Magnification in photography is the ratio between the size of a subject’s image on the sensor and its real size. A frame-filling bird photo might have a magnification of only 0.01 (1:100), while a true macro lens reaches 1.0 (1:1) – a 10 mm insect becomes 10 mm wide on the sensor. It is different from “zoom”: zoom compares focal lengths, magnification compares sizes.
Cameras and lenses specify the focus distance from the sensor plane (the ⦽ mark on the body) to the subject. That distance is the sum of the object distance g (lens to subject) and the image distance h (lens to sensor). For a thin lens these satisfy 1/f = 1/g + 1/h, so a known focus distance gives both and the magnification h/g. The closest focus is 4f for a simple lens, where magnification reaches 1:1.
Extension tubes are hollow spacers that move the lens farther from the sensor. They add nothing optical, but the extra image distance raises the magnification – by e ÷ f at infinity focus – which is why a 25 mm tube on a 50 mm lens gives about 0.5× even when the lens is set to infinity.
How to use the Camera Lens Magnification Calculator
- Focus distance (sensor to subject)
- Lens focal length
- Extension tube length (0 = none)
- Sensor width (for the area covered)
- Subject size (optional)
Results update as you type, every field has its own unit menu, and everything is calculated in your browser.
How it works
Enter the focus distance (sensor to subject) and the focal length. The quadratic from the thin-lens equation gives the object and image distances and the magnification. An extension tube adds to the image distance; the new subject distance and magnification follow from the same equation.
Worked example
A 50 mm lens focused at 1 m: r = √(1000²/4 − 50 × 1000) = √200,000 = 447.2 mm, so the image distance is 500 − 447.2 = 52.8 mm and the object distance 947.2 mm. Magnification m = 52.8 ÷ 947.2 = 0.0557 (about 1:18), and a full-frame sensor covers 36 ÷ 0.0557 ≈ 646 mm of the scene.
Add a 25 mm extension tube: the image distance becomes 77.8 mm, so m′ = 77.8 ÷ 50 − 1 = 0.556 (about 1:1.8) and the subject must now be 50 × 77.8 ÷ 27.8 ≈ 140 mm from the lens.
Magnification and reproduction ratio
| Magnification | Ratio | Typical use |
|---|---|---|
| 0.01× | 1:100 | Full-length portrait |
| 0.1× | 1:10 | Flower, product shot |
| 0.5× | 1:2 | Close-up / “half macro” |
| 1× | 1:1 | True macro – insects, coins |
| 5× | 5:1 | Extreme macro, e.g. the Canon MP-E 65 mm |
Inputs explained
| Field | What to enter |
|---|---|
| Focus distance (sensor to subject) | Units: m, cm, mm, ft, in, km … |
| Lens focal length | Units: mm, cm, in |
| Extension tube length (0 = none) | Units: mm, cm, in |
| Sensor width (for the area covered) | Units: mm, in, cm |
| Subject size (optional) | Units: m, cm, mm, ft, in, km … |
What each result means
- Magnification = [h]÷[g]
- Reproduction ratio 1 : x, x = = [g]÷[h]
- Object distance (lens to subject) = [g]
- Image distance (lens to sensor) = [h]
- Width of the scene covered = [Sensor width (for the area covered)]÷[Magnification]
- Size of the subject on the sensor = [Subject size (optional)]×[Magnification]
- Magnification with the extension tube = ([h] + [Extension tube length (0 = none)])÷[Lens focal length] – 1
- Subject distance from the lens with the tube = [Lens focal length]×([h] + [Extension tube length (0 = none)])÷([h] + [Extension tube length (0 = none)] – [Lens focal length])
- Tube magnification with the lens at infinity (e ÷ f) = [Extension tube length (0 = none)]÷[Lens focal length]
- Closest possible focus distance for a simple lens (4f) = 4×[Lens focal length]
Formula
r = √(d²/4 − f·d) · image distance h = d/2 − r · object distance g = d/2 + r · m = h ÷ g · with an extension tube e: h′ = h + e, m′ = h′/f − 1, g′ = f·h′ ÷ (h′ − f)
Worked example
Inputs — Focus distance (sensor to subject): 1 m; Lens focal length: 50 mm; Extension tube length (0 = none): 25 mm; Sensor width (for the area covered): 36 mm; Subject size (optional): 0.01 m.
Results — Magnification = 0.05573 ×; Reproduction ratio 1 : x, x = = 17.94; Object distance (lens to subject) = 0.9472 m; Image distance (lens to sensor) = 52.786 mm; Width of the scene covered = 0.646 m; Size of the subject on the sensor = 0.557 mm; Magnification with the extension tube = 0.55573 ×; Subject distance from the lens with the tube = 0.14 m; Tube magnification with the lens at infinity (e ÷ f) = 0.5 ×; Closest possible focus distance for a simple lens (4f) = 0.2 m.
Key terms
- Magnification
- Image size on the sensor divided by the real subject size.
- Focus distance
- Distance from the sensor plane to the subject.
- Extension tube
- A hollow spacer that moves the lens away from the sensor to focus closer.
- Working distance
- Space between the front of the lens and the subject.
Frequently asked questions
What magnification counts as macro?
1:1 (1.0×) or more is true macro. Lenses that reach 1:2 (0.5×) are often sold as "macro" too.
How do I get 1:1 with a normal lens?
Add extension equal to the focal length: about 50 mm of tubes on a 50 mm lens gives roughly 1:1.
Why is the magnification of a 55 mm lens at 100 m so small?
The image distance is about 55 mm and the object distance 100 m, so m ≈ 0.00055 – a 2 m subject is only 1.1 mm tall on the sensor.
Does an extension tube reduce light?
Yes. The effective f-number rises by (1 + m), so at 1:1 you lose two stops; through-the-lens metering compensates automatically.
Is focus distance measured from the front of the lens?
No – from the sensor plane mark on the camera body. The distance from the front of the lens is the working distance, which is shorter.
Why does the calculator say no result for very close distances?
A simple lens cannot focus closer than four times its focal length (2f object distance plus 2f image distance).