Re: Fresnel Lens

[Leigh Palmer <palmer@SFU.CA>]

At 7:47 AM -0700 4/15/00, John Denker wrote:
> At 12:15 AM 4/15/00 -0700, Leigh Palmer wrote:
> > John Denker supposes that Fresnel lenses are figured to within a
> > wavelength of light or fraction thereof. That is not the case.
>
> OK, I'll bite.  Yes, I thought all lenses and mirrors were figured to
> within a fraction of a wavelength.  Multiply by (n2/n1 - 1) if you're picky.

That's not the case. Lenses for application to collimation, as is
the case with most Fresnel lenses, are frequently cast instead of
being ground.

> 1) Suppose we have a lens, any lens that deserves the name lens.
> 2) Suppose we divide it into a hundred or more regions (Fresnel-like rings
> or otherwise) and introduce random phase shifts between regions.  The phase
> shifts are large compared to 2pi.
>
> Could somebody please explain the _physics_ of how this works?  Why do not
> the randomly-phased contributions add to near-zero?

The amplitudes of andomly phased contributions usually add to the
square root of N, where N is the number of mutually coherent but
randomly phased rays of unit flux. That is the case here as well.

The physics is straightforward. Consider monochromatic light which
originates in a point source. The light is incident upon the lens
in question. Each small portion of the light is bent by a small
prismatic portion of that lens. It must then head in more or less
the direction indicated by Snell's law.

Each small prismatic reqion of the lens may be considered to be
optically ideal to, say, 1/10 wavelength for our purposes. Any
old piece of lumpy glass with a smooth surface (a cast drinking
glass, for example) will satisfy this condition over a sufficiently
small region which is still many wavelengths in extent in the
direction transverse to the direction of propagating of the
transmitted light. After all, one can still see an image through a
cast drinking glass, though a distorted one. The case of a drinking
glass with a frosted surface is quite another matter, of course.

When the various small portions of light arrive at the spot
preordained by Snell's law they will, of course, interfere with one
another. A complex interference pattern will be formed near the
point designated by Snell's law, but the lateral spread of the
pattern will be governed by diffraction related to the size of the
transverse extent of the portions of the lens which can be
considered to be ideal. If those regions are smaller, the lateral
confusion in the image will be relatively larger. If the lens is
perfect over its entire surface (a criterion which must be related
to the positions of the image and object points as well) the one is
said to have achieved the diffraction limit of imaging. Even with
an optically perfect aplanatic image formed by a lens of finite
aperture ther will still be an interference pattern at the image.

> > M. Fresnel invented these devices to collimate lighthouse beacons,
>
> OK.  But he is also credited with greatly advancing the wave theory of
> light, and with inventing Fresnel zone plates (which absolutely depend on
> proper phasing).  Why would such a person pass up the opportunity to
> properly phase the rings in his lenses?

The hard fact is that there is no way to phase the rings as you
suggest without reference to a specific imaging application. It
was not his intent to do so.

> > not to make high resolution optical images, and they don't.
>
> OK, but that doesn't prove that the rings are randomly phased.

You are being unreasonable. I can't be expected to prove that
proposition. The rings aren't phased and they can't be, even in
principle. Isn't that sufficient?

> Suppose we have one of these alleged lenses with randomly-phased
> rings.  What happens to the energy that (because of destructive
> interference) does not go into the main image?
>   a) Does it go into "nearby" locations in the image plane, which might be
> somewhat useful, or
>   b) Does it go into random far-away side lobes, which are useless for
> lighthouses and for every other purpose I can imagine?
>
> I suspect (b).

As I have already explained, a) is a pretty good answer.

> > For a plastic lens this means it can be formed from sheet stock.
>
> If the plastic is good enough to implement micron-scale figuring _within_ a
> given ring, is it not good enough to implement micron-scale control over
> the step height?

Well, that is a question you can think about if you know something
about creep or thermal expansion, or if you have ever seen a
Fresnel lens made of a flexible sheet material that, because it is
nonrigid, can't possibly be held to micron tolerances across its
lateral extent.

Leigh