How to calculate your telescopes magnification (power)
The magnification produced by a telescope is determined by the focal length of the telescope divided by the focal length of the eyepiece used.
So to determine the magnification for your telescope and eyepiece combination first determine the focal length of your telescope. This will usually be found on a label on the side of the telescope or around the outside of the lens. If it is a refractor, it will say something like f=800mm. Now find the focal length of the eyepiece, which will be marked on the side in mm.
Divide the telescope focal length by the eyepiece focal length of the eyepieces you are going to use. For example, a 10mm focal length eyepiece used with a telescope with a focal length of 800 will give 80X magnification.
How much magnification can I use with my telescope?
A good rule of thumb is 50x per inch of aperture or approx 2x the aperture in mm. So if I have a 6” telescope, then 6 x 50 = 300. that is 300x magnification under good seeing conditions.
Remember when you are looking at astronomical objects, you are looking through a column of air that reaches to the edge of space and that column of air seldom stays still. Under steady conditions, the 300x of your 6” telescope may be possible but in less than ideal seeing conditions better clarity will be achieved by reducing the magnification. Always use the least magnification needed to see the objects you are observing. When no further detail can be seen, there is no point increasing magnification.
Calculating Field of View
The amount of sky that you see through your telescope is called the true (or actual) field of view and it is determined by the design of the eyepiece. Every eyepiece has a value, called the apparent field of view, which is supplied by the manufacturer. Field of view is usually measured in degrees and/or arc-minutes (there are 60 arc-minutes in a degree).
The true field of view produced by your telescope is calculated by dividing the eyepiece’s apparent field of view by the magnification that you previously had. Using the figures in the previous magnification example, if your 10mm eyepiece has an apparent field of view of 52 degrees, then the true field of view is 0.65 degrees or 39 arc-minutes.
To put this in perspective, the moon is about 0.5ｰ or 30 arc-minutes in diameter, so this combination would be fine for viewing the whole moon with a little room to spare.
Remember, the more magnification the smaller the field, this will make finding objects very difficult, so better to use low power to find objects then change magnification as required.
Calculating the Exit Pupil
The Exit Pupil is the diameter (in mm) of the narrowest point of the cone of light leaving your telescope. Knowing this value for a telescope-eyepiece combination tells you whether your eye is receiving all of the light that your primary lens or mirror is collecting.
The average person has a fully dilated pupil diameter of about 7mm. This value varies a little from person to person and decreases with age. Your eyes will need to become fully dark adjusted for best results. This normally requires 10 to 15 minutes in dark conditions to achieve.
To calculate exit pupil, divide the diameter of the primary mirror or lens of your telescope (in mm) by the magnification in use.
A 200mm f/5 telescope with a 40mm eyepiece produces a magnification of 25X and an exit pupil of 8mm. This combination can probably be used by a young person but would not be of much value to the older adult. The same telescope used with a 32mm eyepiece gives a magnification of about 31X and an exit pupil of 6.4mm, which would be fine for most dark-adapted eyes.
A 200mm f/10 telescope with a 40mm eyepiece gives a magnification of 50X and an exit pupil of 4mm, which is fine for everyone.
Normally expressed f/n for example f/8, what does this mean and how do we calculate it?
Focal ratio = telescope focal length divided by telescope aperture
So a telescope with 200mm aperture and focal length of 2000mm will have a focal length of f/10. An example of this would be the popular 8” (200mm) Schmidt Cassegrain telescopes made buy Celestron and Meade.
A typical fast newtonian telescope of 8” (200mm) might have a focal length of 1000mm and so give a focal length of f/5. A typical example would be the Skywatcher Explorer 200p
What does this mean?
Photographically, if you were to fit your camera to both of the above example telescopes, the image produced in the f/10 SCT telescope would be twice as large as the f/5 telescope.
For visual purposes it would make the f/5 instrument better suited to low power observing of say deep sky objects, whilst the f/10 telescope would be better suited to high power use; for example moon planets, solar double stars etc