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Theoretical limits are probably very high. For one, collecting area can be pretty large. Obviously you don't want a square meter lens, but a foot-wide one wou

Theoretical limits are probably very high. For one, collecting area can be pretty large. Obviously you don't want a square meter lens, but a foot-wide one would probably be acceptable for a specialized camera. Second, nocturnal animals like howls see very well in the dark. Granted, they don't see colors but they still show that in theory seeing in the night is physically impossible. PS. regarding the collecting area, something I've been wondering for a bit : isn't the number of photons a lens can capture proportional to the square of the surface and not the surface itself? I know that sounds counter-intuitive but with interference, quantum mechanics and stuff, I checked the math once and I could not see where I was wrong.

greglindahl 10y ago
I don't know what you mean by "surface", but the number of photos collected depends on the area of the lens. No quantum mechanics or interference involved.
- grondilu 10y ago
  I did mean area. And I thought the number of photons was proportional to the square of the area because the probability amplitude is proportional to the area. Therefore the probability should be proportional to the square of the area, shoudn't it? It also seemed to make sense to me otherwise we would not need to build large telescopes, we could just build lots of small ones and fuse the images.
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  - titanomachy 10y ago
    > the probability should be proportional to the square of the area, shoudn't it? No.
    - grondilu 10y ago
      What's wrong with the reasoning, then? Isn't the probability amplitude proportional to the area?
      - analognoise 10y ago
        It's proportional to the area, which is already a term encompassing length^2.
        grondilu 10y ago
        If the probability amplitude is proportional to the area, it doesn't matter if that term already encompasses a length² term. QM tells you then that the probability should be proportional to the square of that amplitude, even if that means it would encompass a length^4 term.
    - candiodari 10y ago
      And if we change to "proportional to the area of the lens, not the sensor" ?
  - davrosthedalek 10y ago
    No, the probability is proportional to the area, not the amplitude. If a sensor of size A sees N photons, a sensor of the same size next to it sees another N. Fusing these two sensors together see 2N, not 4N. Otherwise, you violate energy/momentum conservation.
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    - TeMPOraL 10y ago
      I love this kind of explanations. You can skip a lot of hard math with a simple counterexample that shows the absurdity of proposed theory.
  - greglindahl 10y ago
    We do build small(er) ones and fuse the images (using interferometry), that's what many large telescopes are these days. In the radio, we've been mostly doing it that way since 1978 (VLA).
    - grondilu 10y ago
      Interferometry, as its name indicates, uses interferences. So it adds probability amplitudes.
      - greglindahl 10y ago
        ... we use interferomers because we want the added resolution, not just the additional photons. If you don't want extra resolution, you can just add up the images from all of the smaller telescopes. Light is a wave and a particle, and if you are getting wildly different answers from thinking about it as a wave and as a particle, and you're looking at a macro and not micro scale, then you're doing it wrong. That's why I answered your wave question with a photon count answer.
        grondilu 10y ago
        > That's why I answered your wave question with a photon count answer. But how do you count photons without using probability amplitudes? If you count them by using a classical reasoning of photons being small particles falling from the sky, I'd say you're doing it wrong, because photons are not classical particles.
        greglindahl 10y ago
        CCDs count photons in a particular fashion, and it happens to involve individual photons doing things. You might think of it as photons knocking electrons off of atoms, but it's actually semiconductors with a narrow bandgap, so CCDs work at much lower energies than needed for ionizing radiation.
- tigershark 10y ago
  Nope, nothing to do with that. A lens with a 1m diameter f/10 will be less bright than a lens with a 10 cm diameter f/1.4
  - greglindahl 10y ago
    You might want to read https://en.wikipedia.org/wiki/F-number#Focal_ratio_in_telescopes https://en.wikipedia.org/wiki/F-number#Focal_ratio_in_telesc...
    - tigershark 10y ago
      That is valid only for point sources. Stars are not points when seen in a telescope because of airy disks, atmospherical scattering and imperfections in the optics. If you have perfect optics, you are outside of the atmosphere and the only limit is the airy disk then yes, for point sources you don't have that problem. Now, are you saying that a valley is a point source?
