Does the pinhole size affect the Nyquist rate?
See Pinhole And Bandwidth.
Old information
In principle, the Nyquist rate is independent of the pinhole size. This is due to the choice to relate the Nyquist rate to the theoretical bandwidth of the system: the spatial frequency beyond which *nothing* is transferred by the microscope.
It would be a different story if we would have used a criterion based on attenuation of spatial frequencies below a certain factor. (Larger pinhole sizes attenuate higher frequencies more, but still are not zero). Although a practical approach (because about last third of the band has so low intensities that they can be considered zero most of the times) this involves an arbitrary choice, so therefore we base the Nyquist rate on the well defined theoretical bandwidth.
A problem occurs with extremely large pinholes like those used in two-photon systems. In these cases the optical properties are practically identical to a widefield system whereas due to the presence of the pinhole the theoretical bandlimit is still that of a confocal microscope. In the two-photon case it is best to set the microscope type to 'widefield' when doing deconvolution with the Huygens Software, since this will result in the same optical properties but with a more practical Nyquist rate.
Note however that with single photon confocals even a very large pinhole will still have a noticeable effect on the blur contributions of far off focus regions, thus improving resolution along the optical axis.
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Keywords: pinhole size Nyquist rate two-photon<br>
Categories: Faq Deconvolution, Faq Microscopy, Huygens Faq, Imported Faqs<br>
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