why does high NA excitation illumination give better resolution in fluorescence microscopy?

[Daniel White-2]

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Thanks Arne, Guy, Retro et al. for all the great answers and comments.

Now we get to the CRUNCH:

We agree that the NA of illumination has no effect on epi wide field
fluorescence resolution.

But for confocal, we hear very strong claims about NA of illumination
strongly affecting resolution, both lateral and axial.
Here is where my understanding needs some support....

Can we say this statement is true?:

As a form of structured illumination microscopy, scanning point confocal
can benefit from a high illumination NA and proper back aperture beam
filling, because the resolution is determined by the product of the
excitation and detection point spread functions.

In other words,
Because in confocal, the illumination pattern is itself high spatial
frequency, contrasty, focused and small, rather than featureless and large
as in widefield, this illumination contrast over space multiplies with the
detection point spread function to increase resolution, and is also the
source of the optical sectioning power of a confocal.

If that's the case, then it begins to make more sense, to a stupid
biochemist like myself with a very tenuous grasp on maths.

I understand at least that the square of a Gaussian function is a bit
sharper than a Gaussian. But not that much.

One last thing:
If the axial response of a confocal has real detectable power, where does
that come from?
The product of nothing and nothing is nothing. Widefield detection psf has
a nothing response axially, right?  Does the illumination psf in confocal
somehow have some non zero power axially,  in z direction?  How is the
confocal really able to fill in the missing wedge in the axial direction in
Fourier space... when zero times zero is zero? Or are my zeros an over
simplification?

Just looking for plain natural language explanations instead of maths and
equations that pretty image fixated cell biologists and dummies like me are
unlikely to comprehend.

Cheers

Dan

[Andrew York]

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I've done some thinking about how the resolution of a confocal depends on
the excitation and the emission, maybe it's relevant/helpful here.

As a form of structured illumination microscopy, scanning point confocal
> can benefit from a high illumination NA and proper back aperture beam
> filling, because the resolution is determined by the product of the
> excitation and detection point spread functions.
>

 I agree that the resolution of a confocal is the product of the excitation
and the emission PSFs. I also agree that a high effective excitation NA is
crucial for getting a nice tight excitation PSF. Of course, a tightly
closed pinhole is crucial for getting a nice tight emission PSF, and in
practice, this is often a bad idea (low signal levels). If you open your
pinhole, it blurs out your effective emission PSF in the same way that
using too-big pixels on your camera blurs out a widefield image's effective
emission PSF.


> Because in confocal, the illumination pattern is itself high spatial
> frequency, contrasty, focused and small, rather than featureless and large
> as in widefield, this illumination contrast over space multiplies with the
> detection point spread function to increase resolution, and is also the
> source of the optical sectioning power of a confocal.
>

 Because confocals are often operated with pinholes that are comparable to
the size of the diffraction-limited emission PSF, the effective emission
PSF ends up much bigger than the diffraction-limited emission PSF, and
therefore also much bigger than the excitation PSF. The lateral width of
the product of a tight PSF with a fat PSF is dominated by the lateral width
of the tight PSF.

 The axial width of the product of a tight PSF with a fat PSF, however,
still falls off nicely in Z, giving sectioning.


> If that's the case, then it begins to make more sense, to a stupid
> biochemist like myself with a very tenuous grasp on maths.
>

You seem to be doing just fine.


> I understand at least that the square of a Gaussian function is a bit
> sharper than a Gaussian. But not that much.
>

Root(2) sharper, if the Gaussians are the same size (excitation aperture
filled, detection pinhole tightly closed, excitation and emission
wavelengths fairly close together). Of course, the "frequency support" of
this microscope doubles (easier to justify/explain if I'm allowed to talk
about Fourier transforms, but easy enough to just assert), so you expect
deconvolution to recover a full factor of 2 transverse resolution
improvement compared to widefield (just like regular stripe-pattern SIM,
although it took me a while to realize this).


> One last thing:
> If the axial response of a confocal has real detectable power, where does
> that come from?
>

Power?


> The product of nothing and nothing is nothing. Widefield detection psf has
> a nothing response axially, right?


The widefield detection PSF has (roughly) constant total intensity vs
defocus, but a highly variable shape vs defocus (of course). That means
axial resolution varies depending on the presence or absence of sharp
lateral structures in your object, and of course axial resolution vanishes
if lateral structure vanishes.


>  Does the illumination psf in confocal
> somehow have some non zero power axially,  in z direction?


