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Hello All,





How does OpticStudio calculate polychromatic rms wavefront error? I tried my own formulation and couldn’t get a match.







Derived from Born & Wolf, the sum of the Strehl weighted peak irradiances divided by same in absence of aberrations.  u is the slope of the marginal ray or a/f in the Born & Wolf formulation.





Thanks, BJ Housand

The difference is lateral color. The wavefront at any wavelength is referenced to the primary wavelength chief ray. There used to be a Knowledge Base article on this but I couldn't find it when I searched.



- Mark



 


Mark,



I get that and for investigative purposes I removed that complication by only doing an on-axis comparison of a rotationally symmetric system. Interestingly, my answer is only 3.38% different from ZOS. Close enough for engineering decisions but not close enough to indicate matched formulation.



-Brien


Did you remember to adjust the non-primary wavelengths to use wavefront error in primary wavelength units?



Zemax team: could you find the old KB article on this and republish please? It was called something like 'Why is the polychromatic wavefront error not the same as the average of the monochromatic error' or similar. I couldn't find it when I looked, but this is a freqeuntly asked question and that article answered it.


Hi Brien and Mark



We recently updated the help files to give a better description of the calculation. It is at the bottom of this section: The Analyze Tab (sequential ui mode) > Image Quality Group > RMS > RMS vs. Field.







  • If a polychromatic computation is done, the RMS polychromatic is done for all wavelengths at the same time and for all the pupil.


    RMS polychromatic = sqrt(sum ((optical path difference for each ray of the pupil for each wavelength)^2 x weight of the ray)/ sum(weight of the ray)).


    Again, this same method is used when computing RMS wavefront 'Ignore Lateral Color' unticked' by the default merit function, as all wavelengths will refer to the reference sphere generated for the primary wavelength.







And Mark is right. We used to have an article but for some reasons, it is not published. I'll check and see if we can get it published again.



Sandrine


I think the article was replaced by this forum thread: Why the polychromatic MTF is further degraded than the MTF at the individual wavelengths?


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