exceeded with image processing.

The Rayleigh criterion is indeed based on diffraction - the far-field diffraction pattern of a circular aperture (the entrance pupil). The Rayleigh criterion is a statement about how far two Airy functions must be separated before you can tell there are two peaks. The Rayligh criterion fails for many cases- aberrated beams, noncircular entrance pupils, mutually coherent sources...

There are other resolution limits besides Rayleigh as well: Sparrow, Dawes, Johnson...
 
  • #8
sophiecentaur
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" tell" is the right word. No measurement involved. I think you have stated it the 'wrong way round'. If the signal processing in our eyes was very different then we could 'easily tell' with that sort of separation. In the limit, you could be talking about the Shannon Limit - involving signal to noise ratio (how dim the stars were).
Yes, the Raleigh Criterion is based on ideas of diffraction but I still say it is, essentially a practical rule. If you had a dodgy telescope and were attempting to analyse a photograph of two stars, you would say that distinguishing them was possible / not possible, based on the brightness level in the middle of the black splodge on the emulsion. If you just had the photo and didn't know what telescope it had been taken on, that's all you could go on; you wouldn't know whether the fuzziness was due to diffraction or some other aberration. Your densitometer would dip to 1/root 2 (or whichever it is) and you'd say there were two stars and not just a single blob.
But I'm not an astronomer, so I may have it the wrong way round, myself.
 

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