naive question about FOV

I understand magnification is determined by objective & eyepiece focal
lengths. So doesn't a given magnification "lock you in" to a certain field
of view? Looking through any two eyepieces that give 40x magnification
should show the same swath of sky, right?
Of course I know this is not the case as there are some really nice wide FOV
eyepieces out there, so my question is, what am I missing (or how are they
beating the mathematics ?

I understand magnification is determined by objective & eyepiece focal
lengths. So doesn't a given magnification "lock you in" to a certain field
of view?

Looking through any two eyepieces that give 40x magnification
should show the same swath of sky, right?

Of course I know this is not the case as there are some really nice wide FOV
eyepieces out there, so my question is, what am I missing (or how are they
beating the mathematics ?

Field of View and magnifaction are two different things.

Each eyepiece has a different Apparent Field of View (AFOV.) To a first order
this is determined by the focal length of the eyepiece and the diameter of the
field stop of the eyepiece.

The field stop is ring at the focal plane of the eyepiece which limits the
field of view. When you look into an eyepiece, and look at the edge of the
Field of View, you are seeing the field stop. Looking into that eyepiece, it
hopefully seems reasonable that the field of view could be made larger or
smaller simply by changing the diameter of the field stop.

The simple approximations a

Apparent Field of View = (57.3 radians/degree) x Field stop diameter/Focal
length eyepiece

True Field of View = Apparent Field of View/Magnification...

jon

how are they
beating the mathematics ?

They are not beating the mathematics. These lenses are complex and are able to
maintain a flat FOV over a greater angle.

This page on the Tele Vue site shows how field of view, magnification, =
etc, are calculated...

http://televue.com/engine/page.asp?ID=3D107


-Florian

IRR wrote:
I understand magnification is determined by objective & eyepiece focal
lengths. So doesn't a given magnification "lock you in" to a certain field
of view? Looking through any two eyepieces that give 40x magnification
should show the same swath of sky, right?
Of course I know this is not the case as there are some really nice wide FOV
eyepieces out there, so my question is, what am I missing (or how are they
beating the mathematics ?

From MOPFAQ (link below):

Q. What is the distinction between true field of view and apparent
field of view?

A. In short, the true field of view tells you how much of the sky you
see, and the apparent field of view tells you how big that much sky
looks when magnified by the telescope (or binoculars).

If you look through a telescope at the Moon, say, at 100x, you'll find
that the Moon more or less fills the field of view. In wide-angle
eyepieces, there may be quite a bit of dark space around the Moon, and
in other eyepieces, you may not be able to squeeze the whole Moon in at
100x, but we can ignore these differences for the moment.

If the Moon exactly fills the eyepiece field of view, the true field of
view is 30 arcminutes, since that is the angular size of the Moon, and
the Moon is filling the field of view. That's how much of the sky you
can see through the eyepiece. It doesn't make any difference that the
patch of sky you happen to be looking at is all Moon--if you rotate the
telescope over to a featureless expanse of sky, the true field of view
is still the same old 30 arcminutes.

However, the Moon certainly doesn't look only 30 arcminutes across. It
looks that big to the unaided eye, but the point of using a telescope
on the Moon is to make it look bigger. How much bigger? Well, 100x
bigger, and 100 times 30 arcminutes, or half a degree, is 50 degrees.
That's how big the Moon, or whatever patch of sky you're looking at,
seems to you, and that therefore is the *apparent* field of view.

That's also the relationship in general:

apparent FOV = true FOV * magnification

This is only an approximate formula, and to be more precise is affected
by a number of optical vagaries, but we don't need to worry about that
here.

Incidentally, the apparent field of view is inherent in the eyepiece.
In fact, if you just look through the eyepiece, without inserting it
first into any telescope, you'll see a sharply defined circle. That
circle's angular size is the apparent field of view. The true field of
view is related to the angular field of view by the magnification, and
so it is *not* inherent in the eyepiece--it also depends on the
telescope you're using the eyepiece with.

Q. How can I measure the apparent field of view of my eyepiece?

A. Here's a method that's worked for me. You need a tape measure and
the ability to use your two eyes for slightly different purposes.

As mentioned in the answer to the preceding question, if you look
through an eyepiece just on its own, without a telescope, you'll usually
see a sharp edge to the field of view. That's because you're seeing the
field stop, which lies at the focal plane of the eyepiece. If the
barrel itself is being used as the field stop, the edge might not be as
clean, but you can still use this method.

Set up in front of an object of known height (or width) h. I use a
doorway that's 80 inches tall. Holding both eyes open, look through
the eyepiece with one eye. Be sure to keep the eyepiece level with
the midpoint of the object. You should see an indistinct field with
the eyepiece eye, and the object with your other eye.

