I'm working in computer vision and choosing the right lens is obviously as critical as having a good sensor if you're going for quality inspection (for example).

Sadly I have very little experience in lenses and what I've found doing some web search is not really clear to me: I'm talking about lens resolution.

Lens specifications are often expressed as LP/MM, or micron resolution at 550 nm wave lenght. But what does that mean?

As an example, for an application where a FOV of 600x500 mm is needed with acquiring distance of 500 mm I was told by a few sellers that a 10 Mpx camera with a telecentric lens would be fine to be able to clearly see 1 mm details on object surface. The lens (a very expensive one at $6000 where camera cost is $1000) has 6 micron resolution.

Since pixel on camera sensor are 1.6x1.6 micron that sound to me like I will not be able to use the full resolution of the sensor, so I'm not getting an effective 10 Mpx image.

Doing some easy math:

  • lens resolution of 6 micron correspond to 166 LP/MM

  • camera sensor with pixel size of 1.6 micron can reach 299 LP/MM

What am I misunderstanding?

As a rough choosing criteria, does the lens resolution have to be equal or greater than the camera resolution if you need as much details as possible?

  • \$\begingroup\$ UPDATE: following link to a related question ( photo.stackexchange.com/questions/59415/… ) I've found out that current sensors with more than 16 Mpx outresolves any (practical affordable) lens on the market in terms of LP/MM. So it seems like the lens is really the bottle neck in this case, even if I'm considering 1/3" sensor with C-mount lensens for industrial application. Not sure if I'm still missing something.. \$\endgroup\$ Mar 10, 2015 at 13:38
  • \$\begingroup\$ With 16 Mpx and a 600 mm FOV width I'm getting more or less 6 pixels/mm..Is that the maximum resolution I can go for? The only way to increase image quality would be working on luminance? I'm trying to understand what are the limits and what's the best I can expect \$\endgroup\$ Mar 10, 2015 at 13:45
  • \$\begingroup\$ It is unlikely that any current sensor outresolves modern, current lenses that are on the market. The problem with a lot of lens tests is that they are tested on current cameras. That imposes a convolution problem, where the output resolution of the system as a whole is less than either the resolution of the sensor or the resolution of the lens. Resolution is not some concrete thing either...it doesn't just stop...it fades. At MTF50, most modern lenses at diffraction-limited apertures are good, at MTF10 they are phenomenal. \$\endgroup\$
    – jrista
    Mar 11, 2015 at 1:30

2 Answers 2


Resolution is a complex thing. For one, there is a LOT of misinformation about resolution floating around the net, and many photographers do not quite understand it. First, I believe it is incorrect to say "A outresolves B" when talking about lenses and sensors. Sensors do not outresolve lenses. Neither do lenses outresolve sensors. As a matter of fact, the two work together in concert to convolve an image...and that image has a resolution of its own.

(NOTE: A 1.6 micron pixel can resolve 1/(1.6 / 1000) lines per millimeter, which comes out to 625 l/mm. Divide by two, for pairs of lines, and you get 312 lp/mm (line-pairs per mm). Similarly, a lens that can resolve a 6 micron spot can resolve 1/(6 / 1000) lines per millimeter, which comes out to 166 l/mm. Again divide by two and you get 83 lp/mm. Whatever this $6000 lens is, it is probably diffraction limited, but at an aperture of around f/8 (assuming MTF50). The price tag seems rather high for that...many lenses can get quite close to a resolution of 83 lp/mm at f/8...as that aperture is usually very close to fully diffraction limited on a majority of modern lenses.)

Resolution can be reported in line pairs per millimeter (lp/mm), or as a spot size (microns). The spot size, which is ultimately what a point of light resolves to, is the result of the light from the original source of that point being "convolved" as it passes through all the things between it and the sensor's pixels (which themselves are part of the convolution). That includes the air between the light source and the lens, each and every element in the lens, as well as all the air gaps between them, the air between the back of the lens and the sensor, the cover glass or filters over the sensor, the microlenses on the sensor, and the nature of the sensor's pixel layout itself (i.e., their size and whether they have a CFA or not). All of that works in concert to produce the spot size of that original light source in the final image.

The original light source may be infinitely small, however as its light passes on towards the sensor surface, it is being spread out. You don't end up with a mathematically infinitely small point in your final image, you end up with a point that is many pixels in size.

The lens in question may be able to resolve a diffraction limited (maximum potential) point of light at 6 microns in size. The sensor's pixels are 1.6 microns in size, and let's just assume for now it's monochrome. The size of that point of light in the final image is going to be the RMS of the components involved in making the image:

imageSpot = SQRT(lensSpot^2 + sensorSpot^2)

If we plug our numbers into this formula, we get:

imageSpot = SQRT(6um^2 + 1.6um^2) = SQRT(36um + 2.56um) = SQRT(38.56um) = 6.21um

The resolution of an image output by using this lens, which can resolve a 6 micron spot, with a sensor that has 1.6 micron monochrome pixels, is a 6.21 micron spot. That comes to an output resolution of 80.5 lp/mm. That is slightly less than the 83.34 lp/mm the lens is capable of delivering itself, however it is still very close. That is thanks to the fact that the sensor has such tiny pixels.

