If so in what conditions (ie day/night, does it matter). I'm talking about 950nm of light.
206667
No, humans can never see infrared.  It's just a limitation of our evolution.  We evolved to see in the visible spectrum of light, not the infrared.
265213
All natural light sources generate a "spectrum" of frequencies that approaches the "black body" radiation curve. The peak of that curve denotes the temperature of the emitting object. Certain frequencies can be affected in transit via interaction with materials as in emission and reflection nebula (ie. various places within the Orion constellation). The sun emits UV radiation commensurate with its black body ideal.\n\nUNLESS someone can show that the moon absorbes ALL UV frequencies on reflection then by definition:\n\nMOONLIGHT DOES INDEED CONTAIN UV FREQUENCIES\n\n===============================================\nSome of you people amaze me, you answer questions that you know NOTHING about. Answers like "No because it is reflected light" are a case in point. Is your blue shirt invisible in the mirror because the light coming from your shirt is lost as it reflects off of the mirror. I THINK NOT! Reflected light IS LIGHT. Some of it is absorbed, reflected, refracted but it is light. YES; the UV portion in moonlight is very weak, hey guess what -ALL THE LIGHT frequencies in moonlight are weak; THATS WHY MOONLIGHT IS DIM.\n==============================\nUV is NOT produced only from objects above 24000 degrees. It is produced at ALL temperatures, but the lower the temperature the lower the amount of UV that is produced. PLEASE Google "black body radiation"
55561
yes because the light will take some of the vitamin power.
185962
There are no materials that are 100% reflective.  Silver and aluminum are in the 90% range, but nothing is 100%.\n\nIf you had a room covered in 95% reflective materials, it would still dim almost instantly after turning off the light.  Light travels about 186,000 miles in one second.  If your room were 10'x10', a light beam about be hitting the walls around 10,000,000 times in the first second, losing 5% of its energy each time.  In the first 1/100,000th of a second, its energy would be reduced to about 0.6% of its orignal value, much faster anyone would notice.
79883
I like it when it stays lighter later. Who care if its dark in the morning? I would rather have an extra hour of light in the evening!!!
502602
The amount of light will not double, since mirrors only reflect about 80% of light that hits them (absorbing 20%).  It would be brighter in the room, though.  However, if you had a room full of mirrors on all surfaces, the room would be much brigher than a room full of black surfaces, since the light rays will be reflected rather than being absorbed by the dark color.
655302
If the deflection surface has any diffraction or diffusion to it, the light may just be redirected imperfectly. 12kV? Jumpin Gigawatts!
12922
depends on the climate. Certain weather conditions such as cloudy days, rain and light and dark hours affect how effective the photoelectric cell collectors are. Usually such collectors are built in places with very low probability of adverse weather and very high number of intense sunny days. Countries such as Norway may not be able to harness such energy as they have very little sun light days and countries in the equator fare better in utilizing such technologies.
541587
I would also pick A as the correct answer.\n\nB. the camera might or might not have to be adjusted depending on how much light is being reflected into the picture.\n\nC. A reflector may or may not prevent one side othe image from appearing lighter or darker depending on the actual lighting of the scene.\n\nD. because B & C are not right
518993
A lot.  Only 1% or less of the light actually makes the image.  This is why digital photography is so much better when it comes to astronomy.
219490
It is theoretically possible, but very, very, very difficult!\nL2 is roughly 1,500,00 km (or a million miles) from the earth, on the far side from the sun.  So when the moon is on the sun side of the earth, with it's dark face toward the earth (a new moon), the distance from the L2 point to the moon would be 1.5M km (to earth), plus 375,000 km  (to the moon), then bounce back to earth, for a total of 2.25 million km (or 1.5M miles).  Since the moon only reflects about 1/10 of the light that hits it (its albedo is .12), the light source at L2 would have to be 10 times as powerful as if you were reflecting off a mirror (or 6-8 times what it would be off a good screen).\n So let's figure how powerful that light would have to be.  Let's say that you want to just barely be able to see this projection, say, about as well as you could see a candle on a perfectly dark night at a distance of a kilometer. Well, light power varies inversely as the square of the distance.  So (simplifying a lot to just get a rough idea) that light would have to be 2.25M, squared, times 10 (for the moon's reflectivity) candlepower to hit your eye with that same power.  So you'd need a light of 50,625,000,000,000,000,000 candlepower.  By one account, the most powerful light currently existing on earth is about 1,000,000,000 (one billion) candlepower.  \n I've simplified quite a bit; you could reduce the power requirements many orders of magnitude by focusing the beam or using a laser, but the answer still comes out to be that you'd need a light millions of times more powerful than the brightest lighthouse.  And you'll need thousands of square kilometers of solar cells to power this super projector.  And thousands of rockets to get all this equipment out to L2!\n\n So to answer your question - is it THEORETICALLY possible?  Sure!  But there is a huge world of difference between theory and practical application on this one!
264524
if light is thr then where is dark?
404282
No.  The opposite (emitting visible light after absorbing UV) is possible.  When light is emitted, it must have the same energy as the light absorbed, or it must have a lower energy.  UV light has higher energy than visible light, and so a material can't absorb visible and emit UV.
84781
depends on the shade.
451197
You need to know the luminous efficacy to convert lumens into watts: http://en.wikipedia.org/wiki/Luminous_efficacy
481510
Light IS energy...sorry I can only think of one...:(
127599
yes, it's a method to use space light, am sure you are a good researcher, you can try this,\nhttp://uttoransen.blogspot.com/