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Marcel asked-|
"I don't really understand why George thinks that this function can't be
quadratic."

Here's where I need a blackboard, but this attached sketch will have to do.
Sorry about its crudity.

If we accept that a ray that's on the optical central axis will produce a
spot in the centre of the array, at (0,0), and that the response must be
anti-symmetric about that centre, it should be clear that any quadratic
component, added in to the basically linear variation with angle, creates a
curve that cannot meet that requirement for anti-symmetry. On the other
hand, a cubic term, being in itself antisymmetric about the origin,
presents no such problem. It's as simple as that.

George.

contact George Huxtable, at  george@hux.me.uk
or at +44 1865 820222 (from UK, 01865 820222)
or at 1 Sandy Lane, Southmoor, Abingdon, Oxon OX13 5HX, UK.

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