# Understanding the Method behind Huygens PSF

• Student
• 4 replies

Hello everyone,

I'm interested in finding out more details on the Huygens PSF method.

The OpticStudio Help claims that 'a grid of rays is launched through the optical system'. Here I would like to know, what algorithm is used for the actual propagation. How are the abberations of all optical components accounted for? Are they propagated as rays, by paraxial propagation or by actually using Huygens formula?

It would be very helpful if anybody could clarify this detail.

Thanks a lot in advance, best wishes

### 13 replies

Userlevel 5
+3

Hi Bianca,

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 FFT (Fraunhofer) PSF then takes a FFT of this data to give the FFT PSF.

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. There is a nice example in Documents\Zemax\Samples\Non-sequential\Coherence Interference and Diffraction\Interference Example 4- Diffraction Limited Imaging.ZMX of a Huygens integration done in non-sequential mode where you can see the exact calculation working one ray at a time.

- Mark

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

Userlevel 5
+3

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

Userlevel 5
+3

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

Userlevel 5
+3

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

Userlevel 5
+3

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 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,

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

Userlevel 5
+3

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

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