Posted by
Vitaly Boyko on
URL: http://confocal-microscopy-list.275.s1.nabble.com/why-does-high-NA-excitation-illumination-give-better-resolution-in-fluorescence-microscopy-tp7581531p7581535.html
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