How is the resolution of a Laser Scanning Microscope defined? I am reading about the image formation process in fluorescence microscopy, and I am puzzled by a question. Why the axial resolution of a Laser Scanning Microscope depends on excitation wavelength, but not emission wavelength of fluorophores?
The formula to compute the axial resolution is given (e.g. here):
$$
\textrm{Axial resolution} = \frac{0.88 \lambda_{\mathrm{exc}}}{\mathrm{NA}} \ ,
$$
where $\mathrm{NA}$ is the numerical aperture of the objective, and $\lambda_{\mathrm{exc}}$ is the excitation wavelength of the laser.
However, the laser beam excites the fluorophores and they emit photons with lower energy and larger wavelength $\lambda_{\mathrm{em}}$. It is not clear for me why we do not use the formula:
$$
\textrm{Axial resolution} = \frac{0.88 \lambda_{\mathrm{em}}}{\mathrm{NA}} \ .
$$
I would highly appreciate if you:


*

*could help me to clarify this question,

*give me some concise references on the topic of image formation in fluorescence microscopy (not like Born and Wolf textbook which has 952 pages :).

 A: Your formula for axial resolution has the wrong dependence on the numerical aperture. It is the correct dependence for the lateral resolution. The exact values that come into the formulas will depend on exactly which criteria are being used to define resolution. In the following, I shall use full width half maximum for axial resolution and $1/e^2$ for lateral resolution, but the ideas are the right ones.
In general, both excitation and emission wavelengths strictly come into the equation, because the excitation wavelength defines the shape and size of the focussing excitation beam, and the excitation fraction of fluorophores is defined by the intensity of the excitation beam. The fluorescence wavelength comes into the equation because, by the reciprocity theorem, the plot of the sensitivity of the imaging optics to a particular fluorophore's emissions as a function of the fluorophore's position is proportional to the plot of the focussing field from the imaging optics at the fluorescence wavelength. Generally, however, some average of the two is used because the he difference between the fluorescence and excitation wavelength is not great, so little error is incurred by use of the excitation wavelength alone.
In a fully confocal system then, the probability of fluorescence photon capture is proportional to the product of the fluorescence and excitation beam intensities, as I discuss in more detail in my answer here.
From the Wikipedia article for "Gaussian Beam", in particular the section "Complex Beam Parameter" we have the expression for the transverse, linearly polarized electric field as a function of position:
$$E_x(r,\,z) = \frac{1}{z+i\,z_R}\exp\left(-i\,k\,\frac{r^2}{2 (z+i\,z_R)}\right)\tag{1}$$
where the Rayleigh length $z_R$ for a beam of wavelength $\lambda$ is defined by:
$$z_R = \frac{\pi\,w_0^2}{\lambda}\tag{2}$$
and $w_0$ is whilst the numerical aperture (defined as the sine of the half vertex angle of the cone containing $1/e^2$ of the farfield radiation) is:
$$\eta = \frac{\lambda}{\pi\,w_0}\tag{3}$$
so that:
$$z_R = \frac{\lambda}{\pi\,\eta^2}\tag{4}$$
From (1), the intensity of the beam decreases to half its focal center value at $z=\pm\,z_R$, therefore it is natural to cite some multiple of the Rayleigh length as the axial resolution. The lateral resolution, however, is proportional to the square root of the Rayleigh length. For a fully confocal system, the full width half maximum axial resolution is often cited and this is the distance between the points where the intensity in (1) falls to $1/\sqrt{2}$ of its peak value, because the sensitivity is proportional to the product of fluorescence and excitation beam shapes, i.e. approximately to the excitation beam intensity squared, so the axial resolution is:
$$\Delta z_{FWHM} \approx 2\,(\sqrt{2}-1)\,\frac{\lambda}{\pi\,\eta^2}\approx 0.263\frac{\lambda}{\eta^2}\tag{5}$$
The product of fluorescence and excitation intensities on the focal plane is proportional to $\exp\left(-\frac{4\,\pi^2\,\eta^2\,r^2}{\lambda^2}\right)$ so that the $1/e^2$ diameter of the product is:
$$\rho = \frac{2\,\sqrt{2}\,\lambda}{\pi\,\eta}\approx \frac{0.9\,\lambda}{\eta}\tag{6}$$

Questions from OP

Thank you for your reply! Could you please also add some details/formulas about: "In a fully confocal system then, the probability of fluorescence photon capture is proportional to the product of the fluorescence and excitation beam intensities." If it is a product of intensities, how can be sure that it integrates to 1? 

Firstly, I added the link to the Physics SE post I wrote on confocal imaging; this is probably the quickest explanation I can give. Secondly, about the integral: you can't be sure that it integrates to 1 because in general it doesn't and in many cases it can't equal unity. The less than unity value means that some of the fluorescence is lost and does not register at the photodetector. Indeed, in confocal imaging, this is exactly what you want. If the product integrated to unity, this would have the following physical meaning: every photon launched by the excitation system would cause the registration of one fluorescence photon at the photodetector. That is, all the out of plane noise would be picked up and added to the pixel in question. This fundamentally contradicts a confocal microscope's designed for, theoretical and observed behavior. 
A: I don't do confocal microscopy, but I guess I can answer

Why the axial resolution of a Laser Scanning Microscope depends on
  excitation wavelength, but not emission wavelength of fluorophores?

In epi-fluorescence, where you make the image of the emitted light, the emission wavelength sets the resolution. But in scanning microscopy you don't look at the emitted image, you measure point by point how much light is emitted in the excited volume.
The focused excitation laser makes a certain volume to become fluorescent. The size of this confocal volume in $\mu m^2$ depends on the excitation wavelength (it is basically the PSF of the objective at the laser wavelength) and does not depend on the properties of the fluorophores.
Once emitted, you'll see in the microscope a (generally different) PSF, now at the emission wavelength, which depends on the emission wavelength of the fluorophores and can be bigger than the excitation PSF. But what is done is to collect all you can of this emitted light (through a pinhole or not, on a detector -not a camera) and measure its intensity. Then you move the excitation volume and measure again repeating all across the sample. As you collect always the same percentage of it through your emission optical path, the size of the emission PSF is not a problem, it could be huge. What matters for the resolution is the physical volume where the emission is generated, which depends only on the excitation properties (optics, wavelength ...).
