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Re: Cartesian Sign Convention for Optics



On Friday, Mar 28, 2003, at 12:28 US/Eastern, Tina Fanetti wrote:

>> How many of you use the Cartesian sign convention for optics?
>> I am used to the sign convention that has virtual images being
>> negative and object distances being positive.
>>
>> I am wondering which to use with my calculus physics students.
>> Their book uses the Cartesian sign convention....

I vote for Cartesian. More discussion below.

On 03/28/2003 01:03 PM, Ludwik Kowalski wrote:
>
My suggestion is to use the same convention as in their textbook.

That's one consideration (but not a hard-and-fast rule).
Fortunately in this case, the stronger arguments
point in the same direction.

For reasons to be discussed below, even if the book
didn't use the Cartesian representation, I would be
tempted to introduce it anyway.

I usually spend
some time to show that conventions are arbitrary.

Yes, conventions are arbitrary.

But sometimes when choosing a representation or a
notation, one faces deeper choices. Sometimes one
representation is objectively much better than
another for capturing the physics.

One possibility is to keep all distances positive.

Aha. That gets right to the point. In my book,
!!all!! distances are positive. That's part of
the definition of "distance". Similarly all
lengths are positive.

But the point is, the quantities that appear in
the thin-lens formula are not distances. They
can't be distances, because they have nontrivial
signs. Similarly, the "focal length" isn't
properly a length, because it has a nontrivial
sign.

In fact, the quantities of interest are vectors.

For thin lenses and paraxial rays, they are
vectors in D=1, which is such a simple vector
space that we tend to not pay much attention to
the vector properties, but maybe it's time to
look more closely.

In the thin-lens formula
1/a - 1/b = 1/f (1)
people may squirm because there are vectors in
the denominator, but it is easy (particularly
in D=1) to define
1/x := x / |x|^2 for all x (2)
as you can easily verify by multiplying both sides
by x. The RHS is a vector divided by a scalar,
which is a vector, so clearly 1/x is a vector, too.

The minus sign in equation (1) is not negotiable.
This is easy to see by considering the case of
a trivial lens with no thickness and no curvature,
i.e. 1/f = 0. Then clearly the image and object
must be at the same place, and must be described
by the same vector: a = b.

I emphasize that a and b are not distances but
rather vectors describing the location of the
image and object. You are free to write
equations involving |a| and |b| if you like,
but they are more complicated and less useful
than equation (1).

==============

Tangential remark: If the dimensionality D is
arbitrary (not limited to D=1) you can still take
the reciprocal of a vector using equation (2).
Clifford Algebra and all that.