I know there are other questions about this but I couldn't find the answer I was looking for...

I am developing a control system using a camera and I need to know theoretically the precision/resolution (I am not sure what term to use for this particular case) I can get with this camera. So I need to calculate the precision of what my camera can see with a given distance.

Despite searches I really can't find a formula to know how precise I would be with my camera. Can anyone help on that? Thank you in advance.

 - Sensor Size  : 3.7 mm x 2.8 mm
 - Resolution (HxV) : 659 px x 494 px
 - Pixel Size (H x V) : 5.6 µm x 5.6 µm
 - Focal length : 16.0 mm
 - Distance from the measured surface : 90 mm to 110 mm

EDIT: I am trying to measure variability on a surface with a contrast analysis after a machining. So I need to know how much/small I can measure with the camera in order to know the precision of my system.

  • It also depends on the sharpness of the lens. – Kai Mattern May 18 '20 at 9:21
  • Rather than asking the question this way, tell us what you're trying to get a photo of, and we can then help you work out what you need to do it. – Philip Kendall May 18 '20 at 10:01
  • @KaiMattern Yes I can imagine, I am trying to search for this information but apparently the manufacturer of the lens didn't include this information transparently – JackA May 18 '20 at 10:13
  • 2
    This appears to be a question about using cameras as measuring devices, not about photography, per se. – xiota May 18 '20 at 15:36
  • 1
    The information you desire is usually determined experimentally by photographing resolution charts. You'll need a chart that is the same size as the surface you wish to work with. – xiota May 18 '20 at 15:42

The magnification of an imaging system is the ratio of the size of an object on the sensor to the real-world size of the subject. This relation, when combined with the thin lens formula, and solving for subject size yields:

enter image description here

The smallest detail you can resolve corresponds to a single pixel in your resultant image, so the "image size" should just be your pixel size. Thus your minimal resolvable subject size is 5.6 µm x (100 mm – 16 mm) / 16 mm = 29 µm, or about 0.03 mm.

In order to resolve down to 0.01 mm, you need a magnification ratio of 5.6 / 10 = .56 or greater. With a 16mm lens, this means a subject distance of no more than 25 mm.

  • Thank you for your answer, that fits what the manufacturer just sent me ! However, I just found on the data sheet of the lens that there is a distortion of 0.96%, although insignificant, how can I integrate it in the model ? Should I just take 0.96% of 0.03mm? – JackA May 22 '20 at 10:20
  • Also why do we not take the pixel surface ? – JackA May 22 '20 at 14:46
  • @JackA You’d have to be more specific about the lens (i.e., post a link to the data sheet), but lens distortion specs are usually given as a graph vs distance from optical center, or as a combined spec number, such as “< 0.5% at 4mm”, or similar. This characterizes the lens as having pincushion, barrel, or even mustache distortion. Also, this should be a signed number, so positive distortion indicates pincushion distortion, negative indicates barrel. For lenses with distortion, the further away from the center of the image, the greater the distortion, as characterized by the spec. – scottbb May 22 '20 at 16:32
  • @JackA let’s assume the distortion is “0.96% at 1mm” from the center of the image (i.e., getting close to the top and bottom edges of the image). Then that means pincushion distortion has “squeezed” your image slightly towards the center, which reduces your resolution even more. Your magnification ratio at 1mm away from the center of the image has been reduced by 0.96%, which is the same as multiplying M by (1 – 0.0096). The magnification ratio is M = ƒ/(subj. dist. – ƒ). – scottbb May 22 '20 at 16:38
  • @JackA Also why do we not take the pixel surface? I’m unclear what you’re asking, but it sounds like you’re asking why the resolution is a distance metric, and not an areal one? That’s simply because the measure is fundamentally a radial measurement from the optical axis, that is, from the center of the image. Because pixels are an orthogonal grid, we can talk of horizontal and vertical spatial resolutions. On film, which doesn’t have natural “x” and “y” coordinates, everything is just some distance at an arbitrary angle from the center of the image. – scottbb May 22 '20 at 16:47

You will eventually need to physically test candidate systems to determine which lens-sensor combinations meet your needs. The lower bound of the size of what you can measure is limited by pixel size, as described by scottbb. However, there are other factors that further limit the detail you can reliably record.

  • Resolving power of the lens – The sensor cannot record what the lens cannot see.

  • Distortion, aberrations, and other optical imperfections – The resolving power of the lens varies across the frame. One solution is to use multiple images taken with a thin slice of the imaging circle. This is how some scanners work.

  • Diffraction and aperture size – See Diffraction-limited system.

  • Antialiasing filter – Many cameras have built-in filters that intentionally blur the image to prevent moiré. See Nyquist Frequency.

  • Color Filter Array – Provides data needed to reconstruct color, but reduces resolution as a side effect.

  • Pixel binning – Improves low-light performance at the expense of resolution.

  • Sensor bias and noise.

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.