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Equal-area detector sampling for a circular detector?


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By definition, nonsequential rectangular/square detectors have uniform detector area sampling via a Cartesian grid.  However, Detector Surfaces can only be partitioned by angular and radial zones in polar coordinates, which poses great and almost useless difficulty in determining uniformity over the circular detector aperture.  Detector pixels near the origin are much smaller in area than pixels out around the circumference.  Can we have a Detector Surface option that tesselates its pixels to give an approximation to Arthur Cox’s recipolar (termed hexapolar in Zemax) equal-area distribution?  Each detector pixel could have approximately the same surface collecting area, without fractional pixels around the edge.

The Detector Surface pixels could be hexagons clipped off around the periphery, but that doesn’t solve the unequal area sampling issue. 

I suppose in a pinch an absorbing annular aperture can be placed over a rectangular detector to get answers, but again, that doesn’t solve the unequal area sampling issue.

Mike

 

 

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Best answer by Jeff.Wilde 22 May 2022, 05:30

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Userlevel 7
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Would a polar detector do what you want?

Userlevel 6
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Hey Mike,

Have you looked into converting a Standard Surface to a STEP file, then re-importing it with CAD Part: STEP/IGES/SAT using the ACIS library?  A new option for ACIS was Maximum edge length which limited how big a triangle could be; ACIS will attempt to make all the triangular facets the same area.  You of course will need to manually parse the Detector Data Text tab to get any useful numeric data out.

Userlevel 7
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Hi Mike,

It seems like the Detector Surface Object should be okay for sampling purposes.  See the help documentation regarding pixel size:

Here’s an example:

 

Regards,

 

Jeff

Userlevel 5
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Thank you Jeff!  I had somehow not read down into the “Comments about detector triangles” text.  It appears this has been well thought out in software.  I had noticed that flat circular detector surfaces were giving fairly uniform intensity when illuminated uniformly, but when you display the surface intensity, you see a misleading diagram of highly non-uniform pixel sizes.  Apparently the triangles are not shown in the plot.

Problem solved!

Mike

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