Now, move back and forth until the object is just as tall (or wide)
as the field of view. Measure the distance d between your *eye* (not
eyepiece) and the object. The apparent field of view is then measured
directly as

h
aFOV = 2*atan ---
2*d

Here's a rough ASCII diagram:

___
/ ^
/ |
/ |
/ |
/ |
/ |
/
eyeEP h
\
\ |
\ |
\ |
\ |
\ |
\_v_

|-- d --|

For example, my 6 mm Radian yielded a distance d of 69 inches, so the
aFOV was 2*atan(80/138) = about 60 degrees. I estimate the error on
this method to be on the order of 1.5 degrees, plus or minus.

Brian Tung
The Astronomy Corner at http://astro.isi.edu/
Unofficial C5+ Home Page at http://astro.isi.edu/c5plus/
The PleiadAtlas Home Page at http://astro.isi.edu/pleiadatlas/
My Own Personal FAQ (SAA) at http://astro.isi.edu/reference/faq.txt

Only the mechanical diameter of the field stop in the intermediate
image plane determines the actual FOV. The intermediate image plane
usually is located between the prisms and the eyepiece (in
binoculars). In telescopes it is located inside the eyepiece holder.
With astro eyepieces, the stop is usually part of the eyepiece.
Stop size often is used to restrict the FOV to clip off outer parts of
an image, where it usually gets worser.
Think of the eyepiece as a loupe with which you look at the
intermediate image. The optical design of the eyepiece might include
some (wanted or unwanted) amount of distortion, so the usual formula
FOV x magnification = apparent FOV
is only an approximation.

Hope this helps a bit...

Infinity

"IRR" wrote in message .com...
I understand magnification is determined by objective & eyepiece focal
lengths. So doesn't a given magnification "lock you in" to a certain field
of view? Looking through any two eyepieces that give 40x magnification
should show the same swath of sky, right?
Of course I know this is not the case as there are some really nice wide FOV
eyepieces out there, so my question is, what am I missing (or how are they
beating the mathematics ?

Consider looking at an outdoor senic vista through a cardboard plate
with a 2" hole in the middle. If you hold the plate at 12" distant
you observe a small senic view. If you hold the plate at 1" from your
eye, you see a very large vista. Both vista are at the same magnification.
What is different is the total angle through which light can enter your
eye.

"IRR" wrote in message .com...

I understand magnification is determined by objective & eyepiece focal
lengths. So doesn't a given magnification "lock you in" to a certain field
of view?

No, having the same magnification only means that any given object
you're viewing will appear to be the same size in any eyepiece design.

Looking through any two eyepieces that give 40x magnification
should show the same swath of sky, right?

Eyepieces with wider FOVs will show you a larger portion of the sky
around any given object, although everything in that wider FOV will be
the same size as it would appear in an eyepiece with a narrower field
of view. In other words, you can see more things within a larger
circle of visibility, without making any of the objects smaller.

Of course I know this is not the case as there are some really nice wide FOV
eyepieces out there, so my question is, what am I missing (or how are they
beating the mathematics ?

As you're aware, they're not beating the mathematics, nor is anyone
trying to "pull a fast one" on us. An analogy would be looking
through a hole in a thick fence--the larger the hole, the wider your
FOV will be, even though everything you see through the hole would
appear to be the same size, regardless of the size of the hole.
Instead of seeing only one person's head framed by the edge of a
narrow hole, you could see several people's heads and bodies in a
wider hole, even though their heads are the same size as they would
appear in the smaller hole.

By the way, I've deliberately traded off technical completeness for
the sake of clarity, and I hope this has helped. If you want to go a
little deeper, "apparent" FOV is how wide the picture in an eyepiece
appears to your eye, while "true" FOV is how much of the sky you are
actually viewing ("swath of sky" as you put it). While these two
types of FOV are different things, for a given magnification, an
eyepiece with a wider apparent FOV (specified by its design) will also
give you a wider true FOV.


- Robert Cook

Thanks for all the great answers and links, I think every one helped me
understand from a slightly different perspective -- or field of view .

I understand magnification is determined by objective & eyepiece focal
lengths. So doesn't a given magnification "lock you in" to a certain field
of view? Looking through any two eyepieces that give 40x magnification
should show the same swath of sky, right?
Of course I know this is not the case as there are some really nice wide FOV
eyepieces out there, so my question is, what am I missing (or how are they
beating the mathematics ?

It took me a while to get this too. The key is understanding this:

Stand in front of an object and look at it. Now cover your eyes with
your hands. Slowly move your hands apart. You are increasing your
field of view without changing the magnification.

That's all there is to it. All eyepieces have a certain "field stop",
where the light is cut off by an actual physical barrier (your hands
in the previous analogy). The field stop is put there because the
view is not well enough corrected outside it. In reality, when you
look into any eyepiece, your eye is seeing over 180° apparent field of
view. But the eyepiece's field stop forces the great majority of this
to be black space. Widefield eyepieces simply have a larger field
stop; they use a complex configuration of lenses to allow a larger
area of good correction.

If you were to take your Plossl eyepiece apart and enlarge the
standard 50° field stop to say 80°, the view through much of the field
would be pretty poor. So the manufacturers of Plossls set the barrier
to not allow light past 50°; the rest of the 180°+ field of view that
your eye can see is thus filled with black space.

Cheers,
Ritesh