Re: inverse squared law (lighting)
- From: Kaba <none@xxxxxxxx>
- Date: Sun, 11 Nov 2007 00:53:38 +0200
On Nov 4, 12:29 pm, Hans-Bernhard Bröker <HBBroe...@xxxxxxxxxxx>
wrote:
dogbert1...@xxxxxxxxx wrote:
I have a question about the inverse squared law for lighting:
(*) I(d) = I_0 / d^2
This seems to make sense for d >= 1, but what 0 < d < 1, I(d) will
blow up to be greater than the initial intensity I_0.
You're misreading the symbols. I_0 is no "initial" intensity. It's the
intensity at d=1 (in the chosen coordinates). Which is why you'll often
see it expressed differently:
I(d) = P / (4 * \pi * d^2)
Where P is the power of the source.
Ok, so then if we take d = 0.5, the intensity will increase by a
factor of 4. If we take d = 2 the intensity decreases by a factor of
4.
What about when d-->0? Is the intensity really infinite at the point
of the light source? What does that mean physically?
The I(d) is the formula for the radiant exitance of a point light source
at distance d. It's units is W/m^2.
http://en.wikipedia.org/wiki/Radiant_exitance
Compare this pressure: pressure is N/m^2. By making the area you apply
the force to arbitrary small, and keeping the force constant, the
pressure tends to infinity! But the important thing is that while you
can take arbitrary small area, zero area makes no sense.
Note that for every d > 0 the I(d) is finite, no matter how close to
zero you go. At d = 0 the irradiance is undefined.
--
Kalle Rutanen
http://kaba.hilvi.org
.
- References:
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- From: dogbert1793
- Re: inverse squared law (lighting)
- From: Hans-Bernhard Bröker
- Re: inverse squared law (lighting)
- From: dogbert1793
- inverse squared law (lighting)
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