I will describe this only in terms of a very simple refractor.

A reflector operates virtually identically, except that the light
is reflected

off of a curved mirror instead of being bent through a curved lens.

But all other relevant factors- angles, lengths, etc., are the
same.

Schmidt-Cassegrains, Maksutovs and other similar designs also work
the same from the focal plane to the eye- the only difference being
that
to get the right numbers, we consider the *effective* focal
length
of the system (primary, corrector plate, and secondary) rather than the
simple focal length of the main optic. These complications will
not
be dealt with here.

*(This shows a very simple single-lens
objective.
In reality, to avoid chromatic aberration, any useful telescope requires*

*a more complicated objective. But the basic
concepts remain the same)*

The light focuses at a point A', a certain distance on the other
side
of the objective lens. This distance is the "focal length" of the
objective, which I will denote f*o*. By symmetry, it is
obvious
that the focal point lies on the center line of the scope, marked in
red
above.

Now let's see what happens if the star is "off-axis" by an angle **a
**(marked
in green). I call this star "star B". The light again
focuses
at a point B' on the other side of the objective lens. The most
useful
formula in optics is that the angle of this focus point, relative to
the
centerline, as measured from the objective, is the same angle **a**.
This is not obvious, but if you follow the path of the ray marked in
red,
which passes through the exact center of the lens, then it is obvious
by
symmetry that the angle of this ray after going through the lens is the
same as it was going into the lens. So if the light focuses at
all,
it must focus on this red line, and thus be at the same angle. I
won't bother proving that the light focuses- that is a matter of basic
optics. In fact, off-axis rays do not *quite* focus at a
point,
leading to distortions in the off-center portion of the field, but
these
distortions are usually small.

The diagram below is just a combination of the first two. Note
that the two focal points are both at the same distance from the
objective
lens. The collection of all such points is the *focal plane*.
In fact, the distances aren't all quite the same, so this "plane" is
slightly
curved, which leads to more off-axis distortions. But if the
angle
**a**
is small, as it is in telescope use, the "plane" is close to flat.

If we put a piece of film or a CCD chip at the focal plane, the light concentrated at A' and B' would register as point images on the sensor. Any "extended" object can be thought of as just a collection of points, so the film/chip records an image of anything at a large distance on the other side of the objective (if the distance was smaller, we could still focus, but the focal plane would be further from the objective).

Note the dashed red line marked "x", denoting the distance in the
focal
plane between the centerline and the image at B'. This distance
is
important in figuring out the image scale on a chip or film, or in
figuring
out how large of a field of view can be seen through 1.25" or 2"
eyepieces,
or in issues of vignetting. Calculating x is easy. Since **a**
is a small angle, we can avoid any trigonometry- x is very nearly the
same
as the length of the grren curve marked "a" on the right side of the
objective.
If we measure **a** in radians, this length is **a** times the
focal
length f*o:*

It turns out to be the same as the telescope objective, except that
the off-axis angles are often large, so that the curvature of the focal
plane is evident. Fortunately, the retina is curved too, so that
the images still focus properly.

The incoming rays are shown as parallel, as they would for a star at
very large distance. For closer objects, small muscles change the
shape of the lens so that the image still focuses on the retina.

The eyepiece is also the same as the objective, except that we will
be using it in reverse.

The focal length of the eyepiece is denoted f*e.*

(*As with the objective, we show only a simple
eyepiece
made from a single lens. Any modern eye piece*

*is made from a combination of individual lenses, but
the basic concepts remain the same.)*

The eyepiece is actually just a mgnifying glass. Let's see how that works:

Imagine an object (in blue) at point P. Light (yellow arrows)
shines on it. The scattered light from this (black arrows) goes
in
all directions. Some of this goes through the eyepiece. If
P is at the focal length f*e* from the lens, then it is following
the same type of path as with the objective above, except in reverse,
so
it comes out as a bundle of parallel rays. From there, it goes to
the eye:

Note that the angle **b** is the same on either side of the
diagram.
This is also the same as the angle of the rays relative to the
centerline,
between the eyepiece and eye. I just left that out of the diagram
to avoid cluttering the picture.

While it's not critical to understanding scope optics, let's go ahead and finish understanding how a magnifying glass magnifies: First, let's fill in the picture with an on-axis object. Specifically, put an arrow in the picture, with the foot of the arrow on the centerline, and the head at the point P:

A simple definition of magnification is that it is the ratio of how large something appears, compared to how large it really is. In astronomy, the distance, and hence size, of objects is often hard to determine, so for the purposes of defining magnification, we define "size" as the angular size of an object, or the angle that it subtends, as viewed from the eye (and we'll use an even better definition later).

