If I look at the space I can see the Andromeda Galaxy as a small dot in the sky. If I look through a telescope, I can see the spiral shape. If the saw right through a huge telescope, I could see a solar system inside and if the telescope were larger, could see the topology of one of its planets. Is that correct? If is correct then I wonder if a photon which was reflected by the surface of a planet in the Andromeda galaxy coming straight at us with other reflected photons: how much information can fit in a beam of light? or "rays" are mixed and there comes a point where, no matter how big the telescope, the information is lost. As when we expand an image on the computer and looks pixelated.
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$\begingroup$ Why the downvote? This seems to me to be a perfectly good question. $\endgroup$– John RennieApr 17, 2015 at 7:25
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$\begingroup$ Full Width Half Maximum = 1.02 * (wavelength) * (Focal Ratio) and Star Size = (Seeing * Focal Length)/206.3 just to share some facts if you are into astro-photography. $\endgroup$– user6760Apr 17, 2015 at 7:29
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2 Answers
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Suppose you are using a CCD or a photographic plate to record your image. The interaction with the light occurs when the detector absorbs a photon, and this happens at a point. So the image is built up from a collection of points - one for each photon that is detected.
In everyday life, e.g. taking pictures with the CCD in your phone, the intensity of the light is high enough that vast numbers of photons are captured. The intensity of direct sunlight is around 1 kW per square metre, and one visible photon has an energy of around $4 \times 10^{-19}$ joules, so in direct sunlight there are about $2.5 \times 10^{21}$ photons per square metre per second. Your camera captures far less than this because it captures reflected light and uses a short shutter time. However your picture is made up of so may dots (i.e. captured photons) that it can capture very fine detail.
But when photographing very distant galaxies the light intensity can be exceedingly low. For example in the Hubble Deep Field survey (here I quote from the web page):
Photons of light from the very faintest objects arrived at a trickle of one photon per minute, compared with millions of photons per minute from nearer galaxies.
In other words, if you used a shutter speed of one minute your picture would be just one dot. Obviously that wouldn't be a very useful picture! The Hubble Deep Field used a total exposure time of around 35 hours, so the pictures of the faintest galaxies would be made up of 2,100 dots.
answered Apr 17, 2015 at 9:32
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If a single red photon hits your telescope from the direction of a planet in the Andromeda Galaxy, then all you know that the planet emitted a red photon. Was it caused by a fire? A scattering of starlight through its atmosphere? An Andromedan with a laser pointer? A single photon of light from an unknown source has about as much information as a random letter from an unknown document. If I tell you that the letter "g" is in a certain book, can you determine anything about the book? Can you even determine what word the letter came from?
We can gather meaningful information from light by looking at the distribution of colors in a large sample of photons. For example, run sunlight through a prism and you'll see this:
http://en.wikipedia.org/wiki/Fraunhofer_lines
From these lines, you can tell what elements are present in the Sun. If you look at a similar star in the Andromeda galaxy, not only can you tell what the star is made of, but how fast it's moving based on the shifts of the lines compared to the Sun's spectra (the Doppler effect for light).
Just about the only way for a single photon to encode a significant amount of data is for that photon to be a part of a communication protocol, like Morse code, between two people who agreed on the protocol beforehand. This is not too far from how fiber optic communication works, and it doesn't require much light; in fact, in quantum cryptographic key exchanges, each photon carries one bit of information (compared to almost zero for light from an unknown source).