No, but the product of the illumination and emission PSFs does. (Assuming
by power you mean resolution)


>  How is the
> confocal really able to fill in the missing wedge in the axial direction in
> Fourier space... when zero times zero is zero? Or are my zeros an over
> simplification?
>

Multiplication in real space is convolution in Fourier space. You multiply
your effective excitation PSF by your effective emission PSF to get your
confocal PSF; this is in real space. In Fourier space, you don't multiply
your two OTFs, you convolve them. Does this resolve the apparent
contradiction?


> Just looking for plain natural language explanations instead of maths and
> equations that pretty image fixated cell biologists and dummies like me are
> unlikely to comprehend
>

I used concepts like Fourier space and convolution, which are pretty mathy,
but you seem to grasp that stuff, so maybe I can get away with it.

-Andrew

[Arne Seitz]

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Hi Dan,

I directly move on to your "one last thing":
In a previous post Jim Pawley pointed out the zero frequency support along the z-axis would only be true in a aperture free system. I'm too lazy to find the post in the archives but I remember the publication which demonstrates this fact:

Biophys J. 1990 Feb;57(2):325-33.
Determination of three-dimensional imaging properties of a light microscope system. Partial confocal behavior in epifluorescence microscopy.
Hiraoka Y, Sedat JW, Agard DA.

Nevertheless there is still the problem why zero fold zero is not zero.....
If I remember it correctly you have to do a convolution of the emission and excitation OTF. Thus it is not a multiplication. You can come to the resulting OTF of a confocal microscope if draw the excitation OTF and you draw the emission OTF on a different piece of paper an cut it out. Now you place the center of the cut out OTF on every point of the first OTF and you combine the outlines of the overlaid OTFs. This results in the OFT of the confocal microscope.  

I found a nice illustration of what I tried to explain  in the following lecture from Kai Wicker (page 15):
http://www.iap.uni-jena.de/iapmedia/Lecture/Advanced+Optical+Microscopy1365976800/AOM_Slides+AOM+121203.pdf

This is clearly not a math-free explanation but at least something that can be derived graphically.

Hope that helps.

Cheers
Arne
   

-----Original Message-----
From: Confocal Microscopy List [mailto:[hidden email]] On Behalf Of Daniel White
Sent: mercredi 5 février 2014 20:53
To: [hidden email]
Subject: why does high NA excitation illumination give better resolution in fluorescence microscopy?

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Thanks Arne, Guy, Retro et al. for all the great answers and comments.

Now we get to the CRUNCH:

We agree that the NA of illumination has no effect on epi wide field fluorescence resolution.

But for confocal, we hear very strong claims about NA of illumination strongly affecting resolution, both lateral and axial.
Here is where my understanding needs some support....

Can we say this statement is true?:

As a form of structured illumination microscopy, scanning point confocal can benefit from a high illumination NA and proper back aperture beam filling, because the resolution is determined by the product of the excitation and detection point spread functions.

In other words,
Because in confocal, the illumination pattern is itself high spatial frequency, contrasty, focused and small, rather than featureless and large as in widefield, this illumination contrast over space multiplies with the detection point spread function to increase resolution, and is also the source of the optical sectioning power of a confocal.

If that's the case, then it begins to make more sense, to a stupid biochemist like myself with a very tenuous grasp on maths.

I understand at least that the square of a Gaussian function is a bit sharper than a Gaussian. But not that much.

One last thing:
If the axial response of a confocal has real detectable power, where does that come from?
The product of nothing and nothing is nothing. Widefield detection psf has a nothing response axially, right?  Does the illumination psf in confocal somehow have some non zero power axially,  in z direction?  How is the confocal really able to fill in the missing wedge in the axial direction in Fourier space... when zero times zero is zero? Or are my zeros an over simplification?

Just looking for plain natural language explanations instead of maths and equations that pretty image fixated cell biologists and dummies like me are unlikely to comprehend.

