Moon's Graden

> What about it?  The inverse square law applies to all
> electromagnetic radiation-- something twice as far away
> receives 1/4 the light, something three times as far away 1/9 the light
> etc.

> It's the reason deep space probes
> can't rely completely on solar power and have to carry nuclear
> power supplies.

> To calculate the contribution of reflective surfaces you have to factor
> in the efficiency of the reflector as well as the inverse square law.

> J. Del Col

> In a greenhouse, the diffusion of the "glass" creates a nearly infinite
> number of points, and under a fluorescent light bulb, each atom of phosphor
> is a point source.  The end result is reinforcement of light intensity by
> nearby points, which makes the intensity calculation a bit more complex.

> In a greenhouse, the diffusion of the "glass" creates a nearly infinite
> number of points, and under a fluorescent light bulb, each atom of phosphor
> is a point source.  The end result is reinforcement of light intensity by
> nearby points, which makes the intensity calculation a bit more complex.

> > In a greenhouse, the diffusion of the "glass" creates a nearly infinite
> > number of points, and under a fluorescent light bulb, each atom of
phosphor
> > is a point source.  The end result is reinforcement of light intensity
by
> > nearby points, which makes the intensity calculation a bit more complex.

> As far as I know the Inverse Square Law applies to all light sources,
> point or diffuse. The plants may receive light from several directions,
but
> the intensity of the light still obeys the Inverse Square Law.

> For a fresh fluorescent tube, the light output should be fairly
> uniform along the length of the tube.  Use of reflectors isn't really that
> complicating.  If you treat the tube/reflector combination as a unit, the
> calculation is straightforward.

> For a shelf of plants arrayed along the length of a fluorescent tube
> light intensity is, of course, greatest directly under the tube with
> some light striking the plant at an angle from either end of the tube.
> A plant placed directly in the middle of the tube will receive the
> same light from both ends  Those plants closer to one end or the other
> will, of course, receive more light from the closer end.  However, the
> inverse square still applies, whether to light from directly above
> or at an angle.

> Calculating the intensity of light reflected from walls, etc. is
> a little trickier.  You must know the efficiency of the reflector,
> the intensity of the light striking the
> reflector and the distance from the reflective surface to the plant in
> order calculate the intensity of the reflected light and add it to
> the intensity of the direct light.

> J. Del Col

> If you put a plant with a 6" totally flat leaf spread right up against a 48"
> bulb it gets a certain amount of light - considering the potential angles of
> incidence, let's say it's all straight down from the bulb at that point.  It
> obviously isn't only straight down, but for all practical purposes the light
> emanating from the other 42" of the bulb is not hitting the leaves to any
> significant degree.

> Now, move it farther away from the bulb.  The light from the 6" part of the
> bulb can still go straight down - and will be reduced in intensity in
> relation to the inverse square - but now the leaves can "see" the light
> emitted by more of the bulb, which means that the total amount of light
> reaching the leaves is not reduced the full extent of the inverse square.

> I'm sure that someone could do the simple calculus calculation and plot the
> net intensity, but it just ain't that important.

> > If you put a plant with a 6" totally flat leaf spread right up against a
48"
> > bulb it gets a certain amount of light - considering the potential
angles of
> > incidence, let's say it's all straight down from the bulb at that point.
It
> > obviously isn't only straight down, but for all practical purposes the
light
> > emanating from the other 42" of the bulb is not hitting the leaves to
any
> > significant degree.

> > Now, move it farther away from the bulb.  The light from the 6" part of
the
> > bulb can still go straight down - and will be reduced in intensity in
> > relation to the inverse square - but now the leaves can "see" the light
> > emitted by more of the bulb, which means that the total amount of light
> > reaching the leaves is not reduced the full extent of the inverse
square.

> > I'm sure that someone could do the simple calculus calculation and plot
the
> > net intensity, but it just ain't that important.

> You have to know the  inverse square result for the slant range
> to different areas of the bulb and add them to the result for the light
> from directly above, but, as you say, it isn't really worth the trouble.

> This is where a good incident footcandle meter is invaluable.

> J. Del Col