From: Jonathan Kirwan
Subject: Re: Make infrared goggles inexpensively (like $10!!!)
References: <firstname.lastname@example.org> <4Aoh9.15391$1C2.email@example.com>
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Date: Tue, 17 Sep 2002 10:30:51 GMT
On 16 Sep 2002 19:50:14 GMT, Mike Poulton
>At 1mW/cm^2 full spectrum, the sun is not that much of a hazard to
The incident flux is some 1387 watts/m^2 across the Earth. This is
the solar radiation constant, last I remember seeing it. I also
recall that about 31% of the power or so is reflected or diverted
away, which means about 960 watts/m^2 at the equator. (Near the
poles, the incident is spread over more curved-away land.) Let's go
with that figure, for now, knowing that elsewhere will be somewhat
less at any given moment, depending on season and time of day.
[By the way, as a sidebar, for Earth's temperature to remain in long
term equilibrium, it must radiate away as much as it absorbs. Since
the face of the Earth presents a PI*R^2 area, while a spherical
surface has 4*PI*R^2 area, the Earth needs to radiate away 1/4th as
much energy per unit time per unit area as it takes in. This means
that the Earth radiates away about 240 watts/m^2.]
Just by way of cross-verification of the above, I just checked and
found an actual reported measurement at München (Munich) of about 295
watts/m^2 in the visible range from 400nm to 700nm. This reading
includes all the various peculiarities of the atmosphere, as it is an
empirical, ground-level measurement. Roughly speaking, that range of
from 400nm to 700nm accounts for about 39.3% of the sun's spectrum.
This suggests that the actual total was about 750 watts/m^2. This is
a useful reasonableness check against the above numbers. It shouldn't
be surprising that Munich would have a somewhat lower value than the
equatorial estimate, given its latitude and some guesses about the
orientation of the measuring instrument and the time of year.
Anyway, 960 watts/m^2 is 96 mW/cm^2. This includes the full spectrum.
I've no idea where you got the figure of 1mw/cm^2 for the full
spectrum, but it doesn't seem reasonable to me. It seems off by
perhaps as much as two orders of magnitude. Can you tell me where
that figure came from?
>Once you cut out all the visible and UV light, you're left with
>a small fraction of the original irradiance in the near IR.
The sun's color temperature is estimated at 5800 Kelvin. Here's the
breakdown of the percentages:
Shorter than 400nm: 13.4%
From 400nm to 700nm: 39.3%
From 700nm to 1600nm: 40.1%
Longer than 1600nm: 7.2%
The reason for using the 1600nm figure is that the cornea passes
wavelengths shorter than that, approximately. I used the 700nm figure
because of the filters being discussed, which appear to pass much of
the light longer than 700nm (including some, shorter than that.)
Once again, I don't know what you imagine "small fraction" to be, but
the IR portion of sunlight getting through the filters, through the
cornea, and to the retina is closer to 40% or so of the available
solar radiative power.
With Munich (48.150 N latitude) measuring in the neighborhood of some
75mW/cm^2 and with the IR portion somewhere in the area of 40% of
that, this means something more like 30mW/cm^2 of IR.
Anyway, this "left with a small fraction" comment seems to underplay
the reality, a bit.
>glances of the sun without a filter aren't too damaging, brief glances
>with the filter can't be very bad, even if your pupils are a bit bigger.
A comparison I'm uncomfortable with, but I'll leave this one for
others to think about.