      - greglindahl 10y ago
        If you look back, I made a statement about the number of photons collected. I totally understand that camera people, who have a shitload of photons available most of the time, don't think about them like astronomers do. But if it's very dark, it becomes astronomy.
    - tigershark 10y ago
      Not sure why I cannot reply to your other comment. Anyway, astronomy has nothing to do with the scarcity of photons. If you take a picture of M31 you have a relative abundance of photons, with most of them not being point sources. If you look at any planet or globular cluster or nebula is the same. Only when you are observing single stars you are having that. And single star observations are for sure not the whole astronomy.
      - greglindahl 10y ago
        HN requires a certain delay before responding, to encourage people to think first. Many distant things, like galaxies, are very dim. M31 is almost visible to the naked eye, which is not dim to an astronomer. You might want to study astronomy before having strong opinions about it.
grondilu 10y ago
Erratum: instead of "they still show that in theory seeing in the night is physically impossible.", read "they still show that in theory seeing in the night is NOT physically impossible."
tigershark 10y ago
The resulting brightness has nothing to do with the lens size. The important factors are the focal ratio and the sensor size. Obviously the sensor should have an appropriate size to receive all the focused light otherwise it's just wasted.
- scythe 10y ago
  C'mon HN this guy is right, you can't apply Gauss's Law with a sphere when you have photons from every angle. However I'm not sure whether the total number of photons collected is the right thing to be measuring. Aren't there serious aberrations at high magnifications in practice? And aren't there, y'know, glowing objects in the night sky? Stars I believe? The moon? I don't know how many photons there are but a single green photon carries a miniscule 376 zeptojoules and I don't think my eye responds to anything below the picojoule range. Counting photons and looking for the information limit seems a little extreme in this, er, light. I wonder whether you could use this technology for medical imaging if you shield the camera from any light that isn't being transmitted through the body. The possibility of recovering color information is exciting.
- banaanbanaan 10y ago
  Surely 'nothing' is an exaggeration. In the limit of the lens size going to zero, the brightness also goes to zero. (Ah, you're also scaling other things in your mind I guess, so then this is not true.) If you keep the setup such that all the light that comes through the lens will get to the sensor, and keep the sensor the same size, a larger lens would mean a higher brightness, right? Maybe in practice you're limited by other factors, but 'nothing' can't be right. Especially when we're talking about theoretical limits. Edit: I think what you're saying is that the lens doesn't matter for the brightness if you scale the focal length and sensor size in such a way that the brightness stays the same. Which is obviously true, but if you leave out the last part, it isn't.
  - tigershark 10y ago
    A lens with a smaller focal length will produce a projected image smaller than the same lens with a longer focal length. A lens with a focal length of 1m and an aperture of 50 cm will produce a projected image that will be 4 times bigger than a lens with a focal length of 50 cm and 25 cm of aperture. So even though the amount of light captured in the first case is 4 times, the effective brightness is exactly the same because it is spread in an area 4 times bigger compared to the second lens. For this reason the only way to increase the brightness is to reduce the focal ratio, simply increasing the aperture maintaining the same focal ratio won't help. Obviously if you have an aperture of size 0 then all the light is wasted because you can't have a sensor with size zero. In an empirical way you can see this concept in the real world with the abysmally small lenses in the smartphones that have the same brightness of the much bigger lenses in SLR cameras or in refractor telescopes.
candiodari 10y ago
An important difference between animal (and our) eyes is that the biological systems use logarithmic intensity in signaling, not linear. 2 pulses encode a signal that's a factor higher than a 1 pulse signal. You see this limitation in the video as well. That CCD video, the one with some vision and color but huge amounts of noise, I feel you see that it's clamping to some limited domain. Increasing it's range would improve things a lot.
duncan_bayne 10y ago
> Second, nocturnal animals like howls see very well in the dark. True, but that's because as far as we know they evolved on a very low light world orbiting a red dwarf star. They're not actually nocturnal on their homeworld. They also supplement their incredible night vision with sonar that's in the audible range for humans, hence their name.

