Newbie Eyepieces 101

Stephen Paul wrote:
To say that the angular size of an object in the eyepiece is 50x larger than
the angluar size of the unaided eye is also very useful, but I'm now
wondering just what is the correct relationship between the image at the
focal plane and the image "in" the eyepiece.

The linear size in mm of the image in the focal plane, divided by the
angular size in radians of the image in the eyepiece, approximately
equals the focal length of the eyepiece in mm. That's because radians
are dimensionless (except for those eyepieces made by Tele Vue, of
course!)

Essentially, the eyepiece allows you to view the image in the focal
plane as though your eye were at the distance from it equal to the
focal length of the eyepiece (with respect to angular size), except
that naturally, your eye cannot focus at a distance of (say) 6 mm. A
6 mm eyepiece collimates presents a virtual image that is placed at
infinity, so that your eye has no problem focusing on it.

Brian Tung
The Astronomy Corner at http://astro.isi.edu/
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"Brian Tung" wrote in message
...
Stephen Paul wrote:
wondering just what is the correct relationship between the image at the
focal plane and the image "in" the eyepiece.

The linear size in mm of the image in the focal plane, divided by the
angular size in radians of the image in the eyepiece, approximately
equals the focal length of the eyepiece in mm.

Essentially, the eyepiece allows you to view the image in the focal
plane as though your eye were at the distance from it equal to the
focal length of the eyepiece (with respect to angular size), except
that naturally, your eye cannot focus at a distance of (say) 6 mm. A
6 mm eyepiece collimates presents a virtual image that is placed at
infinity, so that your eye has no problem focusing on it.

So, magnification is really a function of the distance of the eye, or film
plane, from the focal plane. In the case of the eyepiece and magnification,
it makes intuitive sense to me that the closer I am to the image the larger
it will appear. However, with imaging, the claim is that the further you
move the film plane back from the focal plane, the greater the magnfication.
(I guess it's time to hit the books.)

With Bill's presentation that a 1200mm focal length objectve presents an
image that is three times larger than that presented by a 400mm focal length
objectve, it is now obvious why the image is larger for a given eyepiece
when used in a longer focal length telescope. That is, because the image you
start with is larger in the longer focal length objective's prime focal
plane. (It is also apparent that my original idea that the eyepiece provides
a fixed magnification is correct, see below.)

I am still missing how we actually determine the "cardinal" (thank you)
ratio of the size of the image in the eyepiece to the size of the image
against the naked eye sky. It seems to me that we must first understand how
we derive the linear size of the image at the focal plane, and then how, and
why the distance from the focal plane changes the angular size of that image
(again I'll need to hit the books to fully comprehend this one).

Be that as it may, allow me to continue to expound upon these ideas. I am
informed that the prime focus photographic magnification is calculated from
the focal length of the lens/mirror, where 50mm is "given" to be 1x, hence
2000mm is considered to be 40x. Does this mean that at 50mm, the image scale
at the focal plane is equal to the naked eye image scale, or is this just an
arbitrary standard value?

If the former, can we then say that a 40mm eyepiece in a 2000mm focal length
telescope, which provides 50x, is magnifying the image at the focal plane by
50x/40x, or 1.25x??

Let's consider a second example. If the focal length of the objective is
400mm, the prime focus magnification is 8x, and that same 40mm eyepiece
yields 10x, which is 10/8, or 1.25x. Based on these two examples, it seems
to hold that a 40mm eyepiece magnifies the focal plane image by 1.25x. :-).

Now, given a 20mm eyepiece in the 2000mm scope, we would have 100x
magnification of the naked eye image, which is 2.5x the photographic
magnification at the focal plane. For the 400mm scope, we would have 20x
magnification which is also 2.5x the photographic magnification at the focal
plane.

So, where are we? The image scale at prime focus = focal length of objective
/ 50, hence the magnification of the eyepiece = ((focal length of objective
/ focal length of eyepiece) / (focal length of objective / 50). So, if my
algebra doesn't fail me here, (x/y)/(x/z) = z/y, and we can determine the
magnification of any eyepiece by dividing 50 by the focal length of the
eyepiece.

Not that I'm sure _what_ we gain from this, but is it right?

-Stephen Paul

On Sat, 19 Jul 2003 21:11:09 -0400, "Stephen Paul"
wrote:

So, magnification is really a function of the distance of the eye, or film
plane, from the focal plane.

