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Point spread function (PSF) describes the volumetric image of a point light source in the object space. But in practice, we measure the PSF by moving the light source (the fluorescence bead) and fixing the CCD. Are these two equivalent?

A simple illustration of my question is described as follows. This first is the designed object and detector. The second one means the object is shifted(which causes defocus). And the third one is the shift of the detector. My question can be simplified to, can it be proved that in the second and third case, the detector get the same image?

Complemented on PSF

I think the PSF of the a 3D imaging system can be describe as the images that a point source creates on different layers that the detector take as different depth rather that the real 3D image that the point source create in the image space(these two cases correspond to the second and the third case in the figure). In this sense, despite the fact that maybe there is no equivalence of the second and the third case in the figure. The formation of the imaging and the process of measurement of the PSF do be the same(the case two)

Hence, it is absolutely OK to use the measured PSF to do deconvolution.

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They are equivalent.

Moving the source and fixing the CCD, or moving the CCD and fixing the source, will result in the same image, the PSF. It is only the relative positions of the CCD and the source that matter, and the relative position can be obtained by moving either the CCD or the source.

Each relative position will yield its own PSF. If they are all the same then the CCD camera is isoplanatic (spatially invarient).

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  • $\begingroup$ how to prove that with a graph. I think it is not obvious $\endgroup$ – Rickyim Feb 27 '18 at 15:29
  • $\begingroup$ I changed my answer to make it more clear. Its true that the previous reciprocity argument was not obvious. $\endgroup$ – user45664 Feb 27 '18 at 16:22
  • $\begingroup$ In an anisplanatic(spatial variant) system, I think they are not the same. Let's say the system has depth variant PSF. When the source is fixed and the CCD is moving, then what the CCD get is the PSF at the corresponding depth, let's say at $z_0$. But when the CCD is fixed and the source is moving up and below $z_0$, the PSF changes, let alone the image the CCD get. This time, the PSF that CCD takes surely differs from the previous one. $\endgroup$ – Rickyim Mar 4 '18 at 11:20
  • $\begingroup$ What you say is correct, but the distance z is a RELATIVE distance which you can get either by moving the source or by moving the CCD. So my answer is still also correct--they are equivalent. $\endgroup$ – user45664 Mar 5 '18 at 20:22
  • $\begingroup$ Did your answer take the lens set between the source and CCD into account? $\endgroup$ – Rickyim Mar 6 '18 at 5:51

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