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Let's suppose the ultimate disaster occurs... you've blown several thousand dollars on travel & lodging to view an eclipse, and now... minutes before the big event... the sun is behind a cloud.

Could a video camera capable of sensing UV-A (behind a suitable neutral density filter to avoid frying the sensor itself) potentially save the day & keep it from turning into a total loss?

I know "clouds don't protect you" from UV-A,and to some extent UV-A passes through clouds... but I haven't been able to find a single example of what a UV-A imaging sensor actually "sees" when looking at the sun through a cloud (or, maybe, what it would see indirectly if focused on a surface where solar energy is being focused through a pinhole). I assume there's both some degree of attenuation, and probably diffusion as well... but how much are we talking about? Like, an amount that might still allow you to at least see a fuzzy, blocky black disc moving in front of a bright one?

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    $\begingroup$ The shorter the wavelength, the worse the clouds influence visibility. You will see a total eclipse also behind clouds: it gets darker (been there, done that). If you want to really "see" it through clouds: get a radio receiver instead :). Generally plan you trip to be enjoyable also when there should be bad weather during eclipse :) $\endgroup$ Jan 8 at 21:05
  • $\begingroup$ Compare the panorama images from before totality and during, both taken with my phone: photos.app.goo.gl/X8cdzJr4EnerNSoZ7 and photos.app.goo.gl/wHKTZMRLm51dmKJ5A (we did see a bit during the partial phase: photos.app.goo.gl/gRp1DznhYrgXNysc9 but nothing through the clouds during totality while that's usually 99% of the fun) $\endgroup$ Jan 8 at 21:18

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I think the argument with Rayleigh scattering which scales with $\frac{1}{\lambda^4}$ and therefore producing more scattering of shorter wavelengths doesn't hold here because Rayleigh scattering is an approximation for particles (much) smaller than the wavelength. So it describes scattering on molecules, which for example explains the blue color of the sky in the absence of clouds.

Clouds consist of much larger particles, which are made up of a solid core called cloud condensation nucleus, which is surrounded by a water sphere. Usually we simply think of small water droplets. No idea what a typical diameter of water droplets in clouds is, I guess it's a distribution of all diameters between zero until the droplets become macroscopic and result in rain. Read somewhere that typical fog particles are in the range of several microns... Anyways, scattering on particles comparable or larger than the wavelength is described by Mie scattering, which has much less dependence on the wavelength than $\lambda^{-4}$ and therefore clouds are white (all wavelengths in the visible are scattered more or less to the same amount).

The math of Mie scattering is complicated, but fortunately there is a python module called miepython. It needs the complex refractive index of water, which can be found for example here. I downloaded miepython and did some calculations of light at 350 nm (UVA), 550 nm (Visible) and 10 $\mu$m, which is already long wave infrared (LWIR) and in the range thermal cameras work. Different droplet diameters are on the x-axis ranging from 0 to 0.1 mm (which is starting to get macroscopic). The result is this:

enter image description here

On the y-axis is the scattering efficiency per particle. To get a cross section you need to multiply by the geometric particle cross-section. To get then an extinction in unit 1/m you additionally need to multiply by the particle concentration (guess it's a distribution with much more small than large particles). But anyways, for a qualitative comparison this is enough. First thing to observe is that there are many oscillations, which is typical for Mie theory (as it assumes all particles to be round spheres with reflection and interference everywhere...). Apart from that, UVA and Vis are almost the same. However, when zoomed in

enter image description here

the Visible line seems to be on average a tiny bit larger than the UVA line. Which means an image would be a bit sharper in the UV, but this difference would be hardly visible in any real world image.

The real difference, as stated in the answer from uhoh before, comes with LWIR, which shows much less scattering, escpecially when particles are small (and I still think the majority of particles will be small). So this is somehow a proof of what was answered before by uhoh. UV won't work, but maybe give it a try with LWIR... (but as mentioned before the sun is not as bright in LWIR as it is in the Vis. Room-temperature emits at LWIR, so there might be a lot of different drawbacks as well).

