Newbie Eyepieces 101

On Sat, 19 Jul 2003 12:25:05 -0400, "Stephen Paul"
wrote:

I have to admit though, I am a little more confused now about the term
magnification as it concerns the image at the focal plane. I picture the
eyepiece being a fixed microscope under which the image at the focal plane
is presented by the telescope objective.

That is essentially how an eyepiece functions.

Although just now thinking about
it, I can see that this couldn't really be an accurate way to define
magnification, since doing so means that the eyepiece always magnifies the
image at the focal plane by the same amount, and cleary (I think) this isn't
true, since different focal length objectives dictate the magnification
provided by the eyepiece.

Any given eyepiece will magnify the (real) image at any telescope's
focal plane by the same amount; but the size of the image at the focal
plane will be different for telescopes of different focal lengths.
Thus the same eyepiece, even though it magnifies different real images
by the same amount, will result in different magnifications when used
with telescopes of different focal lengths.

IOW, the focal length of the telescope is just as important as the
focal length of the eyepiece when it comes to computing the
magnification of the system.

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 image at the focal plane is a "real" image. The image in the
eyepiece is a "virtual" image. The relationship between the two is
very much like that between a postage stamp (a real object) and the
enlarged virtual image of the postage stamp that a magnifying glass
provides.

Bill Greer

"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

Awesome thanks!

"Stephen Paul" wrote in message
...
"BenignVanilla" wrote in message
...
Does anyone have a FAQ or care to engage in a short Eyepieces 101
thread?
I
have an 8'' Harden Optical Dob. I am very fond of the scope. I bought
the
biggest bucket I could afford, now I would like to learn more about
eyepieces, in the hopes of adding to my default set. I don't know where
to
start.

Short thread on eyepieces? Surely you jest.. :-)

Where to start?

First some (generalized) definitions:
Aperture - the diameter of the telescope's objective lens/mirror (unless
intentionally "stopped" down by a mask of some type, where the aperture is
then the diameter of the stop).

Focal Plane - the point at which all rays coming from the objectve
lens/mirror create a representation of the target image. (A "crossing of
the
beams", so to speak.)

Focal Length - the distance light rays must travel to reach the focal
plane

Focal Ratio - the ratio of a telescope's aperture, to its objective's
focal
length.

Magnification - an ordinal value that indicates the number of times the
image at the focal plane is made larger, measured in diameters. Example:
50x
means that the image "in" the eyepiece is 50 diameters larger than the
image
at the focal plane.

Apparent field of view (rating for eyepiece) - the field of view you would
get on the sky, when (if you could see) looking through an eyepiece
without
a telescope

True field of view - the field of view you get on the sky when the
eyepiece
is in the telescope. (And, not just any telescope, but a telescope of
specific focal length, and all other scopes of same focal length).

Eye lens (eyepiece) - the eyepiece lens closest to the eye.

Field lens (eyepiece) - the eyepiece lens closest to the telescope's
objective lens/mirror.

Field stop (eyepiece) - the diameter to which the field lens is restricted
by the eyepiece manufacturter

Eye relief (eyepiece) - the distance from the eye to the eye lens, where
the
entire field of view is visible in the eyepiece

Exit pupil - the diameter of the light cone at the point where the entire
field of view is visible in the eyepiece (the eye relief point), when in
the
telescope (And, not just any telescope, but a telescope of specific focal
ratio, and all other scopes of same focal ratio).

Eye pupil - the diameter of your eye's pupil, given the ambient light
conditions. Below age 40, consider 7mm to be the maximum, above age 40
consider 5mm to be the maximum, for fully dark adapted eyes. YMMV.

The formulae:
(1) Magnification = focal length of telescope / focal length of eyepiece

(2) Exit pupil = focal length of eyepiece / focal ratio of telescope

(3) True field of view = (diameter of eyepiece field stop / focal length
of
telescope) * 57.3
(3a) True field of view (approx.) = apparent field of view of eyepiece /
resultant magnification of eyepiece in a telescope of specific focal
length

Of primary importance when selecting eyepieces is using the best exit
pupil
for the job. I think it was David Knisely who posted a short document on
useful magnifications, where he lists magnification per inch of aperture,
which in turn dictates exit pupil. This list, as well as others, indicate
what exit pupils are reasonable for the different objects one views in a
telescope.

The five steps to selecting an eyepiece:
1) Determine which exit pupils to work with (find the aforementioned
documents)
2) Based on your telescopes specifications, determine what focal length
eyepieces will give you those exit pupils.
3) Find which currently available eyepieces are closest to those focal
lengths.
4) Beg and borrow eyepieces in those focal lengths, but of different
designs, and try them
5) Select the ones that you can afford, keeping in mind that a premium
widefield eyepiece is worth every penny if you have the money to spare,
but
that the primary goal is the correct focal length.

Final comment: wide field eyepieces are more important in undriven scopes
than in driven scopes. Consider the cost of an inexpensive EQ Platform for
your Dob when considering eyepieces. I have a set of very expensive wide
field eyepieces that I use in my Dob, but there are times when I'd rather
have tracking. In particular, the wide field eyepieces are important at
low
powers, whether the scope is driven or not, as they aid in finding, and
can
frame some of the larger objects. However, at higher powers, even the wide
field eyepieces, leave you with the requirement to nudge your scope along
quite frequently. Since high power views are generally aimed at planets
and
small objects, where a large field of view isn't required, before spending
$900 on three higher power wide field eyepieces, consider spending $600 on
an EQ Platform, and $300 on three narrow, high quality eyepieces of a
simpler design.

If those numbers just blew you completely out of the water, in the
interim,
Plossls are good eyepieces with a 50 degree AFOV. Other than the increased
nudging frequency, I still like the actual images in my 13mm, and 6.4mm
Plossls, which compare pretty favorably to the Naglers.

--
-Stephen Paul