How is light—or actually any frequency in the EM spectrum—self-encoding and self-modulating 3D info in its 1D self, so that it “knows,” once it strikes a surface, to discharge all the data perfectly as a 2D image?? It’s not like FM or AM circuitry is being employed here... light is self-modulating to encode 3D scenery data perfectly without external interference to self-induce a modulation.
Time is the dimension you are missing!
At the end of the day the video signal amplitude at any given moment represents brightness (and colour) at one point, but that point scans across quickly and down the screen slowly.
Both the phosphors decay time and the eyes flicker fusion frequency combine to give the impression of a complete image.
The geometry is encoded in the sync pulses (Negative going pulses that drop below the normal black level) that are detected to reset the scan circuits to the start of a line or top of the screen.
\$\begingroup\$ Thanks for the answer... problem is, the signal is self-organizing 2D data into a 1D stream via time and retains the 1D->2D correspondence as it transposes back to 2D. How? \$\endgroup\$ Sep 27, 2020 at 23:07
\$\begingroup\$ My question applies to the photons in a pin-hole camera .... the photons go through the tiny hole and then splay back into an inverted 2D image. Where’s the transform function that instructs the photons to align in representation of the scene? \$\endgroup\$ Sep 27, 2020 at 23:17
2\$\begingroup\$ I think you need to edit your question to explain clearly what you are talking about. \$\endgroup\$ Sep 28, 2020 at 0:19
\$\begingroup\$ @Transistor ✅ done \$\endgroup\$ Sep 28, 2020 at 0:40
\$\begingroup\$ How is both brightness AND color (2 informational dimensions) being represented by any one 1D amplitude peak? \$\endgroup\$ Sep 28, 2020 at 4:59
light is self-modulating to encode 3D scenery data perfectly without external interference to self-induce a modulation.
The only case I can think of where that is true is in a hologram, where coherent light is split into 2 beams that interfere with each other (so technically not 'external' interference). Different path lengths to and from the 3D object 'modulate' the phases of the beams relative to each other, creating varying brightness when they interfere at the recording surface. this encodes the 3D image as a combination of Fresnal 'zone plates', which reproduce the image when illuminated by a point source.
In most optical imaging systems non-coherent light is focused by some external component which selectively bends, reflects or occludes the beams to produce an image on the 2D surface. This could be a lens, concave mirror, or pinhole. The focusing device uses the spatial directions of the beams to resolve the image. Without this 'external interference' no image is produced because the light is a mixture of photons with random frequencies and phases coming from all different directions.
A 1D image can be obtained by measuring the time taken for a pulse to reach the object and reflect back from it. This was how early RADAR worked, with only distance to the aircraft being shown. To get a 2D or 3D image the antenna must produce a narrow beam and be rotated to sweep across the sky. The other spatial dimensions are again encoded in the direction of the beam.
\$\begingroup\$ Thanks for the answer... “In most optical imaging systems non-coherent light is focused by some external component which selectively bends, reflects or occludes the beams to produce an image on the 2D surface. “ right, so in the case of a lens, it is “intelligently” taking those photons and “configuring them” in 2D such that they two-dimensionally reflect the spatial data encoded in the wave! \$\endgroup\$ Sep 28, 2020 at 3:39
\$\begingroup\$ I wouldn't say 'intelligently' but yes, the lens takes different beams radiating from one place and directs them to converge at another place. When focusing on distant objects the focal distance is proportionally compressed to 'squash' the image into 2 dimensions. A pinhole is less 'intelligent' - It simply discards all beams that are not coming (almost) directly from the object so it doesn't have to do any directing, and objects are always in 'focus' no matter how close or far away. Lenses also use the pinhole effect via the iris, which increases depth of field as it is made smaller. \$\endgroup\$ Sep 28, 2020 at 3:56
\$\begingroup\$ Why not intelligent?? It knows what to do with different photons in different circumstances, fourier-transforming and aligning away! \$\endgroup\$ Sep 28, 2020 at 4:01
\$\begingroup\$ @JordanFine The lens is actually just applying a phase delay to the incoming EM field. Due to how wave equations work, this causes the field to stop diverging and begin converging. Without understanding a lot of math, why a simple phase delay causes this to happen is not going to make sense to you however. \$\endgroup\$ Sep 28, 2020 at 4:05
1\$\begingroup\$ They don't coordinate (one photon per second would produce an image eventually), but they are mathematically dependent - by geometry. The bending may seem 'hella nuts' but is a well understood property of waves. It also occurs in water waves and sound waves. The principles can be proved with basic geometry and physics, and are a direct result of the 3D world we live in. \$\endgroup\$ Sep 28, 2020 at 5:47
photons go through the tiny hole and then splay back into an inverted 2D image
The answer is more readily found when the photons are treated as waves rather than particles. In a real-life scene there are waves (photons) coming from every angle, and those can be treated as mathematically independent, so lets focus (no pun intended) on a wave coming straight on the axis of your pinhole camera. That electromagnetic wave, having come from a far-away point source (say, a star), is essentially a spherical wavefront, but because the hole is so small it can be treated as "flat" or having uniform field strength, and the strength of the wave is uniform across the hole. (You can see this by putting a ring on a soccer ball -- a small ring allows you to see an approximately "flat" surface within the ring, but with a larger ring the curvature of the ball becomes obvious.)