No, not really; though for the eye the effect is "as if" the eye were
a certain distance from the focal plane. For film, the film plane
must be coincident with the focal plane. IOW, the real image must be
placed precisely onto the photographic emulsion (or CCD chip).

It's been a long time since I've dabbled with astrophotography; but
magnification for photographic purposes is *not* the same as
magnification for the visual observer. For photographic and CCD work
image scale is generally more important than any dimentionless
magnification concept. The raw image can always be enlarged provided
it possesses sufficient detail.

Any basic treatise on astrophotography or CCD imaging ought to cover
the simple mathematical details concerning image scale.

I am still missing how we actually determine the "cardinal" (thank you)
ratio of the size of the image in the eyepiece to the size of the image
against the naked eye sky.

(Focal length of objective)/(focal length of eyepiece) *is* that
ratio. Performing the implied division results in the dimensionless
number we call "magnification". That first ratio is equal to this
second ratio: (apparent angular size as seen in the telescope's
eyepiece)/(apparent naked eye angular size).

Be that as it may, allow me to continue to expound upon these ideas. I am
informed that the prime focus photographic magnification is calculated from
the focal length of the lens/mirror, where 50mm is "given" to be 1x, hence
2000mm is considered to be 40x. Does this mean that at 50mm, the image scale
at the focal plane is equal to the naked eye image scale, or is this just an
arbitrary standard value?

To the best of my knowledge the 50mm equals 1x thing is due to
photographic history -- most camera lenses had a focal length around
50mm.

If the former, can we then say that a 40mm eyepiece in a 2000mm focal length
telescope, which provides 50x, is magnifying the image at the focal plane by
50x/40x, or 1.25x??

The standard approach is to treat photographic and visual
magnifications separately.

In the world of astronomy we don't assign a magnification to
individual eyepieces. Instead we label eyepieces by their focal
lengths. The advantage of this comes from the simple formula:
Magnification equals (focal length of objective)/(focal length of
eyepiece).

Nevertheless, there *are* alternative ways of looking at magnification
-- as you've demonstrated in this posting.

Let's consider a second example. If the focal length of the objective is
400mm, the prime focus magnification is 8x, and that same 40mm eyepiece
yields 10x, which is 10/8, or 1.25x. Based on these two examples, it seems
to hold that a 40mm eyepiece magnifies the focal plane image by 1.25x. :-).

Now, given a 20mm eyepiece in the 2000mm scope, we would have 100x
magnification of the naked eye image, which is 2.5x the photographic
magnification at the focal plane. For the 400mm scope, we would have 20x
magnification which is also 2.5x the photographic magnification at the focal
plane.

So, where are we? The image scale at prime focus = focal length of objective
/ 50, hence the magnification of the eyepiece = ((focal length of objective
/ focal length of eyepiece) / (focal length of objective / 50). So, if my
algebra doesn't fail me here, (x/y)/(x/z) = z/y, and we can determine the
magnification of any eyepiece by dividing 50 by the focal length of the
eyepiece.

Not that I'm sure _what_ we gain from this, but is it right?

Your definition of image scale (IIRC) is different from the
traditional (degrees of sky per millimeter at the focal plane)
definition. We don't normally assign magnifications (independent of
any objective) to our eyepieces. Using your formula the magnification
of an eyepiece has the units of "millimeters" -- unless you add
"millimeters" as the unit for the number "50". IOW, your formula is
valid once you add millimeters to 50.

Perhaps on some other planet astronomers have adopted your approach as
their standard for magnification; but on this planet the consensus is
a somewhat different approach.

Now let's drop this and all get outside and do some observing ;-)

Bill Greer

"Bill Greer" wrote in message
...
On Sat, 19 Jul 2003 21:11:09 -0400, "Stephen Paul"
wrote:

I am still missing how we actually determine the "cardinal" (thank you)
ratio of the size of the image in the eyepiece to the size of the image
against the naked eye sky.

(Focal length of objective)/(focal length of eyepiece) *is* that
ratio. Performing the implied division results in the dimensionless
number we call "magnification". That first ratio is equal to this
second ratio: (apparent angular size as seen in the telescope's
eyepiece)/(apparent naked eye angular size).

Thanks Bill.

(Sorry about the silliness that came later on. Although a reasonable mental
exercise for me, I'm not sure what benefit others might have gained.)

FWIW, I did get out and do some observing from 11PM Saturday to 1:30AM
Sunday on deep sky, and then from 1:30AM to 4AM (same session) on Mars,
until clouds cut my session short.

It was the best 5 hours of sky time I've had in over a month.

-Stephen