On 06/05/2003 11:04 PM, Carl Mungan wrote:
>
> Suppose you have a lens with known index and radii of curvatures of
> the two spherical surfaces. An object (in air) at known distance
> forms an image by reflection off the *back* surface of the lens
> (relative to the object). Find the image location.
...
> What stumped me for a while is that the second solution would seem at
> first glance to be wrong because it includes a forward and reverse
> transmission through the back surface, which don't actually occur.
This seems like a nice advertisement for _physical_ optics
as opposed to geometric optics.
My first impulse when faced with a problem in optics
is to ask what the phase is doing. (The same impulse
applies to optics, acoustics, quantum mechanics, et
cetera.)
When traversing a lens, the wave picks up a relative
phase in proportion to how much glass it goes through.
In the case of reflection off the back of a lens, at a
thin part of the lens, the wave makes two trips through
thin glass. At a thick part of the lens, the wave makes
two trips through thick glass. To this way of thinking,
what matters is how much glass there is (and the index
of the glass). The boundary is _usually_ what tells you
how much glass there is, but not _necessarily_, as this
example shows.
Additional buzzwords:
-- Fermat's principle of least time.
-- Huyghens' construction.
-- Method of stationary phase.
-- Path integrals à la Feynman and Hibbs.