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Re: ray tracing assumptions



On Thu, 18 Apr 2002, Wolfgang Rueckner wrote:

I don't agree with this. Mirror ray diagrams have the virtue that
you're always dealing with real light rays traveling in the direction
of the propagation of the light.

So do lens ray diagrams. I don't know what you are implying by
this. It is true that students often get it wrong by drawing
diagrams as if rays head TOWARD the optical element FROM
the location of virtual objects. This is precisely why I like to
emphasize the vergence properties of INPUT and OUTPUT rays over
the LOCATIONS of objects and images.

Thus, even if the image from the first mirror is a virtual
image and resides behind the mirror (where there's really no
light at all), it doesn't matter conceptually because the
light rays you are dealing with have bounced off that mirror
and are heading towards the next mirror in an easily
understood direction -- they're coming from the direction of
that virtual image.

Again, I simply can't understand the distinction you are trying to
make here; it works exactly the same way for lenses.

But in the case of lenses, one runs into a conceptual problem when
trying to draw ray diagrams that you don't run into with mirrors. I
think that you and John Denker are sweeping this issue under the rug
in your reply. Maybe it would help to give a specific lens
configuration:

O.K. Let's see.

When a lamp is placed 42 cm to the left of a particular convex lens,
suppose a real (inverted) image is formed 37.5 cm to the right of the
lens. The lamp and convex lens are kept in place while a concave
lens is mounted 15 cm to the right of the convex lens. A real image
of the lamp is now formed 35 cm to the right of the concave lens.
What is the focal length of each lens?

Applying the lens equation to the convex lens, one obtains +19.8 cm.
Applying the lens equation again to the concave lens (using the image
of the convex lens as the object of the concave lens), one obtains
-63 cm.

All good.

But one runs into a conceptual issue (which I tried to explain
in my first query) if you try to draw the ray diagrams for
this situation. Try it!

I have and I simply don't understand what you are referring to.

The only answer I've come up with is that you can't solve a problem
by drawing ray diagrams when the object of a lens is a virtual object

This is simply not true. You can do it for mirrors and it works
exactly the same way for lenses.

-- that is to say, the object is on the same side of the lens in
which the light is headed.

I don't understand this.

To put it another way, you can only draw the diagrams using
rays which are headed in the real direction of the propagation
of light.

Yes; of course. And if you do so properly, you will get the
right answer.

That conclusion may seem obvious, but I've never heard that
rule nor seen it in any textbook.

?

Can you produce a ray diagram showing the difficulty you imagine
and put it on the web so that we can look at it. In the meantime,
you might want to look at page seven of a talk I gave at AAPT
several years ago. The materials are avilable at

<http://www.csupomona.edu/~ajm/professional/talks/insandouts.pdf>

John Mallinckrodt mailto:ajm@csupomona.edu
Cal Poly Pomona http://www.csupomona.edu/~ajm