Telescope Magnification Calculator
Magnification, exit pupil, and true field for any telescope and eyepiece.
What it does
A telescope's power is not fixed; it lives in the pairing of the tube and the eyepiece. Swap the eyepiece and the same scope can frame a whole galaxy or split a tight double star, all without touching the mirror.
The tool above takes the aperture, the telescope focal length, an eyepiece focal length, and the eyepiece apparent field. It returns the magnification, exit pupil, true field of view, focal ratio, and a practical ceiling for the night's power.
The defaults describe a common 200 mm f/6 refractor with a 10 mm eyepiece: 120x, a 1.67 mm exit pupil, a 400x ceiling, and a 0.42 degree true field. The imperial mirror is an 8 inch f/6 with a 0.4 inch eyepiece, which lands on the same 120x.
The arithmetic follows the Coconino Astronomical Society's eyepiece math and Sky and Telescope's eyepiece guide: power divides the focal lengths, exit pupil divides the aperture by the power, and the usable ceiling caps near two times the aperture in millimeters.
How to use
- Pick metric millimeters or imperial inches.
- Enter the telescope aperture.
- Enter the telescope focal length, printed on the tube.
- Enter the eyepiece focal length, printed on the barrel.
- Enter the eyepiece apparent field, or leave a typical 50 degrees.
- Read power first, then check the exit pupil and ceiling.
Formula
- M
- Magnification, written with an x.
- F
- Telescope focal length.
- f
- Eyepiece focal length.
- A
- Aperture, the diameter of the main lens or mirror.
- AFOV
- Apparent field of view of the eyepiece in degrees.
The exit pupil is the beam of light entering your eye; wider than about 7 mm and the extra light spills past your iris, so low power has a practical floor. The 2x per millimeter ceiling (50x per inch) is the customary limit before the image only gets dimmer and softer.
Worked example
Example: 200 mm f/6 with a 10 mm eyepiece
Defaults: 200 mm aperture, 1200 mm focal length, 10 mm eyepiece, 50 degree apparent field.
- Magnification: 1200 / 10 = 120x.
- Exit pupil: 200 / 120 = 1.67 mm, a good planetary beam.
- Max useful power: 2 x 200 = 400x.
- Focal ratio: 1200 / 200 = f/6.
- True field: 50 / 120 = 0.42 degrees, a bit wider than the full Moon.
Reference table
| Power | Target | Exit pupil check |
|---|---|---|
| 25x to 50x per inch | Double stars, planets at steady air | Exit pupil 1 mm or less |
| Around 100x | Moon, planets, small clusters | A comfortable all rounder |
| 30x to 60x | Large nebulas and open clusters | Exit pupil near 4 to 7 mm |
| Less than 20x | Rich field sweeps of the Milky Way | Watch your iris: 7 mm cap |
Limits
- Assumes optics are collimated and clean.
- Seeing and transparency cap real detail.
- Exit pupil uses aperture, not mirror coating losses.
- True field assumes a simple eyepiece formula.
Common mistakes
- Buying power before aperture; resolution lives in aperture.
- Pushing past the 50x per inch ceiling on a small scope.
- Ignoring the exit pupil when choosing a low power eyepiece.
- Measuring the tube length instead of the focal length.
- Expecting the advertised magnification to look sharp every night.
Try it yourself
Quick checks. Answers sit tucked under each question.
8 inch f/6, 0.4 inch eyepiece?
48 / 0.4 = 120x, exit pupil 8 / 120 = 0.067 in.
1200 mm tube, 25 mm eyepiece?
48x, exit pupil 200 / 48 = 4.2 mm, great for wide fields.
Is 500x fair on a 150 mm scope?
No; the ceiling is about 2 x 150 = 300x, and only in steady air.
Questions
What magnification should I use?
Match it to the target and the night; most deep sky work sits between 30x and 120x, with planets higher on steady nights.
What is the exit pupil?
The light beam leaving the eyepiece toward your eye; it also sets image brightness for extended objects.
Is more magnification always better?
No; past about 2x per millimeter of aperture the image only grows softer and dimmer.
Why 50x per inch?
It is the imperial twin of 2x per millimeter, the customary ceiling in amateur practice.
Does the eyepiece field change power?
No; power comes only from the two focal lengths. The apparent field sets how wide the view looks.
What is a Barlow?
A lens that multiplies the effective focal length, typically 2x, so it doubles every power in your eyepiece set.