# Confocal xy resolution

Hello all. I’d really appreciate it if someone can shed some light on my question about the confocal resolution with different pinhole sizes. Below is a plot I frequently came across illustrating the xy resolution changes when closing down the pinhole. First, does the plot make sense?

The dotted line is flat above 1 AU, but xy resolution increases slightly below 1 AU with significant signal loss. When closing down the pinhole from infinity to somewhere above 1 AU, we are losing the ripples of smaller intensity outside the main peak of PSF.

Those ripples contain spatial information of higher frequency. My question is would we lose xy resolution by closing down the pinhole compared to the wide-field when above 1 AU? Then, how do we gain xy resolution when further closing down pinhole below 1 AU? When the smaller pinhole cutting into the main central peak, aren’t we losing more spatial information and thus lower resolution?

On the other hand, xy resolution is the smallest, resolvable distance between two neighboring point sources. As we’re closing down the pinhole < 1AU, we reject more light from the central peak so that less interfering light of two neighboring point sources reach the detectors. In this way, we can resolve two point sources of smaller distance. This somehow explains why closing down pinhole < 1AU increases the xy resolution. Can someone shed some light on the 1st plot that the dotted line is flat above 1 AU? Wouldn’t the pinhole (from infinity to slightly > 1 AU) removing interfering light from neighboring point source increases the xy resolution as well?

Thanks a lot for taking time to read here and any inputs are greatly appreciated.

Well this question sent me down a rabbit hole because I thought I understood confocal pinholes but apparently I don’t.

The best resource I’ve found around this topic is the SVI website. Here it says that closing the pinhole (from large → 1 AU) decreases the FWHM without improving lateral spatial resolution. So, I think you are right that this is a topic where the two definitions of spatial resolution (single point FWHM vs object separability) diverge and we have to start being really careful about what, exactly, we’re talking about. If any optics person out there can explain better, I’d love to hear it!

To explain the first graph we would need some context because it looks like the result of some particular experiment. However I interpret what you are seeing there as deterioration in XY resolution with increasing pinhole size rather than improvement in XY resolution (beyond the traditional Rayleigh criterion) with decreasing pinhole.
In theory the pinhole size makes no difference to isolated XY resolution - it is only to discriminate in Z. The link provided by @smcardle relates to the XY projection of an XYZ PSF.
I have done work imaging ultra-thin EM tissue sections with a bright field wide field light microscope to investigate resolution issues and, in that setting, there is no structure above or below the focal plane so such composite XYZ PSFs do not apply. So it would be interesting to image such 2D samples with a confocal and see how much deterioration occurs in XY resolution with pinhole size (if any).
Your second points about a smaller pinhole ‘cutting into’ the Airy disc will not work because that pattern is produced by diffraction of light that passes through a pinhole that is the smallest possible sized hole that will allow light through. Any smaller and no light will get past it other than by evanescence. Any larger and the pattern becomes a composite - a larger disc produced by convolution of the Airy pattern with the size (and shape) of the pinhole.

@P_Tadrous It makes sense that the PSF plot is generated using the smallest pinhole possible. Any smaller pinhole, no light goes through. Any larger pinhole the pattern becomes a composite. I think I get it, but not so sure. Can you help me with the following question? For a wide field and a confocal with the same lens and sampling frequency, do they have the same PSF? For a confocal, at different pinhole size, are the composites different than the PSF at smallest pinhole? Only at smallest pinhole, where the MTF curve approaches zero, we get the PSF similar to my 2nd plot? Sorry if those questions don’t make sense, I’m still trying to digest those information.

The first graph you linked is a common depiction of resolution in a point scanner. You can find a similar graph in Colin Sheppard’s chapter in the confocal handbook (actually Colin Sheppard has a lot of good classic texts on confocal theory, you should seek them out):

and it is indeed what one would predict for resolution and signal in a confocal scope: if we ignore stokes shift, lateral resolution will be the same as widefield for pinhole apertures greater than ~1 AU, and increases to a √2 improvement as the pinhole diameter approaches zero.

You can also find a good thorough explanation in the supplement (section 2G) of Betzig’s nonlinear sim paper:

Mathematically, you can think of lateral resolution in a point scanner as the product of the excitation and detection PSFs (equations from the paper above):

where the detection PSF takes into account the pinhole (it is the convolution of the regular widefield detection PSF with the pinhole function … basically just the shape of the pinhole):

The “transverse resolution” plot in the original graph (just like your original graph) shows the relative width of that as the pinhole size approaches zero.

