Which child, in some point of its life, did not want to take a closer look at the planets, stars, galaxies, … in the sky?! At least I did and finally, after taking some pictures of the full moon lately with a telephoto lens I remembered those wishes and bought myself a telescope. Nevertheless, I do not only want to look at the sky but also take beautiful photos, now. So I started my search for a beginner-telescope with the ability to attach my digital mirror less camera (DSLM). Although a telescope is defined by its construction and two numbers and the two numbers of the eyepiece, it took a while to understand all the relevant parameters that could be calculated from those four numbers. So here I write about my way, how I found one.

The first distinction is the construction of the telescope: with lenses the so called refractors or with mirrors the reflectors. Refractors are longer and heavier and beyond a certain diameter they get pretty expensive. So the first decision was to buy a reflector. Exactly a Maksutov, which fits due of its compact construction, on my city balcony.

All telescopes are defined by two numbers, the focal length and the aperture (diameter). The focal length is relevant for the magnification, like with cameras. The aperture (the width of the opening or mirror) the amount of light collected. The ratio of focal length and aperture is called focal ratio and below approximately 10 it is better for not so bright objects like galaxies or nebulas. For bright objects like the moon, Saturn Jupiter a focal ratio higher than 10 is also fine. However, the wider the opening, the more expensive the telescopes get and beyond a certain width the air turbulences (seeing) might disturb the view. I read the aperture of 200-300mm to be maximum for our region. And they get larger and heavier in weight with larger diameters. So I decided to buy a 127mm aperture (5 inches).

## Magnification and exit pupil

Next, one needs at least one eyepiece (ocular lens), also. The ratio between the focal length of the telescope and the focal length of the eyepiece defines the magnification. As an approximation double the aperture (in mm) is assumed to be the maximum magnification of the telescope. But not only that, the ratio between the focal length of the eyepiece and the focal ratio is called exit pupil. If the exit pupil is larger than 7mm (the largest possible opening of the human eye pupil) light will be lost. This defines the lowest suggested magnification. If it is smaller than approximately 0.7mm, one will see the dust in or tiny blood vessels of ones eye (that’s true!). So this defines another maximum magnification.

The following formular offers the possibility to calculate the magnification and exit pupil for a telescope’s focal length and aperture and some values for a predefined set of focal lengths of eyepieces and at the bottom an individual focal length of an eyepiece can be entered.

Ocular | Magnification | Exit pupil | |
---|---|---|---|

minimum | |||

optimum | |||

maximum | |||

for given eyepiece focal length | |||

5 |
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6 |
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8 |
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9 |
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10 |
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12 |
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15 |
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18 |
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25 |
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32 |
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40 |
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indiv. |

## Field of view

I learned about the field of view after I bought the telescope, while playing around with the Opensource astronomy software Stellarium. There I could enter the focal length of the telescope and the eyepiece and the “field of view” for the eyepiece and stellarium showed only the visible subset of the sky visible with the entered values. So I took a closer look and discovered the second parameter of the eyepiece.

### Telescope

But first some theory why the field of view is important: since the moon with a diameter of 3476km is ~370000km away and deep sky objects are even light years away, e.g. the andromeda galaxy is 2.5 Mio light years (ly) away and has a diameter of 140000ly it is hard to imagine how large they will appear through the telescope. And I did not believe that the andromeda galaxy will with a field of view of ~3° appear 6 times larger than the moon with 0.5°.

$$ AOV_{moon} = 2 * \frac{180}{\pi} * \arctan\left( \frac{\frac{3476}{2}km}{370000km} \right) \approx 0.53°$$ $$ AOV_{andromeda\ galaxy} = 2 * \frac{180}{\pi} * \arctan\left( \frac{\frac{140000}{2}ly}{2500000 ly} \right) \approx 3°$$

Therefore the size of a target is given in degree, minutes and/or seconds.

For that the field of view in degree is the relevant number to calculate for the telescope/eyepiece combination. And here the second parameter of the eyepiece is relevant: the apparent field of view, given in degree. The true field of view, what I will be able to see is the ratio between the apparent field of view of the eyepiece and the magnification.

$$ AOV = \frac{field\ of\ view\ eyepiece}{magnification} = \frac{field\ of\ view\ eyepiece * focal\ length\ eyepiece}{focal\ length\ telescope} $$

Here is another formular, which calculates the apparent field of view, or angle of view (AOV) from the relevant values of the telescope and the eyepiece.

### Camera

Since I also wanted to attach my camera to the telescope, I need to calculate the field (or angle) of view for my camera, too. Here the the angle of view (AOV) is based on the cameras sensor dimension and focal length. The sensor dimension of a micro four thirds (MFT) for example is 17.3mm x 13mm. The AOV in degree is defined as:

$$ AOV = \frac{180}{\pi} * 2 * \arctan{\left(\frac{[width|height|diagonal]}{2 * focal\ length}\right)}$$

For my telescope with 1500mm focal length this gives ~40'.

When looking for the right exposure time and aperture, the ‘crop-factor’ is often mentioned. This is the relation of the old 36x24mm film diagonal to the camera sensor diagonal. This factor for example is relevant to calculate the maximum exposure time (E) for a given focal length and sensor to see the stars as points and not as traces.

$$ E = \frac{300}{crop\ factor * focal\ length} $$

Sometimes, it’s also 500 instead of 300 but with the hint that stars far away from Polaris might become traces. With the above entered values this gives:

## Angular resolution

The angular resolution describes how close two stars can be to be still distinguished as individual points. This formula is pretty simple: 138 divided by aperture, gives the angular resolution in seconds. In my case is 138/127 = 1.09". Due to the air turbulences, light pollution (seeing), there is is a limit of the angular resolution, which is mostly between 1" and 3".

angular resolution |
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If I take the above formular for the AOV calculation with an MFT-sensor with 16 MP (4592x3448 pixel) and a focal length of 1500mm (40’x30') this yields already a pixelsize of 0.5".

It is possible to see fine details on Jupiter or Saturn, I just need to find out more here and see if it is worth investing in a barlow lens to get the planets larger on a foto and how long the exposure times can be.

## Neglected parts

I must say that I did not really care about the telescope mount, when I bought it. However, I’m pretty happy that I bought one with GoTo-Function, not to find the objects, but to follow them. At least with the largest magnification I could see Jupiter and Saturn moving through the eyepiece when the tracking was switched off.

## Add-on 2021-11-07: Calculation of camera-sensor/lens parameters

I have created an extra page just with a table calculating the parameters for varying focal length and sensor layouts fitting on a smartphone screen.

## References

- www.astroshop.de: “So finden Sie die richtigen Okulare”
- www.astronomie.de: “Vergrößerung und Grenzgröße im Teleskop – Fallbeispiel M 13”
- Bildwinkel berechnen lassen
- What is a Field of View in Astronomy and Why is it Useful?
- wikipedia.org: “Angular resolution”
- astroshop.de: “Auflösungsvermögen”
- sterngucker.de: “Pixelgröße und Sampling”
- Formel Belichtung Sternenhimmel