Posted by
Andrew York on
URL: http://confocal-microscopy-list.275.s1.nabble.com/why-does-high-NA-excitation-illumination-give-better-resolution-in-fluorescence-microscopy-tp7581531p7581532.html
*****
To join, leave or search the confocal microscopy listserv, go to:
http://lists.umn.edu/cgi-bin/wa?A0=confocalmicroscopy*****
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