Scope Optics

This page describes how a telescope forms an image, and how that image appears magnified to the eye.

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.

Objective Lens

We start with a diagram of light from a distant star going through the main (objective) lens.  The light radiates out in all directions from the star, but the distance is so great that we can assume that the light going into the objective lens is in the form of bundle of parallel rays.  In this view, the scope is aimed directly at the star, so the star is "on-axis".  I call this star "star A".

(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 fo.  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 fo:

 x = a*fo

If the last two sentences confused you, click here.


Before discussing what the eyepiece does, let's take a quick look at your eyeball:

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 fe.

(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 fe 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:

Background light (yellow arrows) shines on the arrow.  Some of the scattered rays from the head of the arrow go through the lens and focus on the retina just as before.  Similarly, some of the rays from the foot of the arrow go through the lens and focus at the center of the retina.  I left those rays out of the diagram to avoid clutter, since it's fairly obvious where they focus.  The image of the arrow focuses on the retina as shown in the diagram.

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.

Now back to telescopes:

The basic story is that the eyepiece magnifies the image at the focal plane of the objective, just as if it were a real object.  The picture of that turns out to be very simple:  just superimpose the diagram of the objective with an off-axis star, with the diagram of the eyepiece with an object at point P:

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.


Recall the discussion of figure 13.  Light from a star at angle a entered the eye at angle b.  In figure 3, there is also light from an on-axis star, which would focus on-axis at the center of the retina.  Thus the real separation between the stars is angle a, while the apparent separation is b.  So the magnification is b/a.  In figure 3, the distance between the center line and the point in the focal plane where the light from the off-axis star focuses is x.  So fo*tan a = x.  But the distance from the eyepiece is fe, and the angle there is b, so we also have x = fe*tan b (see figure 7, where I labelled the off-axis distance in the focal plane with the letter y).

So:  fo*tan a = fe*tan b
or:  fo/fe = (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

fo/fe = (tan b)/(tan a) = b/a

But b/a is the magnification, so:

Magnification = fo/fe

An obvious problem is that, while the angle a may be small enough for the small angle approximation to be reasonably accurate, the angle b often is not.  So our formula is only valid near the center of the field.  But for larger angles, we need only reconsider our definition of magnification.  A real object at a finite distance would not subtend twice the angle if it were suddenly to become twice as large, at the same distance, or if we moved it twice as close.  Our definition of magnification in terms of angles works well when the angles are small, but more generally, we can define it in terms of how close an object appears.  The formula magnification = fo/fe remains valid with this definition.  I'll leave it to you to convince yourself that this is true.

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. 

Image Scale

(especially useful for imaging)

We covered this above (figure 3):  a star at angle a focuses in the focal plane at a distance x = a*fo from the centerline.  For practical purposes, it is more convenient to express the angle in degrees.  The formula then is x = a*fo/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/fo 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/foarcminutes. 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/fo arcseconds.

Field of View/Field Stop

If the eyepiece consisted of only lenses, with no physical structure around them, the AFOV would be very wide, but off-axis aberrations would make the outer part of the field unusable.  So eyepiece manufactures include an annular blocking ring built into the barrel, at the eyepiece's focal plane (which, when the eyepiece is properly adjusted for focus in the telescope, coincides with the focal plane of the telescope's objective lens).  This cuts off the outer part of the field, so that only the sharp inner part of the image is visible.  For a given optical eyepiece design, the choice of field stop size is a judgement call.  Some discount manufacturers may use a larger field stop so that they can claim a wider AFOV, but the image will be soft around the edges.


