Re: reflection from a lens

[Brian Whatcott <betwys@SBCGLOBAL.NET>]

At 11:04 PM 6/5/2003 -0400, Carl, you wrote:
> Here's an item that stumped me for a while. It's based on a question
> from the April 2003 Physics Challenges in The Physics Teacher. I'll
> loosely adapt to bring out the issue of interest.
>
> 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.
> /snip/ find the focal length f for transmission
> through the lens. Also let f' = R/2 be the focal length of the back
> surface of radius R treated as a mirror. Then 1/object-distance +
> 1/image-distance = 1/f + 1/f' + 1/f because you go through the lens,
> reflect off the back, then go back through the lens.
>
> 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.
>
> Nevertheless, if you invent and substitute numerical values, both
> methods agree. It may amuse you (or frustrate you, as it did me for a
> while) to understand intuitively why they agree. Happy trails for
> those who enjoy such puzzles, Carl
> --
> Carl E. Mungan, Asst. Prof. of Physics  410-293-6680 (O) -3729 (F)


This is probably not the level of response that Carl is looking for,
 but to a simple soul, a lens with a back mirror set very close and
evenly distant from the rear surface of the lens will conform to the
simple optical model mentioned, and so one can ask the question:
 Is there a distinct change in behavior of the reflected ray when
the rear mirror grows close enough to be cemented in place?
 One is forced to accept the paradox: the rear surface is still
 as effective. One recalls (vaguely) the interesting process of
 cancelling aberrations by providing a return path through the
same distortion.....

Brian Whatcott      Altus OK