Introduction | Around the Exit Pupil | Exit Pupil and Observing | References | Appendix: Calculation of the Exit Pupil | Appendix: Playing Around with Formulae for the Exit Pupil
On this page, I deal with the exit pupil and try to summarize information I had collected on the exit pupil in one place (it may be more detailed in other places on this Website).
Note: For definitions in a small glossary, see page Quick & Dirty Astronomy Glossary.
The exit pupil is the diameter of the ray bundle that leaves the eyepiece (see Definition). For me, this is a rather abstract quantity that you come across again and again in connection with eyepieces; basically, however, it has a very simple effect: it determines how bright an object appears in the telescope (see Effect). I cover the exit pupil in several places on this Website, but I was missing something like a "context". That is why I created this page, in order to be able to find all aspects of the exit pupil in one place for practical use.
In optical devices for direct visual observation such as telescopes and binoculars, the exit pupil is the diameter of the ray bundle that leaves the eyepiece (according to Wikipedia).
Ultimately, only the exit pupil determines how bright the image of a particular object, for example the moon, appears in the eyepiece. With the same exit pupil, it always appears equally bright, regardless of telescope, aperture and magnification (according to www.hobby-astronomie.com/teleskope_optische_grundlagen.html#austrittspupille).
For a telescope, the diameter of the exit pupil can be calculated by dividing the focal length of the eyepiece by the focal ratio (f-number) of the telescope (see calculations).
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Alternatively, you can calculate the exit pupil from the telescope's aperture and the magnification:
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I usually do not use Formula 2 for calculating the exit pupil, but find this formula more useful for estimating the effects of magnification or aperture on the exit pupil, particularly when comparing different telescopes (or one telescope with and without a reducer).
Instead, I calculate the magnification using the just calculated exit pupil:
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This (formula 2a) is in most cases easier to do in your head than to calculate the magnification from the ratio of telescope focal length to eyepiece focal length. More about magnification below!
In young people, the dark-adapted pupil is 7 mm wide (some sources even write 8 mm) and in adults about 6.4 mm; with increasing age, it becomes smaller and smaller. However, according to the Intercon Spacetec Website, an exit pupil of 6 mm can still be achieved by 70-year-old observers. This source considers an exit pupil of 5 mm as sufficient (if it is not about largest possible field of view). According to another source, even an exit pupil of 4 mm is suitable and sufficient for all observers.
Here are some simple rules about observation and exit pupil:
As already written, the easiest way to determine the exit pupil of a given eyepiece is to divide its focal length by the focal ratio of the telescope.
Various authors give "recommendations for selecting eyepiece focal lengths based on the exit pupil". I myself have consolidated the recommendations of several sources and derived and published my own recommendations from them (see pages Eyepiece Selection (Focal Length Selection), Eyepiece Selection (Focal Length Selection) - Application, Eyepiece Selection (Focal Length Selection) - Application Part 2), which I list again below.
From the values for the exit pupil, as given in the recommendations for eyepiece focal lengths, AND the focal ratio of the telescope in question, one can calculate the corresponding eyepiece focal lengths for a telescope. And not just for one telescope, but for all telescopes with the same focal ratio, no matter how large or small they are. However, depending on the aperture of the telescope used, this leads to different magnifications! These are calculated from the focal length of the chosen telescope and the eyepiece focal length (ratio).