In the case, the angle **b** appears the same on both sides of
the
diagram, so does that mean there is no magnification? No, the
angle
of the arrow on the left is the angle as measured from the lens, not
the
eye. If you remove the lens, the angle subtended by the arrow, as
measured from the eye, is smaller. So it can be said that a
magnifying
glass works by making objects appear closer. Then you might think
that you wouldn't need a magnifying glass, if you just held the object
closer to your eye. In some cases that's true, but often, if the
eye is put at the position of the lens, it is too close to the object
to
be able to focus properly. The lens converts the diverging rays
from
the object into parallel rays. The eye can focus those as if the
object were at infinity. So the lens effectively lets you see
things
as they would appear if you held them very close to the eye, if you
were
also able to focus on them at that distance.

Light from a star goes through the objective and focuses at a point P in the focal plane. But after focusing, the rays then continue, diverging on the other side of the focal plane, towards the eyepiece. Compared this to the eyepiece acting as a magnifying glass, with an object (indicated by the blue dot) located at the same point in the focal point. Rays from an illumination source scatter off of the object and diverge in all directions, but some of this is in the direction of the eyepiece. The important thing to recognize is that the rays from the star, diverging from the same point, coincide with a subset of the rays from the blue dot. Thus they follow the same path through the eyepiece. Now lets simplify this by getting rid of the blue dot:

This is the path of light from an off-axis star through the objective and through the eyepiece. Since the rays come out parallel, they are easily focused by the eye (you may need to scroll to the right to see the whole diagram):

Parallel light rays from off-axis star B, at angle **a,** go
through
the objective and converge to a focus at point B' in the focal
plane.
Then the rays continue, diverging towards the eyepiece, where they are
converted to parallel rays again, to be focused by the eye lens,
converging
at point B'' on the retina. The light enters the eye at angle **b**,
so the observer sees the star at angle **b** (and inverted).
Thus
the magnification is **b**/**a**. I'll explain soon how to
calculate this in terms of the focal lengths of the lenses, but first I
have to clear up another issue-

There is a problem with our sample telescope: the parallel
bundle
of rays going towards the eye is far too wide to fit through the pupil
of the eye, so much of the light is wasted. This is just because
of the dimensions of the optics involved: the focal length of the
objective
is short, so that the diagram would not be too long, and an eyepiece
with
long focal length was used, so that that part of the diagram would be
spread
out enough to avoid clutter. These conditions, combined with the
aperture of the objective, lead to a wide *exit pupil *of the
scope/eyepiece
combination. In a real telescope, this is only a problem when
very
low power eyepieces are used.

The concept of exit pupil becomes more clear if we add a few more stars (scroll to the right again):

Three stars are shown here: one on axis, and two off axis, one above
and one below the axis. The rays focus at three points in the
focal
plane, then go through the eyepiece to form three bundles of parallel
rays,
each at a different angle. The point where all three bundles
intersect
is the exit pupil, shown as a brown line. This is where the eye pupil
should
be placed so that the light from all three stars goes into the
eye.
The distance from here to the rear surface of the eyepiece lens(es) is
the *eye relief* distance.

These are all the pictures needed to understand the basic optical
function
of a telescope.

Now we'll start deriving the basic equations of telescope use from
the geometry in the diagrams.

So: f*o**tan **a = **f*e**tan **b**

or: f*o*/f*e *= (tan **b**)/(tan **a**)

If the angles are small, we can use the small angle approximations
tan
**a
**=
**a**
and tan **b **= **b** to get

f*o*/f*e* = (tan **b**)/(tan **a**) = **b**/**a**

But **b**/**a** is the magnification, so:

For much of the history of astronomy, the apparent field of view (AFOV) of most eyepieces was small enough that the small angle approximations could be used without much quantitative error. That meant you could use the notion of magnification being defined by angular size (in older textbooks, you will hear of an object being magnified a certain number of "diameters"). So, for example, if you wanted to know whether the full moon could be seen in an eyepiece of a given focal length and AFOV, you could just multiply the moon's angular size (about 0.5°) by the magnification, and see if that was smaller that the AFOV. With modern wide-angle eyepieces, it's not that simple. For example, consider viewing the moon through an 8" scope with a focal length of 2000mm, using a 10mm eyepiece with an 84° AFOV. The magnification is 2000/10 = 200, so a simple calculation suggests that the apparent magnified size of the moon would be 0.5° X 200 = 100°, and wouldn't fit in the eyepiece view. But if the moon was actually 200X larger (or closer), it would subtend an angle of only about 82°, so it will fit in the eyepiece view. This issue will come up again when we discuss the field stop and the field of view of the eyepiece.