greglindahl 10y ago

I don't know what you mean by "surface", but the number of photos collected depends on the area of the lens. No quantum mechanics or interference involved.

grondilu 10y ago
I did mean area. And I thought the number of photons was proportional to the square of the area because the probability amplitude is proportional to the area. Therefore the probability should be proportional to the square of the area, shoudn't it? It also seemed to make sense to me otherwise we would not need to build large telescopes, we could just build lots of small ones and fuse the images.
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- titanomachy 10y ago
  > the probability should be proportional to the square of the area, shoudn't it? No.
  - grondilu 10y ago
    What's wrong with the reasoning, then? Isn't the probability amplitude proportional to the area?
    - analognoise 10y ago
      It's proportional to the area, which is already a term encompassing length^2.
      - grondilu 10y ago
        If the probability amplitude is proportional to the area, it doesn't matter if that term already encompasses a length² term. QM tells you then that the probability should be proportional to the square of that amplitude, even if that means it would encompass a length^4 term.
  - candiodari 10y ago
    And if we change to "proportional to the area of the lens, not the sensor" ?
- davrosthedalek 10y ago
  No, the probability is proportional to the area, not the amplitude. If a sensor of size A sees N photons, a sensor of the same size next to it sees another N. Fusing these two sensors together see 2N, not 4N. Otherwise, you violate energy/momentum conservation.
  - deleted 10y ago
    [deleted]
  - TeMPOraL 10y ago
    I love this kind of explanations. You can skip a lot of hard math with a simple counterexample that shows the absurdity of proposed theory.
- greglindahl 10y ago
  We do build small(er) ones and fuse the images (using interferometry), that's what many large telescopes are these days. In the radio, we've been mostly doing it that way since 1978 (VLA).
  - grondilu 10y ago
    Interferometry, as its name indicates, uses interferences. So it adds probability amplitudes.
    - greglindahl 10y ago
      ... we use interferomers because we want the added resolution, not just the additional photons. If you don't want extra resolution, you can just add up the images from all of the smaller telescopes. Light is a wave and a particle, and if you are getting wildly different answers from thinking about it as a wave and as a particle, and you're looking at a macro and not micro scale, then you're doing it wrong. That's why I answered your wave question with a photon count answer.
      - grondilu 10y ago
        > That's why I answered your wave question with a photon count answer. But how do you count photons without using probability amplitudes? If you count them by using a classical reasoning of photons being small particles falling from the sky, I'd say you're doing it wrong, because photons are not classical particles.
        greglindahl 10y ago
        CCDs count photons in a particular fashion, and it happens to involve individual photons doing things. You might think of it as photons knocking electrons off of atoms, but it's actually semiconductors with a narrow bandgap, so CCDs work at much lower energies than needed for ionizing radiation.
tigershark 10y ago
Nope, nothing to do with that. A lens with a 1m diameter f/10 will be less bright than a lens with a 10 cm diameter f/1.4
- greglindahl 10y ago
  You might want to read https://en.wikipedia.org/wiki/F-number#Focal_ratio_in_telescopes https://en.wikipedia.org/wiki/F-number#Focal_ratio_in_telesc...
  - tigershark 10y ago
    That is valid only for point sources. Stars are not points when seen in a telescope because of airy disks, atmospherical scattering and imperfections in the optics. If you have perfect optics, you are outside of the atmosphere and the only limit is the airy disk then yes, for point sources you don't have that problem. Now, are you saying that a valley is a point source?
    - greglindahl 10y ago
      If you look back, I made a statement about the number of photons collected. I totally understand that camera people, who have a shitload of photons available most of the time, don't think about them like astronomers do. But if it's very dark, it becomes astronomy.
  - tigershark 10y ago
    Not sure why I cannot reply to your other comment. Anyway, astronomy has nothing to do with the scarcity of photons. If you take a picture of M31 you have a relative abundance of photons, with most of them not being point sources. If you look at any planet or globular cluster or nebula is the same. Only when you are observing single stars you are having that. And single star observations are for sure not the whole astronomy.
    - greglindahl 10y ago
      HN requires a certain delay before responding, to encourage people to think first. Many distant things, like galaxies, are very dim. M31 is almost visible to the naked eye, which is not dim to an astronomer. You might want to study astronomy before having strong opinions about it.

