Hi Mark,
thanks a lot for the quick response.
How exactly is the pupil function generated though? It's still not clear to me how all aberrations of the optical components can be represented by this function.
Thanks, Bianca

You trace a grid of rays through the whole system, thus picking up whatever aberrations thre are. We then trace the rays backwards to the exit pupil. The ray amplitude and phase there is called the 'pupil function'.

Another way of thinking about it is to think of an optical system as a point source, entrance pupil, optics, exit pupil and image. The point source radiates spherical waves which illuminate the entrance pupil. We differentiate this spherical wavefront to give rays, which are traced all the way through the optical system, picking up optical path length as they go. They are traced to the image surface, then backwards to the exit pupil location where they are integrated to give the wavefront. That wavefront is the pupil function.

Does that help?

- Mark

So do I understand correctly that the tracing uses geometrical optics? And the amplitude and phase at the exit pupil are then derived from the sum of ray vectors?

I'm also wondering why it is necessary to trace the rays all the war from entrance pupil to imaging plane and then back to the exit pupil. Would it not be sufficient to trace the rays from entrance to exit pupil to be able to form the pupil function?

Thanks, Bianca

 

Hi Bianca,

Yes, ray tracing is purely geometric and uses Snell's law at interfaces, and accumulates optical path length as the ray is traced through the system.

I was speaking rather loosely whern I said that rays are traced backwards from the image surface to the exit pupil. They are not traced backwards at all! Instead, we subtract a chief-ray centered reference sphere phase from the optical path lengths. This is of course equivalent to tracing backwards. The net result is that we show the data as Optical Path Difference to a reference sphere. This is how we can trace rays though a huge telescope, and then present the data in terms of lamdba/20 or whatever optical path differences.

In lens design, a 'flat' wavefront is actually perfectly spherical and is converging towards its center of curvature. That's the physically meaningful definition that is useful for considering image degradations due to aberrations and diffraction.

- Mark

Hi Mark,
thanks a lot for the explanations.
The Huygens PSF only considers diffraction from the last surface to the image plane.
If the diffraction-limiting aperture is given e.g. by the first lens of a system of multiple lenses, can the Huygens PSF still deliver a realistic result?
Best, Bianca

Hi,

As far as the Fraunhofer (FFT) and Huygens calculations are concerned, apertures only cast a shadow on the pupil. Rays are either traced or not, and the trajectory of a ray that is traced is unaffected by ones that are blocked. So, the only diffraction that these calculations consider is in the propagation from exit pupil to image. To be fair, this is by far the dominant effect of diffraction on imaging systems.

If you need to account for diffraction surface-by-surface, use Physical Optics Propagation. Check out the Knowledge Base for a lot of useful articles on that.

- Mark

Hi Mark,
thanks for the help. This has clarified a lot!
Best, Bianca

My pleasure!

By the way I want to ask what kind of theories for Huygens PSF?  Can I directly consider the diffraction after stop to image?

Hi Mark,

Your answer gave me a lot of inspiration! In fact, I want to get the wavefront of the exit pupil through ray tracing. Is there a way to get the reference sphere you mentioned? Is this reference sphere the chief ray on the image plane?

Best Nancy

Hey Nancy,

trace the chief ray through the system to the image plane, and then trace it backwards to the exit pupil. This Exit Pupil Position defines a radius we call the reference sphere.

Mark

Hi Mark,

Thanks for the help !

Well, Then can I get the optical path distribution(OPTH) of this Reference Sphere on the exit pupil surface through ray tracing, or do I need to use the Exit Pupil Position as the radius to calculate it through programming simulation?

Best , Nancy

Thanks for this profund discussion. Yet, I would like to clarify on the following point:

It uses the wavefront in the exit pupil. Rays are traced from object to image, then back to the exit pupil. The amplitude and phase in the exit pupil is called the Pupil Function, and it contains all the aberrations introduced by tracing through the optical components. (...)

The Huygens PSF traces rays from the pupil function to the image plane and coherently integrates them directly on the detector surface to give the Huygens PSF.

As far as I understand, a cartesian grid of rays is sampled in the entrance pupil and the OPD in the exit pupil is mapped onto this grid (at least if ray aiming is off). After that, FFT and Huygens calculate the PSF in different ways from this cartesian entrance pupil grid. However, if the sine-condition is not fullfilled and there is no linear relatioship between the grid of the entrance and exit pupil, a slight error is not accounted for.

How is it for systems with very large deviation of the sine condition as it is the fact for axicon/bessel beam systems? The rays that are started as a cartesian grid in the entrance pupil form a ring structure in the exit pupil. As far as I understand neither the FFT nor the Huygens account for this very strong apodization in the exit pupil, correct? I wonder because the results of FFT and Huygens look different though.