# Re: [APD] How much light actually gets down here?

```Ah, it is such a nice feeling to be able to argue about something
other than how to kill a fish!

If you put a point source of light in a parabolic reflector, right at
the focal point, and the reflector is a perfect reflector, reflecting
100% of the incident light throughout the spectrum, there is no drop
off in intensity with distance.  That light would shine as brightly
on an object ten miles away as on your hand right in front of it.
The problem is that there are no point sources of light, so it isn't
possible to locate anything at the focal point.  Such reflectors are
always less than perfect.  But, the light intensity most certainly
does not drop off as the square of the distance, or even directly as
the distance.  A line source, another practical impossibility, can be
located in a linear parabolic shaped reflector, along the focal line,
and be equally perfect.  No  light source we have is even close to
being either a point or a line source, so our reflectors can't be
perfect.  They can be very good, and that allows us to side step the
loss of light with distance to some extent.  That is the only reason
we even use reflectors.

Vaughn

On May 3, 2007, at 8:21 AM, S. Hieber wrote:

> The loss of lightfal over a given size of area decreases as a
> square of the distance from the point or origination --simply put,
> if you draw lines fromthe pint source to the perimeter of the area,
> then lower the area to twice the distance, the line will describe
> an area that increases as a square of the original distance (for
> example, iirc, an 8" diameter comprises about half the area of a
> 10" diameter) . This rule of thumb is often cited and comes from
>
> The square of distance rule yields an easy calculation with a
> single point source of light. However, with a line source, you have
> an infinite number of point sources (all along the length and
> around the tube), so the calculation becomes more complex. For any
> one of these, the square of ditance rule holds, but as the area is
> lowered, more point sources shed light on the area (an
> ovesimplification but that's generally the way to picture it).
> Taking all that into account, the lightfall roughly halves wtih
> distance, ignoring side reflections, and a few other complications.
>
> With an MH, one could rely on the square of distance rule since the
> light emits from a very small source, not quite a point, but close
> enough for gardening purposes.
>
> The reflector won't make a terrific difference -- it does mean that
> lgiht light come from more point sources (from a wider area) but
> not tremendously so.
>
> sh
>
>
> ----- Original Message ----
> From: Vaughn Hopkins <hoppycalif at yahoo_com>
> To: aquatic plants digest <aquatic-plants at actwin_com>
> Sent: Thursday, May 3, 2007 11:06:26 AM
> Subject: Re: [APD] How much light actually gets down here?
>
>
> . . . the bulb sitting above the tank with no reflector loses light
> proportional to the square of the distance the light travels . . , but
> with a good reflector it loses less than that, and with a perfect
> reflector it loses very little.
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