If you have two points, you now have to resolve them. One criteria is the Rayleigh criteria. If the two points of light diffracted through the retina cause two Airy rings, then the minimum of one Airy ring must fall on the maximum of the other ring. It's a pretty straightforward calculation, you can find in any astronomy book, since astronomers use this criterion to resolve two stars seen through a telescope.
Once the size of the object becomes very large with respect to the wavelength, the the eye averages over a range of brightnesses and then takes the log of those brightnesses; actually the brain does all of this, although some initial pre-processing does take place at the retina before the data is passed on to the visual cortex/cortices. So, for a thousand points of light, assuming that they are each large enough and spread out over a distance large compared to the wavelength of the light, it depends on the relative brightness of each of the thousand points and their separation and orientation. Of course, if you were inline to these points, you'd see just one point of light and we're back to a single point of light!
In an 8' foot room filled with objects, the light leaving the room is an average over all the light leaving all the objects, to create a waveform. At this level, you're no longer in the region of Airy rings, the retina captures the entire field, forms the waveform, averages it and processes it as one "scene". The point is that the eye computes averages but you can't average over just one thing and averaging over a limited number of things, like two or three, gives an inaccurate picture. As the number of "things", ie the number of objects in the visual field, increases, the average gets better and better. Also, the eye is both a refractive structure and a diffractive one. At low levels of light for a limited number of "things", the diffractive effects take precedence, but as the number of "things" increases, the refractive effects "take over".Johnfp wrote:How many points are in a 8 foot sqare room filled with objects that can be used as the object beam (of course you would need a nice coherence length) to create a transmission hologram only 2" square. Are the number of fringes in that 2" hologram equal to the number of points in the room?
If you want to make a hologram of the room onto a 2" square, you'd have to place the 2" square far enough away that it would capture the entire waveform. This then brings you back to the formula Tony quoted (why does it look so familiar?). You could, of course, use a lens to focus the light from the room onto a 2" square, if you didn't mind that the longitudinal magnification was not equal to the lateral magnification. In other words, the 3d scene of the room would be squashed along the axis of the hologram.
No. Fringes are not refractive structures, they're diffractive. Think of a single fringe like a line of rocks at the beach; as the wave hits the line of rocks, it diffracts. In other words, the line of rocks don't allow the water "through" them, they divert the flow of water.Johnfp wrote:Couldn't one fringe actually refract the beam like a cylindrical lense and form all 1000 points? The instesity may be weak, but...