It should be noted that modern wide angle eyepieces typically have a
lot of "geometric" distortion, meaning that the magnification is not
uniform
over the whole field. This is probably due to the difficulty of
correcting
for a sharp view over a wide field. The eyepiece has to deal with
significant amounts of "field curvature" of the focal plane, coma, and
other off-axis aberrations, and correcting for that leads to other less
aesthetically important distortions.

As a result of the geometric distortions, manufacturer's specs for
field stop size (see below) may not be quite what the calculations here
would suggest.

We covered this above (figure 3): a star at angle **a**
focuses
in the focal plane at a distance x = **a***f*o* from the
centerline.
For practical purposes, it is more convenient to express the angle in
degrees.
The formula then is x = **a***f*o*/57.3. If you know the
size of the chip or film being used for imaging and want to know the
angular
field of view that can be imaged, invert the formula to get **a **=
57.3*x/f*o *degrees, where the size x of the film or chip is
measured
in millimeters. For CCD, the answer is more conveniently given in
arcminutes, so we multiply the formula by 60: **a = **3438*x/f*o*arcminutes.
It can also be useful to know the resolution of a CCD chip in units of
arcseconds per pixel. Pixel size is generally measured in microns
( =1/1000 mm), so after converting units, the angle per micron is **a**
= 206*x/f*o* arcseconds.

If the eyepice focal length is known, it is possible to calculate
the
field stop size from the AFOV, or vise versa: the maximum
possible
value of **b** would be **b**max =
(1/2)*AFOV,
and the maximum value of x would be xmax =
(1/2)*(field
stop diameter). Basic trigonometry gives x = f*e**tan
**b**.
The small angle approximation is not generally valid here, but for
narrow
field eyepieces, we could try it anyway: x = f*e**tan
**b ~= **f*e****b**.
Then field stop diameter = 2*xmax
= 2*f*e****b**max
=
f*e**AFOV. Here, the angle AFOV is in radians. Using
degrees,
the formula would be:

but keep in mind that this is not very accurate for modern eyepieces, especially wide-angle eyepieces.

This turns out to be very simple: The diameter of the beam at the objective is just the aperture of the objective, which I'll call D. The diameter of the exit pupil is essentially just the diameter of the beam at the eyepiece, which I'll call E. Since the rays converge towards a point in the focal plane, and then diverge at the same angle towards the eyepiece, simple proportions give us

* * D/f*o = *E/f*e*

or D/E = f*o*/f*e*

But f*o*/f*e* = magnification, so:

That version of the formula is easy to remember, but in practice, a
more convenient way to calculate this is to use the *focal ratio*
of the objective, defined as f*o*/D. I'll call it fr.
Then

E = D/(f*o/*f*e)* = f*e**(D/f*o*) = f*e*/(f*o*/D)
, so

Using a very long focal length eyepiece, or a very low focal ratio
scope, or a sufficient combination of the two, can give an exit pupil
wider
than the eye pupil. Simple ray tracing backwards through the
system
shows that this is equivalent to using an on-axis aperture mask, so the
brightness of the image is reduced.

In scopes with a central obstruction, ray tracing shows that each of
the parallel bundles of rays has a hole in it, whose diameter, compared
to the exit pupil size, is the same ratio as the size of the central
obstruction,
compared to the scope's aperture. All of these "holes" overlap in
the center of each bundle when the parallel bundles intersect at the
eye
relief distance behind the eyepiece.

A large exit pupil occurs at low power, and low power requires a large
eyelens in the eyepiece, and typically gives a long eye relief
distance.
In this situation the eye has little physical reference for lining
itself
up properly behind the eyepiece, and as the eye wanders around, odd
effects
can be seen. If the eye is too close or too far from the
eyepiece,
and off-center, it will see more light from one side of the field than
the other, producing an effect commonly known as "kidney
beaning".
When the exit pupil is larger than the eye pupil, and if the eye
centered,
but too close or too far from the eyepiece, it is also possible for it
to see more light from the edge of the field than the center, giving an
effect commonly known as seeing "the shadow of the secondary".

