A black body absorbs all energy. It doesn't reflect or transmit energy. It also absorbs all light and doesn't reflect any light. Why, then, can we see it? For instance, burnt platinum is 98-99% black body and yet is visible.
[Physics] Why is a black body visible
thermal-radiationvisible-light
Best Answer
There are some misconceptions here :
So far so good.
No. Any object that did that would get hotter and hotter as it absorbed more and more energy and just retained it. A blackbody, like any object, will radiate energy away as photons - light at many different wavelengths.
See the previous paragraph.
Because a blackbody does not exist in total isolation and it will try to reach thermal equilibrium with the world around it. To do that it either has to radiate energy away (if it's hotter than the world around it) or absorb energy (if it's cooler).
Even when it absorbs it still radiates some energy, and it simply radiates less than it absorbs.
But when you talk about being "visible" in a human sense, this is a different thing. If you walk into a room with every object and surface emitting light you can see, and then I pop something into it that actually did not emit light at a visible wavelength you would still detect that visually - it would seem like a "hole" in the scene - a silhouette.
For something to be invisible to a human, i.e. not be detectable visually, it would need to be perfectly transparent (at least at visible wavelengths) - all light would have to pass through it without distortion. Alternatively, all light would need to go around it.