acalculator

What is the telescope magnification?

Type the focal length of your telescope and of the eyepiece, and any Barlow lens. The telescope magnification calculator gives the power, and with the aperture and the eyepiece's apparent field, the exit pupil, true field of view, focal ratio and resolving limit.

Your numbers

Units
Magnification
40×

The magnification is 40×.

Magnification (M)
40
Focal length with the Barlow (mm)
1,000
Exit pupil (mm)
2.5
True field of view (°)
1.3
True field of view (′)
78
Focal ratio (f/)
10
Resolving limit at 550 nm (″)
1.38404

Magnification: 40×. The magnification is 40×.

How to calculate

Telescope magnification calculator: M = telescope focal length ÷ eyepiece focal length, with a Barlow, plus the exit pupil, true field of view, focal ratio and resolving limit.

Example with the default inputs (Telescope focal length 1,000 mm, Eyepiece focal length 25 mm, Barlow or reducer factor 1, Aperture (D) 100 mm, Eyepiece apparent field (°) 52): The magnification is 40×.

Method: M = telescope focal length × Barlow factor ÷ eyepiece focal length; exit pupil = D ÷ M; true field = apparent field ÷ M; focal ratio = focal length ÷ D; resolving limit = 1.22 × 550 nm ÷ D radians.

  • The final image is at infinity (a relaxed eye), and the image is upside down; the page gives the size of M.
  • The true field is the apparent field ÷ M, the small-angle rule; the resolving limit is the diffraction (Rayleigh) limit, before the air and the optics blur it.

Machine-readable copies: Markdown, JSON.

Worked examples

Each example is checked against the calculator on every build.

  1. Telescope focal length 1,000 mm, Eyepiece focal length 25 mm, Barlow or reducer factor 1 gives Magnification (M) 40, Magnification 40×.Source: OpenStax, University Physics Volume 3, §2.8 Microscopes and Telescopes (Equations 2.39 and 2.40, M = θ_image ÷ θ_object = −f_obj ÷ f_eye; a 1 m objective gives 40× with a 2.5 cm eyepiece and 80× with 1.25 cm), https://openstax.org/books/university-physics-volume-3/pages/2-8-microscopes-and-telescopes (retrieved 2026-10-19)
  2. Telescope focal length 1,000 mm, Eyepiece focal length 12.5 mm, Barlow or reducer factor 1 gives Magnification (M) 80.Source: OpenStax, University Physics Volume 3, §2.8 Microscopes and Telescopes (Equations 2.39 and 2.40, M = θ_image ÷ θ_object = −f_obj ÷ f_eye; a 1 m objective gives 40× with a 2.5 cm eyepiece and 80× with 1.25 cm), https://openstax.org/books/university-physics-volume-3/pages/2-8-microscopes-and-telescopes (retrieved 2026-10-19)
  3. Telescope focal length 1,000 mm, Eyepiece focal length 25 mm, Barlow or reducer factor 2, Aperture (D) 100 mm, Eyepiece apparent field (°) 52 gives Magnification (M) 80, Exit pupil (mm) 1.25, True field of view (°) 0.65, True field of view (′) 39, Focal ratio (f/) 10, Focal length with the Barlow (mm) 2,000.Source: OpenStax, University Physics Volume 3, §2.8 Microscopes and Telescopes (Equations 2.39 and 2.40, M = θ_image ÷ θ_object = −f_obj ÷ f_eye; a 1 m objective gives 40× with a 2.5 cm eyepiece and 80× with 1.25 cm), https://openstax.org/books/university-physics-volume-3/pages/2-8-microscopes-and-telescopes (retrieved 2026-10-19)
  4. Telescope focal length 57,600 mm, Eyepiece focal length 25 mm, Barlow or reducer factor 1, Aperture (D) 2,400 mm gives Resolving limit at 550 nm (″) 0.057668, Magnification (M) 2,304.Source: OpenStax, University Physics Volume 3, §4.5 Circular Apertures and Resolution (Equation 4.5, θ = 1.22λ ÷ D; Example 4.6: the Hubble Space Telescope, D = 2.40 m at 550 nm, 2.80 × 10⁻⁷ rad), https://openstax.org/books/university-physics-volume-3/pages/4-5-circular-apertures-and-resolution (retrieved 2026-10-19): 2.80 × 10⁻⁷ rad = 0.0577″

How it works

  • Magnification: M = F ÷ f_e, with F the telescope's focal length times the Barlow factor (1 for none) and f_e the eyepiece's focal length. OpenStax writes M = −f_obj ÷ f_eye: the minus sign only says the image is upside down, so the page gives the size, M.
  • Exit pupil: D ÷ M, with D the aperture. The eyepiece forms an image of the objective lens or mirror, at distance f_obj + f_eye in front of it; by the thin-lens equation that image is f_eye ÷ f_obj times as wide, which is D ÷ M.
  • True field of view: the eyepiece's apparent field ÷ M, from M = θ_image ÷ θ_object, in degrees and in minutes of arc (× 60).
  • Focal ratio: the telescope's focal length ÷ D (without the Barlow).
  • Resolving limit: θ = 1.22 × λ ÷ D radians for λ = 550 nm (green light), shown in seconds of arc: θ × 648,000 ÷ π.

