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In this youtube video (which is an except from a documentation I believe) a DARPA development for a high-resolution surveillance camera is presented. This question is regarding the optics that are likely to be required to make it work.

Summary of the specifications given:

  • $1.8\times10^9$ pixel
  • covering $35\,km^2$
  • from an altitude of $5\,km$
  • using $368$ cellphone camera chips

Some back of the hand calculations:

  • pixel-equivalent area: $\frac{35\times10^6\,m^2}{1.8\times10^9\,px^2} = 1.94\times10^{-2}\,\frac{m^2}{px^2}$
  • edge length of pixel-equivalent area: $\sqrt{1.94\times10^{-2}\,\frac{m^2}{px^2}}=1.39\times10^{-1}\,\frac{m}{px}\approx14\,\frac{cm}{px}$
  • pixel count per chip: $\frac{1.8\times10^9\,px}{368\,chips}\approx5\times10^6\,\frac{px}{chip}$

Optics question

Cellphone chips have pixel pitches of the order of a few microns. Say they use a typical, decent quality 5 megapixel chip with $2.25\,µm$ pitch.

How are the optics likely to look like to image 15 cm at 5 km distance onto 2.25 µm on the chip within a camera that is at best a meter in diameter?

I am looking for answers beyond "they use a telephoto lens"... Lens focal lengths, combinations and setup for producing the required image is what I am struggling to work out.

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photo.stackexchange.com –  EnergyNumbers Jan 29 '13 at 17:12
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If the question concerns physical limits on the optics (i.e. Rayleigh criterion and/or super-resolution, design to limit aberration, ...) then it is on topic here, but I think that @EnergyNumbers may be right if the question is of the "select a lens" variety. Care to clarify? –  dmckee Jan 29 '13 at 17:20
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Using the lens equation, I get the system focal length to be $8.036cm = 803.6mm$. Of course that is just a first order calculation, and really not terribly difficult system to manufacture. Aberrations and weight etc. would be the real engineering challenges. –  daaxix Jan 29 '13 at 17:26
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Also, there are probably close to an infinite number of ways to make a system with this focal length, depending on the number of components and the specifications of each component. –  daaxix Jan 29 '13 at 17:34
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"there are probably close to an infinite number of ways to make a system with this focal length," Yes. And the manufacturers of photographic equipment have decades of experience in this art, which is why many such questions are arguably not physics questions at all. They are in the domain of specialized engineers (who, of course, make heavy use of physics as their playing field). –  dmckee Jan 29 '13 at 17:37

1 Answer 1

In first order this is a diffraction problem. No lens design can beat the diffraction limit, but many come close. The finest resolved angle is limited to $$ \theta = \frac{ 1.22 \lambda}{D},$$ where D is the size of your lens aperture.

So to resolve $15 cm=0.15 m$ at $5 km=5000 m$ , the tangent of your angle is about 1/33000. Therefore the size of your lens aperture would be have to be $$ D >= 1.22 \lambda* 33000 = 2 cm$$ for visible wavelengths. That is less than a good SLR already has, so it is doable.

Since, as was pointed out, a telelens can magnify the image, the pixel resolution does not really play a big role for the fundamental resolution limit.

The square root of 368 is about 19, a 5 Mpxl chip is about 2000 pxl across, at 2.5 um pitch that would be 5mm*19 $\approx$ 10 cm squared detector size with 40000 pixels across. For 15 cm resolution that would give a 6 x 6 km field of view.

The real problem is how to get those amounts across the airwaves (1.8 Gpxl* 24 bit color depth * 30 fps $\approx$ 1 Tbit/s before compression), if here back home Netflix can't handle Friday nights.

If you want to use a single lens, then the optics design won't be too hard to do on 'paper' (Zemax), but the lens will be large and expensive. Just compare full frame lenses from Nikon to amateur SLR lenses and think a factor 3 in diameter, 10 in area and 30 in weight.

Now, the not quite seamless edge to edge butting together of the cheapo chips is going to leave some dark lines on the image already. So it would make sense to use 4,9 , 16 or 25 COTS lenses and stitch the image in software. The more lenses you use, the smaller, cheaper and lighter the lenses will be in total. 25 lenses would be the hard limit since each lens has to have the 2 cm aperture to resolve the 15 cm. I'd use 16 lenses with each lens illuminating 21 chips (5x5 with the corners missing).

16*21=336, 16*23= 668, hm.

From a practical viewpoint one would not use cellphone chips, but just bundle the OEM versions of 16 superzoom cameras. Those lenses with their 2cm minimum aperture are the most expensive and the heaviest part of the setup.

The real trick is to do onboard preprocessing and analysis to winnow down the data rates needed. Think of an eagle, who can not likely resolve that mouse on the ground, but has superior data processing for those rare objects that are moving relative to a steady background.

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"No lens design can beat the diffraction limit," Worth looking up "superresolution" which concerns itself with making end-runs around this limit. –  dmckee Feb 9 '13 at 16:47
    
@dmckee: Good point. However, you have to throw away 'in band' information if you use frequencies outside the Fourier limit that are aliased back into the resolved frequencies. It's a pretty advanced topic and in general you'd need to know something about your signal composition (e.g. in the time domain, you sample at 1GHz with an oscilloscope, but the only signal is at 2.5 GHz, you'll see it at 0.5 GHz, but you have to know that there is no 0.5 GHz signal coming in). –  roadrunner66 Feb 9 '13 at 22:52

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