It is actually very rare for camera systems to resolve that close to the "upper limit", which is the resolution of the lowest power component. To get better resolution than 83 lp/mm, you would need to use a lens that is capable of resolving a smaller spot size than 6 microns. If you were able to find a lens that had a 3 micron spot, the lens would be truly capable of resolving 166.67 lp/mm, and your output resolution would jump to 147.1 lp/mm.

Resolution is the result of a convolution, it does not have a hard stop. As such, sensors cannot outresolve lenses, nor can lenses outresolve sensors. The two work in concert to resolve the information in an output image. We are generally quite far from the limits of resolving power with current optics. We are getting close to the limits of resolving power with sensors, which are now down to around 0.95 microns in pitch (950nm, infrared light wavelengths). Your sensor in this situation is actually holding you up; a lens that can only resolve a 6 micron spot is actually holding you back. You could continue to gain resolution well past the point where the lens is resolving a 1.6 micron spot, same as the sensor, since that is:

SQRT(1.6^2 + 1.6^2) = 2.26 microns

Your resolution would be 220 lp/mm, still well below the individual resolving power of each which is 312 lp/mm.

  • \$\begingroup\$ Thanks for your contribution, actually I've ended up into your discussion on canonrumors following the link to a related question here! This is indeed a complex subject, I'm starting to understand basics but still didn't catch the meaning of a couple of things: should camera sensor oversample lens resolution as Michael said? And since I'm not going for microscopy in this example, will I really need a lens this good? I mean, with 600 mm width of FOV I'm not getting more than 6 pixels/mm, so what's the general criteria i should follow to choose the right lens/sensor? \$\endgroup\$ Mar 11, 2015 at 8:03
  • 1
    \$\begingroup\$ Yes, it is better when your oversampling. There is something called the Nyquist Rate, which is the minimum sampling rate at which you can minimally approximate an analog signal with any decent accuracy. For audio signals, the nyquist rate is 2x, or double the frequency of the audio. For spatial frequencies, it's actually 3.3x, when you account for the second dimension. That is, again, the MINIMUM sampling rate to get a decent approximation. You want to sample more than that for best results. Now, oversampling is not the same thing as outresolving, note that. ;) \$\endgroup\$
    – jrista
    Mar 11, 2015 at 8:08
  • \$\begingroup\$ Ok, actually I was thinking that outresolving was the same of oversampling :) thanks again. \$\endgroup\$ Mar 11, 2015 at 8:24
  • \$\begingroup\$ Resolving is what produces the resolution of your output image. Oversampling has to do with how spatial frequencies mesh. Since the output resolution is the result of a convolution, there is no such thing as one thing outresolving another...however it is possible for a sensor to oversample the spatial frequencies of a lens. Also note...oversampling is generally considered for a given contrast ratio...such as MTF50. Lens resolving power increases as contrast drops...at MTF10, lenses resolve FAR higher resolution at most apertures than any current sensor, assuming they are diffraction limited. \$\endgroup\$
    – jrista
    Mar 11, 2015 at 8:33
  • \$\begingroup\$ As I'm working more and more in machine vision I really need to look deeper into this and handle the fundamentals. Now it's clear how to choose the proper sensor resolution depending on the lens I'm using, but still not so clear what I can expect out of a lens with 100 LP/MM instead of a 200 LP/MM one for a defined FOV. Any literature suggestions? :) \$\endgroup\$ Mar 11, 2015 at 16:40

Remember for the camera LP/MM , you can only see those linepairs if they are completely aligned with the pixels. Therefore it is good to oversample like in audio sampling. For example, if you need to sample a 20kHz sine wave, you can be lucky with 40kHz to capture the positive and negative peak, or you can be unlucky and capture only zeroes. So it is good to sample the 20kHz wave with 96kHz.

This aliasing is compensated with a low pass filter glass on the sensor. So the resolution of the camera is quite lower than the 299 lp/mm. Unfortunately, the blur from 166 lp/mm on the lens and the blur from the sensor glass is not a "max" function but a multiplication.

This is why images at 100% always look blurred even though theory should work. The best way to capture more details is to "oversample" your linepairs but getting closer so each linepair is sampled with 4-8 pixels if it's monochrome. If it's a colour chip with bayer pattern which makes an unnatural resolution artifacts, it is good to oversample 8-16 times.

  • \$\begingroup\$ If i've get your point, in the example above i'd be oversampling by a factor 2, right? But how about 8-16 times? It starts from 1300 to 2600 LP/MM, so a 0.3 micron pixel size? I'm not understanding.. \$\endgroup\$ Mar 10, 2015 at 15:15
  • \$\begingroup\$ yes that would be good, hard to do in practice. your lens is still only 166lp/mm so you have to increase your sensor to have more mm's to project on. or decrease your FOV so less linepairs has to be projected on the sensor, and then stitch. So when you cannot increase your lp/mm, you can reduce the need for a high number. \$\endgroup\$ Mar 10, 2015 at 16:27
  • \$\begingroup\$ Thank you. I can't get closer in this case, so i guess that is the limit resolution I'm gonna deal with. "Only" 166 LP/MM? Reading some posts around the web I didn't found any bigger resolution and this one seems to be already quite high (this is not a microscopy application), isn't it? So the basic concept is that with the same lens i'm getting slightly better images with a higher resolution sensor because of this oversampling factor? I thought it was all about the lens, since is that one limiting the amount of light hitting the sensor.. \$\endgroup\$ Mar 10, 2015 at 19:38

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