Re: reflection from a lens

["Carl E. Mungan" <mungan@USNA.EDU>]

Brian Whatcott wrote:

> 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?

Au contraire, this is most definitely what I meant by an "intuitive" approach.

To fill in the details:

Let's ignore the forward and backward transmission through the front
surface, which occurs in both solutions, and concentrate on what
happens at the back lens surface.

The elegant solution took the following approach: We make a
transmission thru the rear surface from glass to air, then we reflect
*in air* off a spherical mirror conforming to the rear surface shape,
then we make a backward transmission thru the rear surface from air
back into glass.

Thus, using the equations for refraction at a spherical surface and
reflection at a spherical mirror, we have:

(n-1)/R + 2/R + (n-1)/R

But this is just equal to 2/R' where R'=R/n. We can interpret this as
a reflection *in glass* off a spherical mirror conforming to the rear
surface shape, in agreement with the first, less compact solution.

That is, as Brian suggested, if you think of the mirror as a separate
metal surface, you get the same behavior (try ray-tracing it!)
whether the mirror is just outside the rear surface of the lens (and
we now ignore the weaker reflection from the glass-air interface) or
is instead embedded just inside the rear surface while the lens was
molten.

Mike Moloney wrote:

> The two reflected images Carl is talking about can interfere, and in a
> dark room using a laser source, one can see quite a nice set of 'Newton's
> Rings' reflected from a meniscus lens.  I wrote up something on this in
> the May 1974 AJP.

Yes, that's one nice extension of this idea.

Another is that if you can measure the reflected image distances off
both the front and rear surfaces, and you can measure the transmitted
image distance through the lens, then from these 3 data values (for
known object distance and assuming a thin lens), you can deduce the
two radii of curvature and the index of the glass. This is closer to
the spirit of the original Physics Challenge.

John Denker wrote:

> 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.

Yes, that's another good approach. Carl
--
Carl E. Mungan, Asst. Prof. of Physics  410-293-6680 (O) -3729 (F)
      U.S. Naval Academy, Stop 9C, Annapolis, MD 21402-5040
mailto:mungan@usna.edu       http://usna.edu/Users/physics/mungan/