anonymous
  • anonymous
Black body any body painted completely with lamb black is 98% perfect black body... it appears black cause it absorbs almost all the light incident on it.. and doesn't reflect any .. then why is it that if i shine a laser on it.. i can easily see the spot???
Physics
  • Stacey Warren - Expert brainly.com
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schrodinger
  • schrodinger
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anonymous
  • anonymous
@rajathsbhat
anonymous
  • anonymous
very good question! I too had the exact same question back when I was studying this. I would guess that it's because soot isn't a perfect absorber of light...but other that, i have no idea.
anonymous
  • anonymous
even if it is NOT a perfect absorber.. it should absorb something right??.. i mean when white light is incident.. it absorbs almost all.. is it because.. the intensity of the laser is very high ( i mean its totally focused).. so it DOES absorb more than when incident on the other surface.. and its just that we can't make out?? but really i can't imagine.. even if we did have a PERFECT black body.. shinning a laser and not see a spot.. !!..

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anonymous
  • anonymous
you know, i do think that it reflects only about 2 percent of the incident light. Because, imagine how intense the laser light would be if you pointed it directly at your eye!!
anonymous
  • anonymous
but. you take the same laser light and shine on something else.. i really don't see any difference.. !!
anonymous
  • anonymous
I am doing it right now .. 5mWatt laser.. no difference :P
anonymous
  • anonymous
really? this calls for an experiment...
anonymous
  • anonymous
on second thoughts.. maybe there is a difference :D.. looking closer i think i see that black indeed absorbs more
anonymous
  • anonymous
shine it into something more reflective... a mirror...
anonymous
  • anonymous
if i shine on mirror.. i ll spoil my eyes :P
anonymous
  • anonymous
if you have optical mouse. you can try with that!
anonymous
  • anonymous
i just tried it. You're right. You see more light than you'd expect. But I still think it's only 2% as intense as the initial light. If only there was a way to measure the intensity....
anonymous
  • anonymous
i used soot from camphor btw
anonymous
  • anonymous
or maybe soot isn't completely amorphous.....
anonymous
  • anonymous
@UnkleRhaukus @Carl_Pham @eashmore @vincent-lyon.fr @CliffSedge
anonymous
  • anonymous
An ideal black body absorbs *and emits* perfectly. If you shine a laser on a spot, it should absorb all that light, then re-emit it again, distributed isotropically and according to a nice Boltzmann distribution of frequencies. So, sure, you should see a spot. You can think of it as representing a small patch of the black body which you have momentarily heated to a significantly higher temperature, and which is therefore glowing. And this is before we even get into the fact that real black bodies have significant non-idealities... You may be confusing a black body with a perfect conductor of energy. You may be thinking that if you shine the laser on some small patch, the energy should be instantly conducted away until it spreads over the entire black body surface, and just raises the overall temperature of the body by some small fraction. But the quality of being a black body doesn't, so far as I know, necessarily encompass being a body that conducts energy without resistance, a sort of heat superconductor.
anonymous
  • anonymous
interesting idea.. so what you are saying is.. the spot that i see on a normal body and the spot that i see on the black body may look the same but are totally different physical phenomena? on normal body its just reflection.. photons bouncing off (normal reflection i mean ,i ll use photon theory cause.. black body radiation :D ) but when it comes ot black body.. its absorbing, getting heat up .. and then starts re-emitting .. which is in total contrast to normal reflection?.. wow interesting indeed
anonymous
  • anonymous
i mean't PERFECT black body!
anonymous
  • anonymous
but one problem with this theory.. if i shine blue laser.. to heat it up a lil bit.. then the spot should look RED.. cause to emit blue light.. it would require significantly high temperature.. and i don't think shining blue laser light is gonna heat it up any more than shinning red laser light.. so what you say??
anonymous
  • anonymous
what you gotta say raj?? did i find a flaw in the theory?
anonymous
  • anonymous
cause if what he says is right.. then .. if i kept on focusing the laser spot... the small patch keeps on heating even more and more and thus should give me different colors!!
anonymous
  • anonymous
infact now i believe that theory is completely flawed.. you can't expect the temperature of the black body to be high enough to start remitting.. for crying it out loud.. you saying.. shinning the laser is actually making it red hot??!!?.. ITS NOT RED HOT!!!!
anonymous
  • anonymous
i think Carl is right. But you have a point too..
anonymous
  • anonymous
no Carl is not right.. its absolutely wrong!
anonymous
  • anonymous
A perfect Black Body absorbs all light period. It radiates light when it is heated to glowing and the rate of emission is given by the Stefan Boltzmann Law and a spectrum given by Planck's black-body radiation curve.
anonymous
  • anonymous
yea i know that.. but that doesn't answer my question.. !!!!.. by pointing laser at it.. i m pretty sure it wouldn't be hot enough to start radiating..!!!
anonymous
  • anonymous
Correct.. the point is the lamp black surface is not a perfect BB. The light you see is diffusely reflected. You think it is too much. What percentage you have to measured. The BB will radiate its spectrum as it is heated externally to a temperature where it glows. I'll bet you will see some significant reflection from any non laser light source if it is bright enough.
anonymous
  • anonymous
but the bb radiating thing doesn't come into the picture.. and yea even i believe the its just a diffused reflection and nothing more. but definitely it doesn't get heated up and start radiating!!!
anonymous
  • anonymous
correct

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