Cheers

Dan

[Vitaly Boyko]

why is this so important?
Would it be not more scientificall
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Daniel,

why is this so important?
Would it be not more scientifically significant to work with physiological concentrations of biomolecules?
When you work with pico- or low nanomolar concentrations, you will likely get something that is close to natural.
Instead, most of us over express biomolecules 1000-fold or more and the need for high axial resolution is obvious (but a good confocal system is ca. $1,000,000.00 in price).
You have probably driven along M5 motorway in the UK many times. Imagine suddenly there are 1000-fold more cars on M5, and all are stubborn and want to travel at 70 miles per hour. Well, is it not obvious that accidents happen, and then all other cars behind would have to take back roads. Exactly the same would happen in a live cell, when biological cargo should, let say, move 1 um per second, but because of 1000-fold over expression it slows down to 0.1 um/sec, and molecules behind pile up around the MTOC, get stuck on membranes, or take back roads. Thus, what we usually report in publications - yes, you are right - ACCIDENTS. If you wanna stay away from accidents, stay away from axial high resolution confocals, and do not get misled by N.A. of the objective as other parameters are also very important in low light microscopy which seems to be possible within wide-field or light sheet microscopy. And there is not much of out-of-focus light under
 physiological conditions. Though experimentally, it might be a bit tough in the beginning but it comes along with experience.

Cheers,

Vitaly

     




On Wednesday, February 5, 2014 5:07 PM, Andrew York <[hidden email]> wrote:
 
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I've done some thinking about how the resolution of a confocal depends on
the excitation and the emission, maybe it's relevant/helpful here.

As a form of structured illumination microscopy, scanning point confocal
> can benefit from a high illumination NA and proper back aperture beam
> filling, because the resolution is determined by the product of the
> excitation and detection point spread functions.
>

I agree that the resolution of a confocal is the product of the excitation
and the emission PSFs. I also agree that a high effective excitation NA is
crucial for getting a nice tight excitation PSF. Of course, a tightly
closed pinhole is crucial for getting a nice tight emission PSF, and in
practice, this is often a bad idea (low signal levels). If you open your
pinhole, it blurs out your effective emission PSF in the same way that
using too-big pixels on your camera blurs out a widefield image's effective
emission PSF.


> Because in confocal, the illumination pattern is itself high spatial
> frequency, contrasty, focused and small, rather than featureless and large
> as in widefield, this illumination contrast over space multiplies with the
> detection point spread function to increase resolution, and is also the
> source of the optical sectioning power of a confocal.
>

Because confocals are often operated with pinholes that are comparable to
the size of the diffraction-limited emission PSF, the effective emission
PSF ends up much bigger than the diffraction-limited emission PSF, and
therefore also much bigger than the excitation PSF. The lateral width of
the product of a tight PSF with a fat PSF is dominated by the lateral width
of the tight PSF.

The axial width of the product of a tight PSF with a fat PSF, however,
still falls off nicely in Z, giving sectioning.


> If that's the case, then it begins to make more sense, to a stupid
> biochemist like myself with a very tenuous grasp on maths.
>

You seem to be doing just fine.


> I understand at least that the square of a Gaussian function is a bit
> sharper than a Gaussian. But not that much.
>

Root(2) sharper, if the Gaussians are the same size (excitation aperture
filled, detection pinhole tightly closed, excitation and emission
wavelengths fairly close together). Of course, the "frequency support" of
this microscope doubles (easier to justify/explain if I'm allowed to talk
about Fourier transforms, but easy enough to just assert), so you expect
deconvolution to recover a full factor of 2 transverse resolution
improvement compared to widefield (just like regular stripe-pattern SIM,
although it took me a while to realize this).


> One last thing:
> If the axial response of a confocal has real detectable power, where does
> that come from?
>

Power?


> The product of nothing and nothing is nothing. Widefield detection psf has
> a nothing response axially, right?


The widefield detection PSF has (roughly) constant total intensity vs
defocus, but a highly variable shape vs defocus (of course). That means
axial resolution varies depending on the presence or absence of sharp
lateral structures in your object, and of course axial resolution vanishes
if lateral structure vanishes.


>  Does the illumination psf in confocal
> somehow have some non zero power axially,  in z direction?


No, but the product of the illumination and emission PSFs does. (Assuming
by power you mean resolution)


>  How is the
> confocal really able to fill in the missing wedge in the axial direction in
> Fourier space... when zero times zero is zero? Or are my zeros an over
> simplification?
>

Multiplication in real space is convolution in Fourier space. You multiply
your effective excitation PSF by your effective emission PSF to get your
confocal PSF; this is in real space. In Fourier space, you don't multiply
your two OTFs, you convolve them. Does this resolve the apparent
contradiction?



> Just looking for plain natural language explanations instead of maths and
> equations that pretty image fixated cell biologists and dummies like me are
> unlikely to comprehend
>

I used concepts like Fourier space and convolution, which are pretty mathy,
but you seem to grasp that stuff, so maybe I can get away with it.

-Andrew