5 ms·

How many frames per second would you need to film at for a photon of light to be in an identical position on 2 frames?

by izzydata 8y ago

jwalton 8y ago
Light travels roughly 300 billion m/s, so at 10 trillion fps, light moves about 3cm.
- rapnie 8y ago
  Joking? Wikipedia: 299,792,458 metres per second
  - caseymarquis 8y ago
    Wouldn't it be fantastic if it traveled that fast though! The benefit to computing would be enormous.
    - rapnie 8y ago
      Wonder how the universe would look like, then, through our telescopes.
refurb 8y ago
Speed of light ~300 million km/s. 10 trillion frames per second. Light would have travelled 3cm on each frame.
- rapnie 8y ago
  If you change the km to m: 299,792,458 metres per second
  - deleted 8y ago
    [deleted]
max68 8y ago
Δd = 1cm = 0.01m v = 3e8 m/s v*Δd = 3e6 frames/s For the camera to pick up 1cm of difference between frames, you have to capture 3x10^6 frames per second. You can do the limit for how much you want your delta to be just by multiplying by the speed of light.
emeraldd 8y ago
For 635nm red laser light, you'd need to be sampling somewhere in the order of 9.4 x 10^14 times a second to get two samples per cycle. Based on roughly 300000000m /635nm x2 but then the question comes down to how many cycles make up a photon and does that question even. Make sense in the first place.
- izzydata 8y ago
  I apologize for my uninformed question.
  - emeraldd 8y ago
    No apology needed. When things get that small and you're dealing with light, it's a legitimate question. See: https://en.wikipedia.org/wiki/Wave%E2%80%93particle_duality https://en.wikipedia.org/wiki/Wave%E2%80%93particle_duality
- abhishekjha 8y ago
  >9.4 x 10^14 Can you explain how did you get this number for the red wavelength? Also can we fundamentally "see" a photon?
  - emeraldd 8y ago
    I was working from the idea of a nyquist or sampling rate needed to "perfectly" reproduce a wave form. I know this works for lower frequencies but I have no idea how well it translates to something like optical frequencies/light. You can find frequency from speed / wavelength: f = v/l or in our case f = c / l since we're dealing with light. I was using 635nm as the wavelength (basically a red laser). That gives you: 3 x 10^8 / 635 x 10^-7 = 4.7 x 10^14 Which should be about 470 terahertz (4.72 x 10^14) give or take a bit. To sample that "perfectly" you'd need to sample at twice the frequency or about 940 terahertz or (9.4 x 10^14). As to your second question, I know that there are single photon detectors. Past that, you've got me. I don't know if that can be classified as "seeing" or not. As to size, there's https://briankoberlein.com/2015/04/14/thats-about-the-size-of-it/ https://briankoberlein.com/2015/04/14/thats-about-the-size-o... but that might or might not make sense. That's about the limit of what I'm willing/able to say on the subject.
cimmanom 8y ago
A photon needs to actually hit the sensor in order to be recorded, so if we're considering the photon to be a particle (rather than a wave) it would have to be in the same position in order to be in two frames at all.

How many frames per second would you need to film at for a photon of light to be in an identical position on 2 frames?

jwalton 8y ago

Light travels roughly 300 billion m/s, so at 10 trillion fps, light moves about 3cm.

rapnie 8y ago
Joking? Wikipedia: 299,792,458 metres per second
- caseymarquis 8y ago
  Wouldn't it be fantastic if it traveled that fast though! The benefit to computing would be enormous.
  - rapnie 8y ago
    Wonder how the universe would look like, then, through our telescopes.

refurb 8y ago

Speed of light ~300 million km/s. 10 trillion frames per second. Light would have travelled 3cm on each frame.

rapnie 8y ago
If you change the km to m: 299,792,458 metres per second
- deleted 8y ago
  [deleted]

max68 8y ago

Δd = 1cm = 0.01m v = 3e8 m/s v*Δd = 3e6 frames/s For the camera to pick up 1cm of difference between frames, you have to capture 3x10^6 frames per second. You can do the limit for how much you want your delta to be just by multiplying by the speed of light.

emeraldd 8y ago

For 635nm red laser light, you'd need to be sampling somewhere in the order of 9.4 x 10^14 times a second to get two samples per cycle. Based on roughly 300000000m /635nm x2 but then the question comes down to how many cycles make up a photon and does that question even. Make sense in the first place.

izzydata 8y ago
I apologize for my uninformed question.
- emeraldd 8y ago
  No apology needed. When things get that small and you're dealing with light, it's a legitimate question. See: https://en.wikipedia.org/wiki/Wave%E2%80%93particle_duality https://en.wikipedia.org/wiki/Wave%E2%80%93particle_duality
abhishekjha 8y ago
>9.4 x 10^14 Can you explain how did you get this number for the red wavelength? Also can we fundamentally "see" a photon?
- emeraldd 8y ago
  I was working from the idea of a nyquist or sampling rate needed to "perfectly" reproduce a wave form. I know this works for lower frequencies but I have no idea how well it translates to something like optical frequencies/light. You can find frequency from speed / wavelength: f = v/l or in our case f = c / l since we're dealing with light. I was using 635nm as the wavelength (basically a red laser). That gives you: 3 x 10^8 / 635 x 10^-7 = 4.7 x 10^14 Which should be about 470 terahertz (4.72 x 10^14) give or take a bit. To sample that "perfectly" you'd need to sample at twice the frequency or about 940 terahertz or (9.4 x 10^14). As to your second question, I know that there are single photon detectors. Past that, you've got me. I don't know if that can be classified as "seeing" or not. As to size, there's https://briankoberlein.com/2015/04/14/thats-about-the-size-of-it/ https://briankoberlein.com/2015/04/14/thats-about-the-size-o... but that might or might not make sense. That's about the limit of what I'm willing/able to say on the subject.

cimmanom 8y ago

A photon needs to actually hit the sensor in order to be recorded, so if we're considering the photon to be a particle (rather than a wave) it would have to be in the same position in order to be in two frames at all.