The page also shows the focal length with the Barlow, in mm. The aperture and the apparent field are optional: without them the page leaves out the results that need them.

Units (exact): 1 mm = 0.001 m; 1 cm = 0.01 m; 1 in = 0.0254 m; 1° = 60′ (minutes of arc); 1 radian = 648,000 ÷ π seconds of arc.

Rules

  • Telescope focal length from 0.0001 m to 1,000 m; eyepiece focal length from 0.0001 m to 10 m; aperture from 0.0001 m to 100 m. Limits are in metres and include their ends; a value typed in another unit is converted first.
  • Barlow factor from 0.1 to 10; apparent field from 1° to 120°.
  • A result too large or too small to show as a number gives no answer. Within these limits it does not happen.

Output format

The headline reads M×, the magnification to 6 significant figures with a times sign (40×). Every number shows to 6 significant figures, rounded half up, with trailing zeros dropped and thousands commas, and below 10⁻⁶ or from 10¹⁵ up as a × 10ⁿ. Typed values are read back as the exact decimals you typed, and every unit size is exact, so M, the exit pupil, the true field and the focal ratio are exact fractions rounded once; the resolving limit is divided by π once, as a 64-bit float.

Worked examples by hand

From OpenStax. A 1 m (100 cm) objective with a 2.5 cm eyepiece: M = 100 ÷ 2.5 = 40, shown as 40×. With a 1.25 cm eyepiece, M = 100 ÷ 1.25 = 80.

With a Barlow, exit pupil and field. A 1,000 mm telescope of 100 mm aperture, a 25 mm eyepiece of 52° and a 2× Barlow: F = 2,000 mm, M = 2,000 ÷ 25 = 80. The exit pupil is 100 ÷ 80 = 1.25 mm, the true field 52 ÷ 80 = 0.65° = 39′, and the focal ratio 1,000 ÷ 100 = f/10.

A large mirror. D = 2.40 m (the Hubble Space Telescope's mirror) with a 57.6 m focal length and a 25 mm eyepiece: M = 57,600 ÷ 25 = 2,304. The resolving limit is 1.22 × 550 × 10⁻⁹ ÷ 2.4 = 2.79583 × 10⁻⁷ rad (OpenStax Example 4.6: 2.80 × 10⁻⁷ rad), or 2.79583 × 10⁻⁷ × 648,000 ÷ π = 0.0576682″.

Other questions people ask

How do you calculate telescope magnification?

Divide the focal length of the telescope by the focal length of the eyepiece: M = f_telescope ÷ f_eyepiece. A 1,000 mm telescope with a 25 mm eyepiece magnifies 1,000 ÷ 25 = 40 times (40×); a 12.5 mm eyepiece gives 80×.

What does a Barlow lens do to magnification?

It multiplies the telescope’s focal length, and so the magnification, by its factor. A 2× Barlow turns the 25 mm eyepiece on a 1,000 mm telescope from 40× into 80×. A focal reducer of 0.63× lowers it instead.

What is the exit pupil?

The width of the beam of light leaving the eyepiece: the aperture divided by the magnification. A 100 mm telescope at 80× has an exit pupil of 100 ÷ 80 = 1.25 mm. A larger exit pupil gives a brighter view, up to the width of your own pupil.

How much sky will I see?

The true field of view is the eyepiece’s apparent field divided by the magnification. A 52° eyepiece at 80× shows 52 ÷ 80 = 0.65°, a little more than the full Moon (about 0.5°).

Does more magnification show more detail?

Only up to the telescope’s resolving limit, θ = 1.22λ ÷ D, set by the aperture D. In green light (550 nm) a 100 mm telescope splits stars about 1.38″ apart and a 2.4 m mirror like Hubble’s about 0.058″. Past that, more power only enlarges the blur.

What is the focal ratio of a telescope?

The focal length divided by the aperture, written f/10 for a 1,000 mm telescope of 100 mm. With the same eyepiece, a higher focal ratio gives more magnification and a smaller exit pupil.