Spring 2012

Spring 2012
by ani

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80

In order to get an image, both the subject and the "camera" must be much larger than the wavelength of the light that you use for imaging. The wavelength of visible light is between approximately 400 and 800 nm, i.e. smaller than a µm. Radio frequencies go up to several GHz, which corresponds to wavelengths of many centimeters. For example, the 2.4 GHz ...


19

I disagree with the answer with many upvotes. Physical lengths can be "swindled" in a number of ways and theoretically it would be possible to build a portable camera that snaps images of a very tiny portion of the electromagnetic spectrum. Plus, you are not considering that there are not only high-band signals, but also ultra-high-band signals that could be ...


12

There are two hard limits on how fast a lens can be: The first is a thermodynamic limit. If you could make a lens arbitrarily fast, then you could point it to the sun and use it to heat your sensor (not a good idea). If you then get your sensor hotter than the surface of the Sun, you are violating the second law of thermodynamics. This sets a hard limit at ...


10

You are focusing on objects, which are reflected. So you are not focusing on reflected surface. You are interested in light rays, which go from object "through reflection" to your camera. Not only from reflective surface to camera. You can try to shot for example puddle - try to focus on ground and you will se that reflection is blurred. Then try to focus ...


8

No. This is like saying you can determine the height of a house by counting steps to get up. The size of steps matters and so does the number. Dynamic-range should be determined by the well-depth (size of capacitors at a photosite) and noise floor (how much noise in the system when these is no signal). Knowing these two, one can compute the dynamic-range of ...


8

Physics plays a role in answering your question and that information is out there. The basics from that linked discussion are that the index of refraction of the lens material will affect the maximum aperture you can achieve, so for pure glass that has an index of refraction of 1.5, the maximum aperture would be f/0.5 or thereabouts. Better substances, such ...


7

Sort of. Not a "camera", but a computational imaging technique. We explore the feasibility of achieving computational imaging using Wi-Fi signals. To achieve this, we leverage multi-path propagation that results in wireless signals bouncing off of objects before arriving at the receiver. These reflections effectively light up the objects, which ...


7

An aperture could be closed which is effectively an infinitely large f-stop number since no light gets through. The fastest possible (smallest f number) is a bit harder. The speed of a lens is limited by the ratio of the entrance pupil to the focal length of the lens. The longer the focal length, the bigger the entrance pupil must be. In theory you could ...


7

Here's a good simulation of the 'perfect' sensor that you describe (one having zero electrical noise, thus recordning every incident photon perfectly) reacting to widely differing levels of light, from 0.001, 0.01 & 0.1 photons per pixel (top row), 1, 10 & 100 photons per pixel (middle row) to 1000, 10000 & 100000 photons per pixel across the ...


5

The focus distance is the distance to the object via the reflecting surface. Try taking a photo of yourself in a mirror at various distances, the distance from the camera to you, via the mirror, is twice the distance to the mirror. Your camera will indicate the focus distance as that.


5

Simple: You are focusing on the reflected subject, not the reflective surface. Ok, I'm not good at explaining this sort of stuff, I just understand how it works, but here's a drawing You see, when you are focusing on a subject, it's a reflection on the reflective surface, but the subject it's not there, it's further away, to explain better, lets say the ...


5

Well, sort of. Think about the sun shining through a lens — it's immediately apparent that the focused spot of light is brighter than the unfocused. However, the catch is that your "real view" also goes through a lens which focuses the light: your eye. So, in a sense, the real comparison is simply "Is there a lens which is brighter than the human eye?" — ...


5

Okay, so, the first thing to understand is that the textbook¹ is trying to get you to understand a theory put forward by Christiaan Huygens in the late 1600s. It turns out he was (generally) right about the wave nature of light, but the actual specifics are iffy. Don't get too bent out of shape trying to make everything make sense, because... well, it doesn'...


3

Another 'sort of' answer: One possibility, more analogous to a traditional camera, is to use a stationary receiver and a strongly directional antenna. If the antenna is directed in the same way that an electron beam moves across a CRT screen, a render of signal strength can be created that can then be overlaid with a photo taken from the same point. While ...


3

Various sources indicate that average reflectance of surfaces is somewhere between 12-18%. Calculating exposure from table values is not very typical. Most photographers would use a light meter, either built in or external.


3

Through @MichaelClark's help to clarify my question and @Achifaifa's hint to look at film curves, I found the answer to my question. It turns out that answer lies in the film characteristics curve. It plots the density of the silver halide in the film (i.e. how opaque the film is) as a function of the amount of light the film is exposed to (measured in log[...


3

When you are calculating the "correct exposure" ("Correct exposure" being the amount of light necessary to achieve a negative with a given density) you are actually playing with shutter speed, aperture and film sensitivity. If you restrict those variables to a fixed value, the only things you can do is overdeveloping or underdeveloping the film after the ...


2

This is physically impossible. Without active amplification, the luminance of the image can not be more than the luminance of the subject. Otherwise you would be violating the second law of thermodynamics. If you try to focus the sun rays on a black body, it can get really hot, but not hotter than the surface of the sun itself. For a mathematical proof, read ...


2

I'd like to disagree with the other answers - we'll maybe just question them. Take your magnifying idea. You are not making the sun brighter! You are just focusing the caught by the lens into a small point, making it appear brighter. Light gets "lost" through every surface it passes through or bounces off of. You may change the appearance of it's luminosity,...


2

This should be possible since the eye samples light from a relatively small surface area where as a lens samples light from a much larger area. The bigger problem is the huge discrepancy between the sensitivity of the eye and the sensitivity of sensors. I can already take photos with my Canon 5D Mark iii with fairly short shutters (sub 1/3 second, ...


2

Theoretically, yes. The human eye reportedly only opens as far as f/3.2, and there are many lenses faster than this. The Canon 50mm f/1.0 for example was marketed as being "faster than the human eye", although the f/3.2 figure suggests it shares that award with most prime lenses. The biggest obstacle is designing a reflex mirror, pentaprism and focus screen ...


2

There is a standard concept in photography called "exposure value". The question and answers at What is the EV scale? go into this in detail, but the quick version is that this is a series of numbers representing the amount of light in different scenes. Different combinations of aperture, shutter speed, and ISO can correspond to the same exposure (see this ...


2

For phase detection AF to work properly, the camera needs more than just a clear glass replacement for the removed filter, it needs a replacement of a different thickness. IR rays don't focus on the same point as visible light, which is why lenses have (used to have?) a red dot to tell you how to adjust focus from the visible light distance to the correct ...


1

Simple answer is no, at least not yet. I say this because if this was possible then equipment would exist in the test & measurement world. and instead we have equipment that can only use calibrated antennas to compute relative strength and frequency. You move a detector around and observe results. I think this is the kind of measurement system out ...


1

As there is currently no such camera known to me, it would be possible to build a quite effective one using an array of patch antennas to form a phased array. As such, a large flat antenna, say 1 by 1m, could be made from printed circuit board. However, a large amount of expensive HF components would be needed to integrate all individual antenna elements ...


1

I think you pretty much answered your own question, there's no hard limit as such. If you really wanted to, you could have a huge aperture and use corrective lenses to bring everything towards the sensors, but you run into two issues: price generally goes up to the square of the size of the glass, having this much would cost a lot image quality would ...


1

It's hard to provide a definitive answer to your question, because it depends on what you subjectively deem “useful”, as well as on many other factors like the strength and quality of the denoising algorithm, the output medium, etc... Thus, this is not really an answer, only some hints to help you find your answer. First, about the parameters to consider. ...



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