grondilu 10y ago

Erratum: instead of "they still show that in theory seeing in the night is physically impossible.", read "they still show that in theory seeing in the night is NOT physically impossible."

tigershark 10y ago

The resulting brightness has nothing to do with the lens size. The important factors are the focal ratio and the sensor size. Obviously the sensor should have an appropriate size to receive all the focused light otherwise it's just wasted.

scythe 10y ago
C'mon HN this guy is right, you can't apply Gauss's Law with a sphere when you have photons from every angle. However I'm not sure whether the total number of photons collected is the right thing to be measuring. Aren't there serious aberrations at high magnifications in practice? And aren't there, y'know, glowing objects in the night sky? Stars I believe? The moon? I don't know how many photons there are but a single green photon carries a miniscule 376 zeptojoules and I don't think my eye responds to anything below the picojoule range. Counting photons and looking for the information limit seems a little extreme in this, er, light. I wonder whether you could use this technology for medical imaging if you shield the camera from any light that isn't being transmitted through the body. The possibility of recovering color information is exciting.
banaanbanaan 10y ago
Surely 'nothing' is an exaggeration. In the limit of the lens size going to zero, the brightness also goes to zero. (Ah, you're also scaling other things in your mind I guess, so then this is not true.) If you keep the setup such that all the light that comes through the lens will get to the sensor, and keep the sensor the same size, a larger lens would mean a higher brightness, right? Maybe in practice you're limited by other factors, but 'nothing' can't be right. Especially when we're talking about theoretical limits. Edit: I think what you're saying is that the lens doesn't matter for the brightness if you scale the focal length and sensor size in such a way that the brightness stays the same. Which is obviously true, but if you leave out the last part, it isn't.
- tigershark 10y ago
  A lens with a smaller focal length will produce a projected image smaller than the same lens with a longer focal length. A lens with a focal length of 1m and an aperture of 50 cm will produce a projected image that will be 4 times bigger than a lens with a focal length of 50 cm and 25 cm of aperture. So even though the amount of light captured in the first case is 4 times, the effective brightness is exactly the same because it is spread in an area 4 times bigger compared to the second lens. For this reason the only way to increase the brightness is to reduce the focal ratio, simply increasing the aperture maintaining the same focal ratio won't help. Obviously if you have an aperture of size 0 then all the light is wasted because you can't have a sensor with size zero. In an empirical way you can see this concept in the real world with the abysmally small lenses in the smartphones that have the same brightness of the much bigger lenses in SLR cameras or in refractor telescopes.

candiodari 10y ago

An important difference between animal (and our) eyes is that the biological systems use logarithmic intensity in signaling, not linear. 2 pulses encode a signal that's a factor higher than a 1 pulse signal. You see this limitation in the video as well. That CCD video, the one with some vision and color but huge amounts of noise, I feel you see that it's clamping to some limited domain. Increasing it's range would improve things a lot.

duncan_bayne 10y ago

> Second, nocturnal animals like howls see very well in the dark. True, but that's because as far as we know they evolved on a very low light world orbiting a red dwarf star. They're not actually nocturnal on their homeworld. They also supplement their incredible night vision with sonar that's in the audible range for humans, hence their name.