Explain a magic upside down rainbow in remedial physics terms
June 4, 2010 10:26 AM Subscribe
[PhysicsFilter] Yesterday I saw what appeared to be an upside-down rainbow, in a part of the sky nearish the sun, but only when I was wearing my polarized sunglasses. This goes against everything I know about rainbows. Can someone explain?
In Seattle, around 2:30 pm. There were cirrus clouds (and some lower cumulonimbus-y things) which led me to the possibility that it could have been a circumzenithal arc. That wikipedia article states that they occur "on the same side as the sun" -- does that mean the same side of the sky as the sun? I would estimate the sun was about 60 degrees away, to the west. The rainbow/arc was due south. So, I almost understand that part. But why was I only able to see it through my polarized lenses, and not when I took my sunglasses off?
In Seattle, around 2:30 pm. There were cirrus clouds (and some lower cumulonimbus-y things) which led me to the possibility that it could have been a circumzenithal arc. That wikipedia article states that they occur "on the same side as the sun" -- does that mean the same side of the sky as the sun? I would estimate the sun was about 60 degrees away, to the west. The rainbow/arc was due south. So, I almost understand that part. But why was I only able to see it through my polarized lenses, and not when I took my sunglasses off?
Best answer: You reminded me of Atmospheric Optics, a nice reference.
The blue light from the sky is strongly polarized (which is why polarizing sunglasses are useful). Your observation suggests the polarization of the rainbow was different from the rest of the polarization of the sky, which is consistent with the rainbow being refracted through ice rather than reflected against droplets. Can you estimate the height of the sun in the sky?
It could also be that the light from the rainbow was not polarized very strongly, but the polarized light from the sky was bright enough to drown it out.
(Why are reflected and refracted light polarized?)
posted by fantabulous timewaster at 11:11 AM on June 4, 2010
The blue light from the sky is strongly polarized (which is why polarizing sunglasses are useful). Your observation suggests the polarization of the rainbow was different from the rest of the polarization of the sky, which is consistent with the rainbow being refracted through ice rather than reflected against droplets. Can you estimate the height of the sun in the sky?
It could also be that the light from the rainbow was not polarized very strongly, but the polarized light from the sky was bright enough to drown it out.
(Why are reflected and refracted light polarized?)
posted by fantabulous timewaster at 11:11 AM on June 4, 2010
I'm not completely understanding the geometry here. The rainbow was concave *up* not concave *towards* (or *away from*) the sun? The sun was off the side, on the "cheek" of the "smile" rather than on the "upper lip" or "chin"?
That's a new one to me.
posted by DU at 11:12 AM on June 4, 2010
That's a new one to me.
posted by DU at 11:12 AM on June 4, 2010
Response by poster: Thanks for that link, fantabulous timewaster. The height of the sun appears to have been about 60 degrees off the horizon, which appears to be the maximum height around this time of year in Seattle. I'd also estimate the arc appeared at about 60 degrees off the horizon, but pretty far to the left of the sun, due south.
The pictures that look closest to what I saw are this one or maybe this one.
It definitely wasn't a sun dog, it definitely wasn't the bottom of a regular circular rainbow (which occur on the opposite side of the sky from the sun), and I was standing at ground level, outside, wearing sunglasses (no car windows or anything in the way).
The explanation that the polarized light from the sky was bright enough to drown it out kind of makes sense, but I guess I'll have to read more into polarization to understand. It's been a long time since college physics.
posted by emyd at 12:33 PM on June 4, 2010
The pictures that look closest to what I saw are this one or maybe this one.
It definitely wasn't a sun dog, it definitely wasn't the bottom of a regular circular rainbow (which occur on the opposite side of the sky from the sun), and I was standing at ground level, outside, wearing sunglasses (no car windows or anything in the way).