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 bmax =  (1/2)*AFOV, and the maximum value of x would be xmax = (1/2)*(field stop diameter).  Basic trigonometry gives x = fe*tan b.  The small angle approximation is not generally valid here, but for narrow field eyepieces, we could try it anyway: x = fe*tan b ~= fe*b.  Then field stop diameter = 2*xmax = 2*fe*bmax = fe*AFOV.  Here, the angle AFOV is in radians.  Using degrees, the formula would be:

Field stop diameter (in mm) = fe*AFOV/57.3

or:  AFOV = 57.3*(field stop diameter)/fe

This formula is not very accurate, especially for wide-angle eyepieces. Not only are the small angle approximations not valid, but off-axis aberrations affect things.  In real-world eyepieces, magnification is not quite uniform across the field (leading to geometric distortions described rather figuratively as either pincusion or barrel distortion, depending on whether the magnification increases or decreases as you move from the center towards the edge of the field), and this is especially true for wide-angle eyepieces (see the discussion at the end of the "magnification" section above).  So in practice, you can either measure the field stop directly, using calipers, or measure it using the true field of view (TFOV), ie, the real (unmagnified) angular size of the field of stars seen through the scope.  With a  computerized scope, you can do this by checking the declination reading, while slewing north or south as a star moves across the field, through the center of the field.  With a non-computerized scope, aim at the celestial equator and turn off the clock drive. Time a star as it drifts across the field, through the center.  1 second of RA on the celestial equator is 15 arcseconds, so either multiply by 15 to get the TFOV in arcseconds, or divide by 4 to get the answer in arcminutes.  Once the TFOV is known, the calculation of field stop diameter is just as before, except that we use the angle a rather than b, the objective focal length fo rather than the fe, and TFOV rather than AFOV (actually, this is the same as the image scale calculation).  a is small enough that the small angle approximations are valid, so, with TFOV in degrees, we have:

 Field stop diameter (in mm) = fo*TFOV/57.3

or:  TFOV = 57.3*(field stop diameter)/fo

Note that if we divide the TFOV formula by the AFOV formula, we get the rather obvious formula

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

Exit Pupil

Roughly speaking, the exit pupil is the diameter of the beam of light exiting the eyepiece.  More specifically, it's the diameter at the point where all of the beams, one from each point in the part of the sky being viewed, intersect (see figure 14) at the eye relief distance behind the eyepiece.  Each beam, once it exits the eyepiece, is a bundle of parallel rays, and the diameter of each bundle is about the same, so we need consider only the beam from an on-axis star:

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/fo = E/fe
or D/E = fo/fe
But fo/fe = magnification, so:

D/E = magnification

(or E = D/magnification)

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 fo/D.  I'll call it fr.  Then

E = D/(fo/fe) = fe*(D/fo) = fe/(fo/D) , so

E = fe/fr

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".

Image Brightness

For imaging, in order to determine exposure times, we are concerned with the brightness per area at the focal plane.  If an object is focused by an objective of aperture D and focal length fo to an image of a certain size at the focal plane, and then we use a different telescope with with the same focal length, but with an objective of aperture D', then the image is the same size (since the focal length is the same), but the amount of light going into the objective from the object (which all makes it to the focal plane) is changed by the ratio of the areas of the objective, which is

[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 fo', the image scale is changed by the factor fo'/fo, so the area of the image is changed by the factor (fo'/fo)^2, so the brightness per area is changed by the inverse of this, (fo/fo')^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 fo with a scope of aperture D' and focal length fo', is (D'/D)^2*(fo/fo')^2 = [(D'*fo)/(D*fo')]^2 = [(fo/D)/(fo'/D')]^2 = (fr/fr')^2, where fr is the focal ratio fo/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.

Eye Relief

For this subject, our highly idealized/simplified scope is not too relevant, because eye relief depends heavily on the particulars of the eyepiece design.  Applying simple geometry and the approximations we've been using, it is not hard to show that the eye relief in our case is approximately equal to the focal length of the EP.  In most eyepiece designs, the eye relief  is proportional to the eyepiece focal length, with the constant of proportionality usually a little less than 1:  in Plossls, orthoscopics, and Erfles, the constant is about 0.9.  In some designs, the constant is as high as 1.25 (some types of Naglers), and some simple eyepieces (Kellners, and the Ramsden and Huygenians sold with department store scopes) have a constant lower than 0.5.  One exception to this "constant of proportionality" is "lanthanum" eyepieces (Vixen Lanthanum, Televue Radians and several others), which use a lens element made of glass containing the rare-earth element lanthanum, in a design which gives 20mm eye relief, regardless of the focal length.