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Astronomy · Observing

Telescope Magnification Calculator

Find a telescope’s magnification from its focal length and eyepiece, plus focal ratio, exit pupil, and the useful magnification limit.

mm
The tube’s focal length.
mm
The main lens or mirror width.
mm
Printed on the eyepiece.
°
Enter the eyepiece’s apparent field of view to also get the true field of view. Leave at 0 to skip.

All lengths in millimetres. Magnification changes only with the eyepiece — a shorter eyepiece focal length gives a higher power.

Magnification
100×Solved

1000 mm scope ÷ 10 mm eyepiece = 100×.

Focal ratio
f/10

Focal length ÷ aperture. Lower is “faster” (brighter, wider views).

Exit pupil
1 mm

Aperture ÷ magnification — the beam of light reaching your eye.

Max useful mag
≈ 200×

Rule of thumb ≈ 2 × aperture (mm); atmosphere and optics often lower it.

A telescope’s magnification is its focal length divided by the eyepiece focal length: M = scope focal length ÷ eyepiece focal length. A 1000 mm telescope with a 10 mm eyepiece gives 1000 ÷ 10 = 100×. Swap in a shorter eyepiece for more power, a longer one for a wider, brighter view.

How telescope magnification works

Unlike a camera lens, a telescope has no single “zoom” number — its magnification depends on which eyepiece you slot in. The telescope’s objective (its main lens or mirror) forms an image, and the eyepiece acts as a magnifier for that image. Divide the telescope’s focal length by the eyepiece’s focal length and you get the power. Because the eyepiece is on the bottom of the fraction, a shorter eyepiece focal length yields a higher magnification. A 25 mm eyepiece in the same 1000 mm scope gives 40×, while a 5 mm eyepiece pushes it to 200×. This is why observers carry a set of eyepieces rather than one — low power for finding and framing large objects, high power for splitting double stars and studying planets.

magnification = telescope focal length ÷ eyepiece focal length

both focal lengths in the same unit (millimetres); a shorter eyepiece gives more magnification

Worked example

Take a common 100 mm (4-inch) refractor with a 1000 mm focal length and a 10 mm eyepiece. What magnification, focal ratio, and exit pupil does it give?

  1. 1
    Note the two focal lengths. Telescope focal length = 1000 mm; eyepiece focal length = 10 mm.
  2. 2
    Divide to get magnification. M = 1000 ÷ 10 = 100×.
  3. 3
    Find the focal ratio. f/number = focal length ÷ aperture = 1000 ÷ 100 = f/10.
  4. 4
    Find the exit pupil. exit pupil = aperture ÷ magnification = 100 ÷ 100 = 1.0 mm.
  5. 5
    Check the useful limit. Max useful magnification ≈ 2 × aperture = 2 × 100 = 200×, so 100× is comfortably within it.

Eyepieces in a 1000 mm f/10 telescope (100 mm aperture)

Swapping eyepieces changes magnification, exit pupil, and how much sky you see.

EyepieceMagnificationExit pupilBest for
40 mm25×4.0 mmWidest, brightest views — star fields, large nebulae
25 mm40×2.5 mmGeneral low-power observing and finding targets
10 mm100×1.0 mmThe Moon, planets, brighter deep-sky objects
6 mm167×0.6 mmHigh-power planetary and double-star work
5 mm200×0.5 mmAt the ≈ 2 × aperture useful ceiling for this scope

Why more magnification is not always better

It is tempting to reach for the shortest eyepiece, but magnification is limited by aperture and the atmosphere, not by arithmetic. A wider aperture gathers more light and resolves finer detail, so a small scope simply cannot support the same power as a large one. The common rule of thumb is a useful ceiling of about 2 × the aperture in millimetres (roughly 50× per inch). Push past it and you get “empty magnification”: the image grows but only becomes dimmer, softer, and shakier. On many nights unsteady air (poor “seeing”) caps you well below the theoretical limit. The exit pupil — the aperture divided by the magnification — is the width of the light beam leaving the eyepiece; when it shrinks below about 0.5 mm the view is dim and floaters become distracting, while an exit pupil larger than roughly 7 mm wastes light your pupil cannot admit.

How do I calculate a telescope’s magnification?
Divide the telescope’s focal length by the eyepiece’s focal length, using the same unit for both. A 1000 mm telescope with a 10 mm eyepiece gives 1000 ÷ 10 = 100×. The telescope’s aperture and focal ratio do not enter this formula — only the two focal lengths set the magnification.
Why does a shorter eyepiece give more magnification?
The eyepiece focal length sits on the bottom of the fraction magnification = telescope focal length ÷ eyepiece focal length. Making that denominator smaller makes the result larger, so a 5 mm eyepiece magnifies twice as much as a 10 mm one in the same telescope.
What is the maximum useful magnification?
A common rule of thumb is about 2 × the aperture in millimetres — roughly 50× per inch — so a 100 mm scope tops out near 200×. It is only a guideline: unsteady air, optical quality, and the target often cap the useful power lower. Beyond the limit the image just grows dimmer and blurrier, not more detailed.
What is exit pupil and why does it matter?
The exit pupil is the width of the light beam leaving the eyepiece, equal to aperture ÷ magnification (or eyepiece focal length ÷ focal ratio). It controls image brightness: a large exit pupil gives bright, low-power views, while one below about 0.5 mm looks dim and reveals eye floaters. Beyond roughly 7 mm the beam is wider than your pupil, so some light is wasted.
What is the focal ratio (f/number)?
The focal ratio is the telescope’s focal length divided by its aperture, so a 1000 mm scope with a 100 mm aperture is f/10. Lower ratios (f/4–f/6) are “fast,” giving wider, brighter fields suited to deep-sky objects; higher ratios (f/10 and up) are “slow,” favouring high-power planetary and lunar views.
Does more magnification always mean a better view?
No. Aperture — not magnification — sets how much light you gather and how much detail you can resolve. Past the useful limit you get empty magnification: a bigger but dimmer, softer, and shakier image. Lower powers often show fainter and wider objects far better, which is why observers keep several eyepieces on hand.
How do I find the true field of view?
Divide the eyepiece’s apparent field of view (AFOV, printed by the maker) by the magnification. A 50° eyepiece at 100× shows a true field of 50° ÷ 100 = 0.5° — about one full-Moon width of sky. Enter the AFOV in the optional field above to have the calculator work this out for you.