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Do you really believe that walls have any effect on the "spread"? Or, do you just like to?
Gotta give ya points for holding your ground, I guess.
Wait, did I get this right? You are claiming that walls do NOT have any effect on the spreading of light?
Less than 10% by my measurements.
I have a tiny space, lined with Reflectix ,and expected to get the same results as fluorescent tubes seem to have from the chart, (2X D = 50%)
But, I don't have tubes so no can measure.
Give him points for being right,
The law is correct but doesn't apply to growing. Is it so hard to grasp?
i dont need to prove it, its simple.
Im folling this ISL, which says that everytime you have 4 times the surface area, you wil have 1/4 the intensity. Thats all it says. there not even a need to mention anything else involving spheres, radius, twice as far away, point source, anything. Thats the law, when things disperse they get less dense. You are saying that you achieve a cetain percentage density (25% for every 2x distance) decrease, without a certain percentage of increase in surface area(4 TIMES). NOT POSSIBLE...
This law is true, and which is widely accepted. Im asking how do you end up with 1/4 the intensity (twice as far away) without 4 times the surface area. It doesnt work if you follow that law....
What part of this law leads to 25% intensity - in the situation I'm describing?
We havent factored in the fact that a photon that has reflected off 2 surfaces and hitting one travelling in the opposite direction that hasnt bounced at all will cancel each other out emitting a combination of heat and other radiating wavelengths of energy, one having bounced off 3 will behave the same way meeting one that has bounced off 1 surface. Which is probably why if you look directly at a bulb and reflector, the reflector surface isnt as bright as the bulb less reflection loss. Therefore while the reflector will be redirecting an great deal of photons, it wont be half of all emitted. So reflector design will have a huge impact on the brightness beneath the bulb.
Light emitting from the bulb will be lost in accordance with the ISL but light reflecting from the reflector will also be lost through various methods. Therefore what we measure at any point below the bulb will contain both the light emitted from the bulb and the light reflected from the reflector that has survived.
I can't see how a photon bouncing off a surface can be said to keep its initial energy twice as efficiently as one being generated at source. I'd like to propose something that I'm just making up. If a photon travels 6", then the wave front around it has a radius of 6" in all directions. If it travels 3" and then bounces 3" back in the other direction, it will still have travelled 6" and have a wave front 6" in all directions other than the direction of its bounce point 3" behind its current direction, and an interference pattern inbetween itself and its rear wave front. In this way the ISL would remain constant, even though what we are measuring has a higher intensity than it would without the reflector. The same would be said of the walls. So I guess what I'm saying is that I'm not sure if I want to change my mind or not. What does anyone else think of that idea? Remember photons are waves as well as particles and can't be considered ping pong balls.
The ISL law only applies to SURFACE AREA, it has nothing to do energy losses/gains from reflection/abosrption. That is a completely different 'law' than we are discussing right now. The only one I'm talking about says that THINGS (not just light) will spread out. When its spread across 4 times the space, it will be 1/4 times as much. Thats as far as the law goes. You start off with a 100% 'relative' value and when its spreads across 4 times the area its 25% of htat original value...You guys misunderstood this law, and tried to say the 25% was because of then ISL EVEN WHEN THE SURFACE AREA ISNT 4 TIMES! Thats against the law...
the best unit for expressing the energy requirement of a specific plant species is the irradiance expressed in milliwatts per sqaure meter (mw/m^2), or milliwatts per sqare foot (mw/ft^2).The way I'm thinking atm, is this, if the photon travels for 1 cm, then it has a wave front of 1cm in all directions, in effect the photon itself does not move, only its wave front. Think of dropping a pebble into the centre of a pond, what we see on the surface of the pond is in effect a 2 dimensional wave front of ripples circulating outwards, and becoming less intense as they travel outwards. A light photon does this in 3 dimensions. In practice this wave front collapses at the point of measurement, (quantum mechanics), and can be recorded as a particle rather than a wave. How it behaves on its travels though relies upon it being a wave up until its point of measurement. In effect, the photon stays stationery until it decides it's destination point, then vanishes from where it was and simultaniously appears at its destination point. Only the wave front takes time to travel, the photon itself doesn't. As the wave front travels, it dissipates. This dissipation is what I was refering to. Like I said, I'm trying to work this out with everyone else, I only half know what I'm talking about. What I don't know is how the energy contained in a single photon is measured. We can count the photons, but I suspect that that is not the same thing as measuring intensity of light or total energy reaching the destination.