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Not UV, but...

We drive cars and play baseball and can even read books on cloudy days, or clear days when the Sun passes behind a thick cloud. There's less light, but still plenty that's been scattered along the way.

And UV light will just as happily scatter from the tiny droplets that comprise clouds as visible light.

So the "UV on cloudy day" argument doesn't mean the clouds have any more transparency there than in the visible.

why not one of the windows in the thermal IR?

  • We know that the Earth's atmosphere has some windows of transparency in the thermal infrared range (say around 10 microns) because we know it's colder on clear nights than cloudy nights.
  • We can also look at plots from Wikipedia like the one below,
  • And we can also note that at high altitude there are even astronomical observatories on Earth that cover various thermal IR windows https://en.wikipedia.org/wiki/List_of_largest_infrared_telescopes#Overall
  • Finally, we can remember that Rayleigh scattering of sub-wavelength particles scales as 1/λ⁴ oops! As @CharlesTucker3 informs us in another answer the water droplets in clouds are way bigger than a wavelenth, not smaller. So Rayleigh doesn't apply at all and we need to invoke Mie scattering instead.

Put all this together and we can surmise that there is a chance that the offending clouds might allow an attenuated but discernible view of a solar eclipse.

Mind you at 5,600°C the Sun is brightest in visible wavelengths and would be a relatively dim bulb in thermal IR by comparison, but if you take your pocket FLIR camera with a suitable telescope having lenses made of germanium or silicon, you just might be able to come home with at least something.


https://commons.wikimedia.org/wiki/File:Atmospheric_Transmission.png

Source and detailed explanation

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  • $\begingroup$ How exotic/expensive are "lenses made of germanium or silicon" for low-end FLIR cameras? Like, are they the NORM, and almost by definition a FLIR camera (even a low end one) uses them, or are germanium/silicon lenses the difference between a semi-affordable FLIR camera that costs hundreds of dollars, vs a FLIR camera that costs thousands of dollars? $\endgroup$
    – Bitbang3r
    Jan 10 at 17:03
  • $\begingroup$ @Bitbang3r well, it's certainly a stretch that one might happen to have a handheld FLIR unless one is in the home insulation business. Those cameras all have either silicon or germanium single crystal internal lenses, there aren't really any good alternative materials. But they have a low pixel count and wide field of view (unless you are rich and have a military-grade mega pixel camera), so to get some useful resolution within a half-degree FOV you need an external telescope. I suppose that if you have a cheaper military surplus thermal IR binocular or telescope that would do out-of-the-box $\endgroup$
    – uhoh
    Jan 10 at 20:19
  • $\begingroup$ The wide field of view + low pixel count is problematic. There are, however, some semi-cheap FLIR cameras matching that description for Android and IPhones. This one, for example, is presently $230: topdon.us/products/tc001 . It's not exactly pocket change, but it's not completely outrageous as an insurance policy against a ruined eclipse-viewing if suitable. There's also a.co/d/8ZHjIru , which I've contemplated buying to use for an AI image-recognizer to sense when I get out of bed, but ignore my cats. Lower-res, but 55deg & possibly useful for future projects :-) $\endgroup$
    – Bitbang3r
    Jan 11 at 3:21
  • $\begingroup$ @Bitbang3r Cool! Remember this is just my hypothesis based on my understanding of the applicable optics and atmospheric science. In addition to water droplets, the precipitable water vapor in a cloud may be far more than in a clear sky. We don't yet have hard data that clouds are sufficiently transparent circa 10 microns. The plot in my answer shows 6% absorption for clear sky i.e. even normal water vapor concentration absorbs a little bit. Is a cloud's integrated column water vapor content 2x or 10x that? 100x? $\endgroup$
    – uhoh
    Jan 11 at 5:27
  • $\begingroup$ @Bitbang3r I've just asked Can terrestrial infrared telescopes see through clouds or haze, sometimes at least? $\endgroup$
    – uhoh
    Jan 11 at 5:45

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