Since the field strength at the focus (the back of the camera) is the 2d Fourier transform of the field at the hole (essentially uniformly flat), the light resolves to a point in the 2d-plane (that is, its a Dirac impulse function in two dimensions). Thus, the light from the star resolves to a single (spatial) impulse, which is a point of light, on the back of the camera.
This math can be applied independently, to every star in the sky, but accounting for the angle of arrival, so the image is created with little dots of light spread out across the back of the camera. As you get further off center, the approximation of "flat" and "uniform" (on the hole) becomes less and less true, so the edge of the image is fuzzy.
With a larger opening, the approximation of "flat" also begins to fail. Using a lens is actually compensating for a specific curvature of the field, so source that are too-far and too-near don't quite fit the equation, giving the lens a "depth of field". Shrink the aperture (make the useful hole in the lens smaller), and depth of field increases. A 3d camera takes advantage of this by effectively measuring the curvature of the field.
\$\begingroup\$ Thanks for the answer... “ In a real-life scene there are waves (photons) coming from every angle, and those can be treated as mathematically independent, lets focus (no pun intended) on a wave coming straight on the axis of your pinhole camera.“ But there are myriad waves require to produce the image... are they not self-organizing to reflect the spatial information? If they’re mathematically independent, how is a mathematically cohesive image formed? The waves somehow “know” to contiguously align ... they bear no resemblance to the image in mid-air—as they don’t as electrons down a wire \$\endgroup\$ Sep 28, 2020 at 3:46
\$\begingroup\$ Actually, in the case of pinhole camera, they do bear a resemblance to the image mid air, if you look behind the pinhole. In front of the pinhole they still do, but are buried among infinite other rays that do not contribute to the image so you don't see an image mid air. \$\endgroup\$– chamodSep 28, 2020 at 4:49
\$\begingroup\$ @chamod I meant in front of the pin hole, yeah... “they do not contribute to the image” implies the refraction process is filtering to a bias of 2D image formation/replication \$\endgroup\$ Sep 28, 2020 at 4:55
My question is applicable to the EM spectrum in general, but more specifically in the case of visible light, we can clearly see a continuous wave refracting into its discrete photons that transform to geometrically reflect what has been modulated when it hits the surface. Radio waves are manually induced modulation (encoding) and visible light waves are self-induced.
Other people have explained image formation abstractly, but I can give you a good example of how a 1D (transverse) wave can be modulated to form a 2D image.
Imagine a LIDAR/SONAR/RADAR scanner composed of a transmitter sending out a beam of radiation (light, sound, whatever) and a receiver (could be a lens, microphone, etc) pointed in the same direction. As the transmitter rotates, it will scan a line of energy across the surroundings. As the scanner passes over an object, it will reflect signal back in proportion to its reflectivity. This modulates the return signal with information about each point's reflectivity. The receiver will then detect this modulated signal, and color in a pixel in proportion to how much signal it receives. Gradually you can built up first a 1D line of pixels and then a 2D image.
This is imaging using raster scanning (1 point at a time), but if you just get two receivers side by side, you can map out two adjacent pixels for each location you point at. Get 4 pixels and you can map out 4 at once. Get 10 million and you can map out an entire 10 megapixel image in one shot. Each receiver is detecting a single 1D signal, and each signal is modulated by the reflectivity of whatever it is pointed at.
Now if you want to know why each receiver can detect a different point in the scene without them all seeing the same point (or maybe the sum of all points at once), you need to understand diffraction and Fourier optics (which other answers have addressed). But conceptually, it should be clear that if you have N receivers each pointed in a slightly different direction, you should be able to receive N modulated signals in parallel.
Thanks for the answer... how do you apply such logic to the image created by a pin-hole camera?
The pinhole camera is the 2D array example above (lots of parallel receivers), each pointed through the pinhole. Since the area around the pinhole completely blocks signal, each receiver can only receive signal from one spot through the pinhole. That the modulation from that spot is what it records.
You can actually draw this pretty easy. Imagine a 2x2 pixel sensor behind a pinhole. Draw a line with a ruler from each pixel on the sensor through the pinhole. You'll see that each strikes a different point in the surrounding scene.
\$\begingroup\$ Thanks for the answer... how do you apply such logic to the image created by a pin-hole camera? \$\endgroup\$ Sep 28, 2020 at 3:55
\$\begingroup\$ @JordanFine Edited. \$\endgroup\$ Sep 28, 2020 at 4:02
\$\begingroup\$ As Bruce said above, a lens is converging those pixels — just the ones needed to create a contiguous 2D replication. Point being, only until refracted, there is no observable mathematical interdependency of the photons... you cannot see the image mid-air until the lens “instructs” the photons to reconfigure into a 2D image... until such time, the photons are independent of each other...there is no explanation for this I can see. \$\endgroup\$ Sep 28, 2020 at 4:09
photons that are aligning contiguously and perfectly to reflect a 2D image... wrong ... a photon that strikes a random point in the scene and reflects randomly, and if by luck happens to travel through the pinhole, that photon can strike only a small area on the viewscreen because of the small size of the pinhole ... this repeats over and over, in huge numbers ... there is no aligning contiguously to reflect the image, in fact, it is quite random ... sort of like rain falling through a hole in a roof ... it is all an exercise in geometry \$\endgroup\$