In the following table with "my" recommendations, I have listed some typical aperture ratios as examples and calculated the corresponding eyepiece focal lengths for them by multiplying the exit pupil values with the aperture ratios (on the pages given above, this is done analogously for concrete telescopes):
Category | Deep Sky Application Area | Exit Pupil (mm) |
Focal Ratio
/Focal Length of Eyepiece (mm) |
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4 |
5 |
6 |
6,3 |
7 |
10 |
12 |
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Minimum Magnification / Maximum/Large Field of View | Search | 7...10 |
28...40 | 35...50 | 42...60 | 44...63 | 49...70 | 70...100 | 84...120 |
Overview, large-area nebulae | 4.5...5...6... 6.5 (7) |
18...28 | 22.5...35 | 27...42 | 28...44 | 31.5...49 | 45...70 | 54...84 | |
Normal Magnification | Optimal for large-area, faint nebulae; nebulae, open star clusters | 3.5...4 |
14...16 | 17.5...20 | 21...24 | 22...25 | 24.5...28 | 35...40 | 42...48 |
Perceptibility optimal for many objects, e.g. for most galaxies, and mid-size deep sky objects | 2...3 |
8...12 | 10...15 | 12...18 | 12.6... 18.9 |
14...21 | 20...30 | 24...36 | |
Maximum Magnification / Maximum Resolution |
Actually, the "normal" upper magnification limit; globular star clusters | 1...1.5 |
4...6 | 5...7.5 | 6...9 | 6.3...9.5 | 7...10.5 | 10...15 | 12...18 |
With perfect seeing, achieves maximum perceptibility of small, low-contrast details; planetary nebulae, small galaxies; maximum magnification for planets that makes sense | 0.6...0.7...0.8 |
2.4...3.2 | 3...4 | 3.6...4.8 | 3.8...5 | 4.2...5.6 | 6...8 | 7.2...9.6 | |
Separation of narrow double stars, small planetary nebulae; at the extreme limit of the telescope, to perceive the faintest details | 0.4...0.5 |
1.6...2.0 | 2.0...2.5 | 2.4...3.0 | 2.5...3.2 | 2.8...3.5 | 4...5 | 4.8...6 |
The exit pupil limits (in the direction of the eye) both the minimum and the maximum usable magnification of an optical instrument, i.e. telescope (if one disregards other effects). According to Stoyan, one should exploit this range as much as possible when selecting eyepieces.
Once you have determined the exit pupil, you can easily determine the magnification for simple values of the exit pupil in your head by dividing the telescope aperture by the exit pupil (formula 2a). You can also can also multiply the focal ratio with the expit pupil to get the focal length of the eyepiece. Here are to tables to simplify or save you from mental arithmetic:
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I own the Celestron C5 (127 mm / 1250 mm) and the C8 (203 mm / 2032 mm), both of which have a focal ratio of f/10. In addition, I own an f/6.3 reducer/corrector that I can use on both telescopes. The intriguing question for me was how both telescopes perform in terms of exit pupil with and without the reducer. The answers may be surprising at first glance, but they are easy to understand using the formulas and eyepiece tables:
Of course, these results also apply to other telescopes and likewise to extender/Barlow lenses.
The exit pupil of an eyepiece is calculated from:
Alternatively, the exit pupil diameter is given by aperture of a telescope divided by its magnification (see Playing Around with Formulae for the reasoning):
In practice, it is often easier to calculate the magnification from the exit pupil, which was already calculated using formula 1:
After I had created an Excel table with magnifications for my telescopes and, using these magnifications, in a second table the focal lengths of the eyepieces for my telescopes, I realized that this was a somewhat cumbersome approach. After some transformations of the formula for the exit pupil, I found out that the focal ratio is actually the "critical" telescope parameter for determining the focal lengths of eyepieces (if it is unknown).
I start from the formula for the exit pupil:
If the formula is solved according to the focal length of the eyepiece, you get:
You get the magnification by inserting the focal length of the eyepiece into to following formula:
Stoyan and other author use a different approach. For illustration purposes, I start from the same formula as above and insert the formula for the magnification:
Solving the formula for the magnification, results in Stoyan's formula:
For determining the focal length of the eyepiece, you can insert the formula for the magnification and solve the formula for the focal length of the eyepiece:
Ultimately, both paths lead to the same end, because, in addition to the focal length of the eyepiece, one needs to know the magnification to be able to determine whether an eyepiece's focal length makes sense (i.e., whether the magnification lies within the range of minimum and maximum usable magnification).
07.10.2024 |