The explanation that the polarized light from the sky was bright enough to drown it out kind of makes sense, but I guess I'll have to read more into polarization to understand. It's been a long time since college physics.
posted by emyd at 12:33 PM on June 4, 2010
I get that effect all the time - usually around dusk, looking at clouds near the sun while wearing my cheapo 7-11 polarized sunglasses. The clouds just burst with rainbow shimmering color, it's like an oil slick across the sky at times.. at first I totally thought this was a byproduct of my youthful amateur neuro-pharmacology, but I gave someone else the glasses and they confirmed it. Totally beautiful if you find some specs with the right properties.
posted by FatherDagon at 12:43 PM on June 4, 2010
posted by FatherDagon at 12:43 PM on June 4, 2010
Best answer:
Polarized sunglasses (when worn the usual way) absorb horizontally polarized light, so that they cut out bright reflections from the ground or from the surface of water. If you rotate your sunglasses 90°, they'll absorb vertically polarized light. So if you go to the lake in the evening with your sunglasses on, they will absorb the glare from the water's surface and you can see the fish underneath. If you rotate your head it's almost like taking the sunglasses off: the glare comes through the lenses so you can't see the fish any more. The fish are still there under the water, and they're sending you the same amount of light, but the reflection of the sky is overwhelmingly brighter.
I think I don't understand your geometry in a way that meshes with the circumzenith arc: if I understand the diagram, that feature goes away when the sun is higher than 30°. Hmmm ... a "Parry supralateral"? That doesn't seem to have its own description page.
posted by fantabulous timewaster at 3:09 PM on June 4, 2010
The explanation that the polarized light from the sky was bright enough to drown it out kind of makes sense, but ...It's the same reason you can't see stars in the daytime. The stars are there in the daytime, of course, and the column of air between you and the star doesn't absorb any more starlight during the day than it does during the night. But during the daytime that column of air reflects an awful lot of sunlight towards you. Suppose the daytime air in front of a star sends you a hundred times more light than a bright star does: then the difference in brightness between the part of the sky with the star and the adjacent part without is only about 1%. Your eye is built to tell "ten times as bright" from "one-tenth as bright," so the small relatives changes in sky brightness due to the stars in the daytime just looks like a uniform blue sky. You have to turn off the air to see the stars.
Polarized sunglasses (when worn the usual way) absorb horizontally polarized light, so that they cut out bright reflections from the ground or from the surface of water. If you rotate your sunglasses 90°, they'll absorb vertically polarized light. So if you go to the lake in the evening with your sunglasses on, they will absorb the glare from the water's surface and you can see the fish underneath. If you rotate your head it's almost like taking the sunglasses off: the glare comes through the lenses so you can't see the fish any more. The fish are still there under the water, and they're sending you the same amount of light, but the reflection of the sky is overwhelmingly brighter.
I think I don't understand your geometry in a way that meshes with the circumzenith arc: if I understand the diagram, that feature goes away when the sun is higher than 30°. Hmmm ... a "Parry supralateral"? That doesn't seem to have its own description page.
posted by fantabulous timewaster at 3:09 PM on June 4, 2010
Response by poster: I don't understand why the sun would be near its zenith at 2:30 in the afternoon, so perhaps I was reading that page wrong. On the other hand, it couldn't have been higher than 30°. It didn't look like a 'parry supralateral,' but I could be misremembering the location of the sun and have seen part of a circumscribed halo (cool animation link!).
But I understand the brightness of the sky drowning out the rainbow now, so thanks!
posted by emyd at 4:14 PM on June 4, 2010
But I understand the brightness of the sky drowning out the rainbow now, so thanks!
posted by emyd at 4:14 PM on June 4, 2010
Response by poster: After looking for more pictures of circumscribed halos, I'm almost positive it was a circumhorizontal arc: pic 1,pic 2, wikipedia link. It matches the sun altitude and the distance from the sun pretty closely. Thanks all!
posted by emyd at 4:22 PM on June 4, 2010
posted by emyd at 4:22 PM on June 4, 2010
Best answer: Coincidentally, I just saw this local news item with a nice photo of a similar effect in north central Oklahoma. This was earlier in the day but like yours, east of the sun.
posted by Snerd at 5:19 PM on June 4, 2010
posted by Snerd at 5:19 PM on June 4, 2010
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posted by ldthomps at 10:43 AM on June 4, 2010