FROM: alanr@rd.bbc.co.uk (Alan Roberts)
SUBJECT: Re: Question about visible light.
DATE: 10 May 1996 08:56:33 GMT
ORGANIZATION: British Broadcasting Corporation, UK

mig@ciao.cc.columbia.edu (Meir I Green) wrote:

>Hi, I was under the impression that the sun is at 5800K.  Does this 
>mean that by the time the light reaches sea level, it is closer to 
>9000K?  In other words, to duplicate natural sunlight at tropical 
>latitudes, would one use a 10,000K Metal Halide lamp?

-- 
******* Alan Roberts ******* BBC Research & Development Department *******
* My views, not necessarily Auntie's, but they might be, you never know. *
**************************************************************************

FROM: will 
SUBJECT: Re: natural sunlight  (was Question about visible light.)
DATE: Sat, 11 May 1996 17:18:52 -0700
ORGANIZATION: Univ. Cal. Santa Barbara ICESS

Here are a few crude, perhaps simplifying, comments on the issue of duplicating natural light.

The sun radiates like a blackbody at about 5800K. There is significant energy from 0.2 to 2.0 microns. We see from about 0.4 to 0.8 microns, which spans the peak of the blackbody curve. When sunlight hits the Earth's atmosphere not much in the visible range is absorbed, but it is scattered, and the scattering is a function of wavelength.

Rayleigh scattering:

Rayleigh scattering occurs when the scatters (air molecules) are smaller than the wavelength of light. It turns out that the phase function of Rayleigh scattering preferentially extracts the blue wavelengths from the collimated sunlight and makes the sky appear blue and the sun appear yellow to an Earthbound observer. So a white piece if paper appears white because it is illuminated by the direct yellow solar beam and the diffuse blue light. The illumination therefore is what you would get if you had a 5800K source. This is the nominal case for clear skies in midday.

On the other hand, if the solar beam is blocked by a sharply-defined cloud (like a thunderhead) the remaining illumination will have a larger proportion of blue. From a radiometry standpoint it is simply a 5800K Planck blackbody curve with the longer wavelengths filtered out. Photometry is a shamelessly autocentric system that weights the radiometric quantities by the human eye's response. Using this system one would need a blackbody illuminant of much higher temperature to get the visible blue-to-red ratio of the blocked beam case, 9000k-40,000K sounds quite plausable. This could be accomplished by filtering a practical source. Note also that since most of the light energy is in the nonscattered solar beam, when it is blocked, the remaining illumination is much dimmer (i.e. shade).

Nonselective scattering:

If it is overcast, then the illumination is pretty much back to the 5800K illuminant situation because the water droplets in clouds are much bigger than the wavelength of visible light, and therefore scatter nonselectively. Cloud cover is illuminated from above and scatters and absorbs the blue and yellow light equally. The light is dimmer, but white.

Mie scattering:

Finally, if there are dust particles in the air that are about the same size as the light wavelengths there will be Mie scattering. The amount of Mie scattering has a complicated dependence on wavelength and can generate the spectacular colors of a sunset: with the long path length through the atmosphere. Sunsets occur more strongly after fires and volcanic eruptions when there is more dust. The equivalent photometric illuminant temperature during a sunset could be lower or higher than 5800K. I have seen many red, but also purple sunsets.

-Will

FROM: crs 
SUBJECT: Re: natural sunlight  (was Question about visible light.)
DATE: Sun, 12 May 1996 22:30:52 +0100
ORGANIZATION: UltraNet Communications, Inc.

William C. Snyder wrote:
[snip]

> 
> Additional comments: Of course, as the atmosphere gets thinner, 
> there will be less scattering. Next time I fly I will check this 
> out. Looking up from an aircraft at 35,000 ft. (~6 miles) one 
> should see a darker color than at the surface. Also, it was 
> pointed out that the wideband flux from the sun at the top of the
> atmosphere is about 1 Kw/m^2 and this is reduced by absorption to 
> approx. 700 Kw/m^2 at the surface on a clear day (of course this 
> depends on latitude, etc.).
> 

Let's outline what one might see from a plane at 35000 feet ~ 10.7 km on a clear day. Look up and away from the sun. One sees a bright blue sky - even though the air is much thinner. To my eyes, the sky is just as bright as it is on the surface - maybe a little brighter. Look down and away from the sun - or any other direction for that matter - again - to my eyes - even at the most oblique angles, there is no bluish tint except possibly over water or above the horizon - even though the air has far greater density below than above.

Molecules of nitrogen and oxygen are not very polarizable and the quantity of blue light scattered through large angles will be small. The usual explanation of why we see a blue sky despite this is that there are a lot of air molecules and the sun is bright. OK - then why do we see the brightest blue when we look up (where the air density is low) and virtually none looking down in any direction (where the air density is far greater)? It would appear that most of the blue light is scattered from the most rarified portions of the atmosphere.

An alternative explanation is that fluctuations in density are the source of a large fraction of the scattering. These become most pronounced at large (delta N)/N. Such a quantity increases at higher altitudes. Light scattering from density fluctuations occurs near critical points. Is it unreasonable to suppose that fluctuations can scatter light from the upper atmosphere? After all, there will be a lot of fluctuations and the sun is bright... :-)

Will, the next time you go up in a plane - check it out. I'll do the same. Let's compare notes at that time...

chuck szmanda
chucksz@ultranet.com

If you have any question, remark, comment, want to share some philosophy or just want to express your opinion about these pages, feel free to send email to: w.j.markerink @ a1.nl