50 ms·

Indestructible light waves travel through opaque material as if it isn’t there

48 points by ithkuil 6y ago

fallingfrog 6y ago
Looks like it works for thin scattering layers, but you have to customize the light pattern for that specific sample and that specific angle and distance. So it’s not something that will allow a flashlight to shine through a wall.
- bozzcl 6y ago
  Yeah, seems it’s purpose-tailored to the material. The title made me think it did not interact with the barrier, which raised a ton of weird questions.
  - other_herbert 6y ago
    Right... it should also not interact with anything you’d be interested in seeing too...
  - doctor_eval 6y ago
    I assumed that the tailoring itself would reveal properties of the material, hence it’s application to imaging etc.
infogulch 6y ago
One step closer to the Tricorder.
- Buttons840 6y ago
  Until then the closest thing we have is dog's noses.
jameshart 6y ago
Opaque / oʊˈpeɪk /, adj. not transparent or translucent; impenetrable to light; not allowing light to pass through. I feel like maybe you need to be more precise in what you mean by 'opaque' before you can make a claim like this...
deleted 6y ago
[deleted]
barbegal 6y ago
Arxiv link to the paper https://arxiv.org/abs/2010.01075 https://arxiv.org/abs/2010.01075
barbegal 6y ago
I really hate when the press releases don't actually reflect the science in the paper. In this case there are no indestructible light waves and there is no opaque material. There is simply a wavefront which passes through free space (air) and through a scattering medium (what you would probably call a translucent material) such that the relative intensity of the two images produced are the same. The phase of the light is not the same and neither is the intensity because some is absorbed and reflected by the scattering medium.
- solipsism 6y ago
  free space (air) Surely that's not the science. Space and air are not the same thing. Air is, of course, a scattering medium.
  - jasonwatkinspdx 6y ago
    Free space is commonly used to refer to EM propagation in air in RF contexts. For example: https://en.wikipedia.org/wiki/Free-space_path_loss https://en.wikipedia.org/wiki/Free-space_path_loss Here the distinction is between propagation models that include ground plain, terrain, or airborne clutter.
    - solipsism 6y ago
      TIL. Still, I think the strongly in strongly scattering medium is important. It's very confusing to leave it out.
cratermoon 6y ago
OK dumb question but if x-rays are a form of light and we use them to see into the body through an opaque material (flesh), how is this news?
- Twisol 6y ago
  The body (at least, the fleshy parts) isn't terribly opaque to X-ray wavelength light -- the point being that opacity depends on wavelength. If you've got a (white) flashlight handy, you can stick your fingertip in front of the front element. A healthy amount of red light will come through. Flesh, as it happens, is also relatively transparent to strong red light... just not as transparent as to x-rays.
Twisol 6y ago
> To do this, the researchers first had to characterize the scattering medium precisely. At Utrecht University, they shone light through a layer of zinc oxide nanopowder deposited on a glass slide. They then used a detector on the other side to measure the light and analyse how it had been scattered by the powder. Once this has been done, Rotter explains, you can calculate “what kind of wave would be transmitted through this medium in the same way as through fresh space”. So it's not just any opaque material (and it's not even "opaque", really), and it's not just any waveform. The headline is pretty darn baity. Nonetheless, the science here is pretty cool. I could have imagined taking a scattering medium and a desired final waveform and computing the initial waveform that would produce the final waveform, but this seems to be saying that there exist waveforms that are essentially fixed points of the transform induced by the scattering medium. I wonder whether how exact that is -- whether the waveform really is identical, or whether any differences are simply hidden below the threshold of measurement sensitivity.
mateo1 6y ago
"Modelling scattering media with a thickness of 1-9μm" would be a better title. It's something, but the layer is so thin that the title looks like a joke.

48 points by ithkuil 6y ago

Looks like it works for thin scattering layers, but you have to customize the light pattern for that specific sample and that specific angle and distance. So it’s not something that will allow a flashlight to shine through a wall.

bozzcl 6y ago
Yeah, seems it’s purpose-tailored to the material. The title made me think it did not interact with the barrier, which raised a ton of weird questions.
- other_herbert 6y ago
  Right... it should also not interact with anything you’d be interested in seeing too...
- doctor_eval 6y ago
  I assumed that the tailoring itself would reveal properties of the material, hence it’s application to imaging etc.

infogulch 6y ago

One step closer to the Tricorder.

Buttons840 6y ago
Until then the closest thing we have is dog's noses.

jameshart 6y ago

Opaque / oʊˈpeɪk /, adj. not transparent or translucent; impenetrable to light; not allowing light to pass through. I feel like maybe you need to be more precise in what you mean by 'opaque' before you can make a claim like this...

deleted 6y ago

[deleted]

barbegal 6y ago

Arxiv link to the paper https://arxiv.org/abs/2010.01075 https://arxiv.org/abs/2010.01075

barbegal 6y ago

I really hate when the press releases don't actually reflect the science in the paper. In this case there are no indestructible light waves and there is no opaque material. There is simply a wavefront which passes through free space (air) and through a scattering medium (what you would probably call a translucent material) such that the relative intensity of the two images produced are the same. The phase of the light is not the same and neither is the intensity because some is absorbed and reflected by the scattering medium.

solipsism 6y ago
free space (air) Surely that's not the science. Space and air are not the same thing. Air is, of course, a scattering medium.
- jasonwatkinspdx 6y ago
  Free space is commonly used to refer to EM propagation in air in RF contexts. For example: https://en.wikipedia.org/wiki/Free-space_path_loss https://en.wikipedia.org/wiki/Free-space_path_loss Here the distinction is between propagation models that include ground plain, terrain, or airborne clutter.
  - solipsism 6y ago
    TIL. Still, I think the strongly in strongly scattering medium is important. It's very confusing to leave it out.

cratermoon 6y ago

OK dumb question but if x-rays are a form of light and we use them to see into the body through an opaque material (flesh), how is this news?

Twisol 6y ago
The body (at least, the fleshy parts) isn't terribly opaque to X-ray wavelength light -- the point being that opacity depends on wavelength. If you've got a (white) flashlight handy, you can stick your fingertip in front of the front element. A healthy amount of red light will come through. Flesh, as it happens, is also relatively transparent to strong red light... just not as transparent as to x-rays.

Twisol 6y ago

> To do this, the researchers first had to characterize the scattering medium precisely. At Utrecht University, they shone light through a layer of zinc oxide nanopowder deposited on a glass slide. They then used a detector on the other side to measure the light and analyse how it had been scattered by the powder. Once this has been done, Rotter explains, you can calculate “what kind of wave would be transmitted through this medium in the same way as through fresh space”. So it's not just any opaque material (and it's not even "opaque", really), and it's not just any waveform. The headline is pretty darn baity. Nonetheless, the science here is pretty cool. I could have imagined taking a scattering medium and a desired final waveform and computing the initial waveform that would produce the final waveform, but this seems to be saying that there exist waveforms that are essentially fixed points of the transform induced by the scattering medium. I wonder whether how exact that is -- whether the waveform really is identical, or whether any differences are simply hidden below the threshold of measurement sensitivity.

mateo1 6y ago

"Modelling scattering media with a thickness of 1-9μm" would be a better title. It's something, but the layer is so thin that the title looks like a joke.