Another drawback to a large exit pupil is that the entire cornea and lens in the eyeball are used, but they frequently have aberrations near the edge, degrading the view.

A very small exit pupil can cause problems, because the parallel bundles of light entering the eye are so narrow that any tiny particles in the transparent vitrous humor in the eye can block them, so you see the shadow of these particles floating around in the view. These particles (some sort of organic debris in the eye, I pressume) like to form long curving chains or clumps, and are commonly known as "floaters".

[pi*(D'^2)/4]/[pi*(D^2)/4] = (D'/D)^2

So the brightness per area at the focal plane changes by that ratio.

Now if we use yet another scope with focal length f*o*', the image
scale is changed by the factor f*o*'/f*o, *so the area of
the
image is changed by the factor (f*o*'/f*o*)^2, so the
brightness
per area is changed by the inverse of this, (f*o*/f*o'*)^2.

Combining this, we see that the change in brightness per area at the
focal plane, when replacing a scope of aperture D and focal length f*o*
with a scope of aperture D' and focal length f*o*', is (D'/D)^2*(f*o*/f*o'*)^2
= [(D'*f*o*)/(D*f*o*')]^2 = [(f*o*/D)/(f*o'*/D')]^2
= (fr/fr')^2, where fr is the focal ratio f*o/*D.

For visual use, assuming that the exit pupil is smaller than the eye pupil, then all of the light from an object (or the part of an object within the field of view of the eyepiece) that goes into the telescope's objective goes into the eye, so the total brightness, compared to the naked eye view, is multiplied by the ratio of area of the objective to the area of the eye pupil. If the eye pupil diamter is denoted by Ep, and the objective aperture is D, this ratio is

[pi*(D^2)/4]/[pi*(Ep^2)/4] = D^2/Ep^2 = (D/Ep)^2

For stars, where the angular size is so small that they appear as points even when magnified, that means that the star is simply brighter by that ratio. Note that this breaks down when the magnification is so high that we see the Airy disc.

For extended objects, ie, those that aren't points, the situation is
a bit different. Now we are concerned with *apparent* *surface
brightness, *which is the* *brightness per apparent angular
area.
Consider an object large enough that it appears as an extended object
even
without magnification, such as the moon. The total brightness of
the part that appears in the field of view is increased by (D/Ep)^2 as
before, but the area of this image is larger by the ratio
(magnification)^2.
So the apparent surface brightness is changed by the ratio
[(D/Ep)/magnification]^2.
But recall that, if the exit pupil diameter is denoted by E, then
magnification
= D/E, so this ratio is [(D/Ep)/(D/E)]^2 = (E/Ep)^2. Surface
brightness
is important for deep-sky objects, which usually have a very low
surface
brightness, so lets see how to increase the apparent surface
brightness:
from the formula brightness ratio = [(D/Ep)/magnification]^2, the two
ways
to do this are to use a larger aperture scope, or to use lower
power.
Either of these choices increases the exit pupil diameter. Note
that
when E = Ep, the surface brightness ratio is only 1, so for smaller
exit
pupils, an object's apparent surface brightness is actually less than
it
is when viewed with the naked eye. If E is increased by using a
larger
aperture, then when it is increased beyond E = Ep, the aperture mask
effect
discussed earlier (when a too-wide exit pupil is used) cancels out any
brightness advantage of using a larger aperture. On the other
hand,
if we increase E by using lower magnification, then the lower
magnification
would compress the image, and increase the apparent surface brightness
by the square of the ratio of the new magnification to the old.
But
from E = D/magnification, we see that if we go beyond E = Ep, this
again
is exactly cancelled by the light loss that results from using an exit
pupil that is too wide.

So the result is that the apparent surface brightness of extended objects is never higher than when the object is viewed with the naked eye. This may seem counter intuitive, because most deep-sky objects viewed through a typical amateur scope are in fact large enough to be resolved with the naked eye. If the telescope doesn't increase the apparent surface brightness, why are they visible through the scope but not with the naked eye? The answer is simply that for very faint objects, the eye does not resolve details well (note that deep sky objects viewed through a telescope are generally seen with much less detail than can be obtained in a long exposure image made through the same scope), so these objects have to be magnified to be detected at all.