So in theory, if you had a wave that is perfectly out of sync with a reflected wave, why would you not be able to match a +1 with a -1 to make a 0 (ie no light) spot?
Because all other light waves can still pass through the region unimpeded. The end result being the canceled out light waves are no longer visible. But this phenomenon does NOT block any other light. any method of creating perfect deconstructive interference would simply make your laser beam vanish, NOT turn black.
I would also like to point out there is no conservation laws broken...
Quite correct. No argument here.
Ok... but using your logic, you are effectively saying that if I theoretically lined up the crystalline structure of an opal so that it perfectly creates deconstructive interference (ignoring the constructive) that it will become see through... or that it will radiate energy.
no no no. I am trying to point out that the color in opal is caused by essentially random conditions. Therefore there will be as much constructive as deconstructive interference. The stone can do both, and normally does both. It's simply a natural structure that creates light interference patterns, and we perceive that as color when viewed in white light. Given this simple fact, if deconstructive interference could create the color black to our eyes, you would see black flecks of color in the opal as well as all the normal color play we do see. Given that there ARE no black flecks of color play in opal, then light interference DOES NOT CREATE THE COLOR BLACK. Other light can STILL pass through an interfered light wave without effect. essentially a -1 +1 state = 0 ass you stated. However add in another +1 state and you get (-1 +1) +x = x where x is the value of any other light passing through. To get the color black you would need (-1 +1) + x = 0 when x itself can not = 0.
To get the effect which nanolaser was talking about, which you were defending....
but not impossible Imagine,... it's a sunny day and sudenly you see a black laser beam coming from the other side of the horizon.
...you would need some method of perfectly canceling out EVERY light wave that passed through the laser beam, which is impossible with deconstructive interference. You would need light with wavelengths that are not constant.
Your argument seems to be stating that when you destructively interfere with light, you create the color black The result is simply that the light is no longer visible. However, just like ocean waves passing through each other temporarily, if the light waves could be separated again, you would then be able to see them again. As you stated, the light isn't gone, the energy isn't gone, the photons themselves are not gone, they are simply a +1 and -1 state that overlaps to appear as a 0 state. but separate those +1 and -1 states again, and you no longer have a 0 state, but two separate +1 and -1 states. Given that the energy is still there, but you can not see it, I think the word invisible applies quite well, don't you?
(there is an oversimplification the conditions that create the color play in opal are not perfectly random, as there ARE conditions that effect the sphere size in opal, and due to such, the color violet is the most common color and red the least common, but this is due to the sphere size is more rare the larger it gets, due to needing longer stability of the initial conditions to form, but this also means that deconstructive interference will be more common then constructive overall in opals, as those spheres are smaller then ones that will produce violet)
-Adrian
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