The reason it approaches a √2 improvement is because that is roughly what you would expect to happen to the width if you took the product of two “regular” PSFs.

Here’s how I like to think of this. Imagine a single fluorophore in point scanning microscope, and we’re thinking about the rise and fall of the emission photons as we scan the beam across:

1. If the pinhole is very large (>1AU), then we are essentially collecting all of the light that is emitted by the fluorophore. We would expect the shape of the PSF (i.e. the shape of the signal as we scan the beam across our fluorophore) to look exactly like the profile of the excitation spot, since it should rise and fall in direct proportion to the shape of the excitation spot as it moves across. mathematically above, this is similar to setting the pinhole to 1 everywhere, reducing the effective detection PSF to 1 everywhere (i.e we collect all the light) and then the final confocal PSF reduces down to just the excitation PSF: (side note here: this is why point scanning microscopes use the excitation wavelength to calculate resolution while widefield systems use the emission wavelength)
2. On the other extreme end, when the pinhole is infinitesimally small, then as we scan our excitation beam across the fluorophore, we are not collecting all of the emission light, but rather just the tiny bit of the emission light that correlates to the point where our excitation beam is. It will rise and fall in proportion to both the excitation intensity provided by the excitation spot, and how far away from the fluorophore the beam is. (this one is hard to describe in words, will try to make a figure). Mathematically here, the pinhole equation P approaches a delta function, which makes converge to the detection PSF … and then we arrive at the final effective confocal PSF that is a “pure” product of the excitation and detection PSFs (which is narrower than either is alone).
3. everywhere in between, the pinhole equation P is a little top hat function of some finite width, and the resulting convolution of the pinhole and the detection PSF leads to an intermediate resolution somewhere between widefield and the √2 gain

This √2 gain in resolution is really what airy scan or instant sim are trying to achieve, but without losing the extra light (which is done in Airy scan, e.g., by simply having multiple adjacent detectors, each of which acts like a 0.2 AU pinhole, some of which just happen to be off-axis from the detection spot.)

Anyway, your original graph is absolutely correct: lateral resolution on a confocal is the same as widefield with a pinhole > 1AU (excepting the fact that we use excitation wavelength rather than emission), and improves to √2 as the pinhole approaches zero. if that didn’t help, I’ll try to put it into figure form

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for lateral resolution, with a wide open pinhole, yes, with the exception that the point scanner will have the PSF width based on the size of the excitation spot (using excitation wavelength) and widefield will use that of the emission spot (emission wavelength).

Thanks @talley That’s really helpful. Talking about Airyscan, it first deblurs the image from individual 0.2 AU detector, then shifts it towards the central detector, and sum the signal. Is it still valid to gaussian fit the Airyscan processed image of 100nm bead to estimate resolution?

The diffraction limit for confocal is 0.61 Ex lamda/NA, for pinhole > 1 AU. But with infinitely small pinhole, we can achieve 0.61 Ex lamda/NA / sqrt2. For Airyscan with 0.2 AU, the resolution should be between the two. But Zeiss claims Airyscan increases the resolution by a factor of two. I guess the signal summing from the off-axis detectors somehow increases the resolution. Can you shed some light on that?

After reading what @talley wrote, I’m not sure this is true. I think that in the conjugate/fourier plane, the light further from the axis has the high frequency information, but the confocal pinhole is in the sample plane, so this might not be a problem?

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here’s one animation I just made that might slightly help. It doesn’t explain why closing the pinhole decreases the FWHM (explained in the math above, but still likely unintuitive), but it does give a better feeling for the effective confocal PSF (in blue) relative to having no pinhole at all (dashed black line)… while the orange (un-normalized) line shows the drop in intensity you get as you close the pinhole. Inset shows relative FWHM and signal as a function of pinhole radius

(code here)

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Thanks for sharing the animation and the code. It looks great. One question, why does the normalized intensity goes to 0.75? Is it because the FWHM goes down by a factor of 1.3 and thus the area under the curve?

where? if you’re looking at the blue line in the inset at the top left, that’s not intensity, it’s the relative FWHM (relative to a wide-open pinhole). Can you clarify what part of the animation made you say the normalized intensity goes to 0.75?

Oh, I mixed y axis with the inlet FWHM y axis. Thanks for sharing.