First of all, remember that any one ray is infinitely narrow, and so has infinite irradiance (W/area) and infinite radiant intensity (W/solid angle). If you have only one ray, and trace it to a detector, as you increase the number of detector pixels, one ray still only lands in one pixel and so irradiance and radiant intensity will increase to infinity as the pixel size goes to zero.

So to compute either irradiance or radiant intensity, we must first trace 'enough' rays to a detector with 'adequate' resolution. In OpticStudio the detector is defined as having nx by ny pixels, and has a spatial extent defined by the x, y half widths. The angular extent is defined by parameters 12-15, the max and min angles which default to +/- 90 degrees in x and y, but can be set lower than that if required.

When a ray is traced to the detector, it hits at some x,y location with some l,m direction cosine. We identify the appropriate spatial pixel and add the ray's energy to that pixel. We then compute the appropriate angular pixel, and add the ray's energy to that angular pixel too, and then loop over all rays, accumulating energy in the various pixels as we go.

So that gives us two arrays of data. One shows us the spatial distribution of rays. No matter what angle the ray makes to the detector, if it lands at some spatial coordinate, it is in the appropriate spatial pixel. The second array is the angular distribution. No matter what spatial position the ray landed at, if it lands at some angle it is binned into the appropriate angular pixel.

Next, we realize that while the spatial data is exactly what we want (power/area), the angular data is not. One angular pixel extends say from 5 degrees in x to 6 degrees in x, and from 40 degrees in y to 41 degrees in y. What solid angle does that angular pixel represent?

To answer that question we project out from the local axis of the detector to construct a sphere some long distance away from the detector, centered on the detector. We then (conceptually) fire rays from this point at x, y angles of (5, 40), (5, 41), (6, 41), and (6, 40). Now we know that patch of area, we can compute the solid angle subtended by that angular pixel.