Extension Tube Magnification Calculator
Enter the lens focal length, native magnification, tube length, sensor size, and aperture to estimate the magnification range and close-up exposure effect.
Maximum magnification0.65x = 1:1.54
- Added magnification
- 0.5
- Magnification range
- Infinity setting: 0.5x = 1:2; closest focus: 0.65x = 1:1.54
- Subject area at maximum magnification
- 55.38 mm x 36.92 mm
- Approximate focus-distance range
- Infinity setting: 225 mm; closest focus: 209.42 mm
- Effective f-number at maximum magnification
- 13.2
- Total macro exposure factor
- 2.72
- Total exposure loss (stops)
- 1.44
- Exposure loss added by the tube (stops)
- 1.04
Uses a thin-lens estimate; internal-focusing and floating-element lenses can differ materially.
Focus distance is sensor-to-subject distance, not clearance from the front of the lens.
Through-the-lens metering may already compensate for the calculated light loss.
How to use this calculator
- Enter the focal length selected on the lens.
- Enter the lens native maximum magnification as a decimal, such as 0.25 for 1:4.
- Add the physical length of every extension tube in the stack.
- Enter the camera sensor dimensions and selected aperture.
- Read the maximum magnification, subject area, focus-distance range, and exposure loss as planning estimates.
How extension-tube magnification is calculated
An extension tube moves the lens farther from the sensor. In a thin-lens model, extra extension adds magnification equal to the extension length divided by focal length. Let f be focal length in millimeters, E be total extension, m0 be the lens native maximum magnification, Sw and Sh be sensor width and height, and N be the marked f-number.
The added magnification is:
a = E / f
At the infinity-focus setting, the tube gives about mFar = a. At the lens's closest native focus, the estimate is mMax = m0 + a. The subject area that fills the frame at maximum magnification is Sw / mMax by Sh / mMax. A full-frame sensor at 0.5x therefore covers about 72 mm by 48 mm, while the same sensor at 1x covers about 36 mm by 24 mm.
The calculator also estimates sensor-to-subject focus distance with the thin-lens relationship:
D(m) = f * (2 + m + 1 / m)
That distance is shown for the infinity setting and the closest-focus setting when magnification is above zero. With zero extension, the far end remains infinity. If both extension and native magnification are zero, subject coverage and the close-focus distance are undefined because the setup has no finite close-up magnification.
Exposure and effective aperture
Close-up magnification reduces the light reaching each unit of sensor area. Using the symmetric-lens estimate, effective f-number at maximum magnification is Neff = N * (1 + mMax). The exposure factor is B = (1 + mMax)^2, and the total exposure loss relative to infinity focus is L = 2 * log2(1 + mMax) stops.
The tube-only loss compares the new maximum magnification with the lens's native maximum magnification: Ltube = 2 * log2((1 + mMax) / (1 + m0)). This is useful when a macro lens already loses light at close focus and you want to know what the added tube contributes.
What moves the result most
Focal length and tube length are the main levers. A 25 mm tube on a 50 mm lens adds 0.5x, but the same tube on a 100 mm lens adds 0.25x. Native magnification matters at the close end: a lens that already reaches 1:2 starts much closer than a general-purpose lens that only reaches 1:8.
What this calculator leaves out
The model treats the lens as a thin, symmetric lens. Internal-focusing, retrofocus, telephoto, zoom, and floating-element lenses can change effective focal length, pupil magnification, and working distance as they focus. The focus distance shown here is measured from the sensor plane to the subject plane, not from the front of the lens, so real clearance can be much shorter.
Worked example
Suppose a 50 mm lens has native maximum magnification of 0.15x, a 25 mm extension tube, a 36 mm by 24 mm sensor, and the aperture set to f/8. The added magnification is 25 / 50 = 0.5x. The range is about 0.5x at the infinity setting to 0.15 + 0.5 = 0.65x at closest focus, or 1:2 to about 1:1.54.
At 0.65x, the frame covers 36 / 0.65 = 55.38 mm by 24 / 0.65 = 36.92 mm. The approximate sensor-to-subject distance is 225 mm at the infinity setting and 209.42 mm at closest focus. The effective aperture is 8 * (1 + 0.65) = f/13.2. The total exposure factor is 1.65^2 = 2.72, or 1.44 stops relative to infinity focus; the tube adds about 1.04 stops beyond the lens's native 0.15x close-focus loss.
Common questions
How much magnification does an extension tube add?
In the thin-lens estimate, added magnification is extension divided by focal length. A 25 mm tube on a 50 mm lens adds 0.5x, while a 25 mm tube on a 100 mm lens adds 0.25x.
Which extension tube length should I use with a 50 mm lens?
Choose the tube length from the added magnification you want. On a 50 mm lens, 12 mm adds about 0.24x, 25 mm adds 0.5x, and 50 mm adds 1x. Real working distance may become tight before the math looks impractical.
How do extension tubes affect exposure and effective aperture?
Extension tubes increase magnification, which raises the effective f-number and exposure factor. At 1:1, a symmetric-lens estimate gives an exposure factor of 4, or 2 stops, and a marked f/8 behaves like about f/16 for exposure and diffraction planning.
What subject area will fill my camera sensor?
Divide the sensor dimensions by magnification. A 36 mm by 24 mm sensor at 0.5x frames about 72 mm by 48 mm, and the same sensor at 1x frames about 36 mm by 24 mm.
Why is the real working distance different from the calculated focus distance?
The calculator reports approximate sensor-to-subject distance from a thin-lens equation. Working distance is the clearance from the front of the lens to the subject, so it also depends on lens length, optical design, hood position, and internal focusing.
Can extension tubes be stacked?
Yes. Add their physical lengths and enter the total extension. Stacking tubes increases magnification and exposure loss, but it can also reduce working distance enough to make lighting and subject clearance difficult.