kelvinA
I blame this confusion on the fact that I couldn't find it mentioned anywhere that the effective focal length is from an image of the same size, and not the first lens. It was all equations without labelled diagrams. The scale is 1mm = 10px and I've got a 52mm beam, F30mm converging and F-10mm diverging lens which is 22mm away. The first lens is at (0,0).For Fusion360, it would be easier to define the resultant top ray as an angle, which would be
arctan(0.5*screen length / effective focal length)
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