Exploring the Logic of Black Absorption: Ryan's Experiment

[RyanXXVI]

In school, I have been told that black appears how it is because it absorbs all frequencies of visible light and reflects no frequencies of visible light. Following that logic, if one were to shine a flashlight in a dark room where the walls, ceiling and floor were all black, one would not see anything (if one pointed the flashlight away from them and the flashlight was cone-shaped) because no light would reflect into your eye. For some reason, it seems to me that that would not happen.I wold try an experiment, but I do not have an all-black room in my house. Could someone please explain why this would or would not happen and even try the experiment if possible?

Thanks in advance,
Ryan

 

[HallsofIvy]

Black is the absence of visible light. A black object absorbs all visible light. Yes, if you were in a room with true black walls, you would not see any light reflected from the walls. Of course, there is no truly black paint so you can't really do that experiment.

 

[sophiecentaur]

The nearest thing (that I know of, in practice) to a truly black surface was used at one time to set up TV cameras - to establish a true 'black level'. It consisted of a rectangular hole, cut in a board, with a box behind it (about 20cm cube). The inside surface of the box consisted of a 'good' black velvet. Any light entering the box would be scattered on the velvet surface and would hit other parts of the inside of the box; rapidly being absorbed before it could get out again. The only light emerging out of the hole would have been reflected off the very ends of the (already black) velvet pile (first reflection) so the hole would be a pretty good 'black' under almost any studio conditions.

'Stealth' vehicles tend to rely on having very flat surfaces, angled up or down, and any radar pulses hitting them that don't actually get absorbed by the paint / surface are reflected off at an angle and not back in the direction of ground. Very high gloss photo prints have good blacks because there is very little light scattered from them and, if you angle them away from any light source, you don't see it - hence the surface looks black. Of course, there are viewing directions where a gloss paper black will look much worse than a matte surface!

 

[A.T.]

For some reason, it seems to me ... Could someone please explain why this would or would not happen...

Can you explain your reason?

 

[Meson080]

In school, I have been told that black appears how it is because it absorbs all frequencies of visible light and reflects no frequencies of visible light. Following that logic, if one were to shine a flashlight in a dark room where the walls, ceiling and floor were all black, one would not see anything (if one pointed the flashlight away from them and the flashlight was cone-shaped) because no light would reflect into your eye. For some reason, it seems to me that that would not happen.I wold try an experiment, but I do not have an all-black room in my house. Could someone please explain why this would or would not happen and even try the experiment if possible?

Thanks in advance,
Ryan

I think you will not be able to see any thing except the path of light before reflection.

 

[RyanXXVI]

Black is the absence of visible light. A black object absorbs all visible light. Yes, if you were in a room with true black walls, you would not see any light reflected from the walls. Of course, there is no truly black paint so you can't really do that experiment.

Can you explain to me the difference between "black" and "true black", I understand that something that is "true black" will not reflect light, but what in a black object prevents it from being "truly black"? Is the space between the stars as I see them with my naked eye truly black?

Despite this confusion your answer was very helpful. Thank you.

 

[RyanXXVI]

Can you explain your reason?

When I was very young I would go around shining flashlights at things an when I would shine the light onto a black surface, a patch on the surface would be illuminated. If black truly absorbed the light no illumination would happen. This contradiction is what led me to starting this thread.

 

[RyanXXVI]

I think you will not be able to see any thing except the path of light before reflection.

Why would I be able to see the light before reflection? The light is traveling away from me and there is a cone around the source so no light would come at me unless through extreme diffraction. The only way for me to see the light in this situation would be for the light to reflect off the walls, although I understand that I could just turn the flashlight towards me and see the source.

 

[RyanXXVI]

On a slightly similar note, can we make the color black with holes? I saw a video by Veritasium awhile ago explaining this phenomenon but it mostly focused on hues such as red and green etc.

 

[UltrafastPED]

Can you explain to me the difference between "black" and "true black

Many dark colors appear black, depending upon the lighting. For example, one of my old uniform coats was "Navy blue", which looks pretty black except when in bright sunlight.

The reference "True Black" means that it appears black to the eye even in bright sunlight.

The closest to "true black" that is seen in the lab is a mat of vertically aligned carbon nanotubes; when I've worked with them in the lab they look just like a patch of darkest night! They absorb almost all of the light across a very broad spectrum, from UV deep into the IR, including all of the visible spectrum. Absorption is better than 99% across the board. See http://en.wikipedia.org/wiki/Optical_properties_of_carbon_nanotubes#Carbon_nanotubes_as_a_black_body

When you used your flashlight there are several effects which can lead to some light reflection:
(1) the paint is slightly reflective
(2) the paint is not "true black"
(3) there was dust or other small particles in the air which caused some light scattering

This last effect, light scattering from dust, is how you can see a laser beam or crepuscular rays from the sun. See http://www.atoptics.co.uk/atoptics/ray1.htm

 

[sophiecentaur]

You must remember that there is nothing that emits absolutely zero EM energy. Objects on Earth are at around 300K and they have a significant emission spectrum. Also, there is radiation (CMBR) from even the deepest regions of space. So there is no such thing as true black; it's all relative.

 

[davenn]

......
The closest to "true black" that is seen in the lab is a mat of vertically aligned carbon nanotubes; when I've worked with them in the lab they look just like a patch of darkest night! They absorb almost all of the light across a very broad spectrum, from UV deep into the IR, including all of the visible spectrum. Absorption is better than 99% across the board. See http://en.wikipedia.org/wiki/Optical_properties_of_carbon_nanotubes#Carbon_nanotubes_as_a_black_body

...

That's some cool stuff, UltrafastPED, thanks for the info and link

another new thing I have learned today, haha

Dave

 

[Meson080]

I think you will not be able to see any thing except the path of light before reflection.

why would i be able to see the light before reflection? The light is traveling away from me and there is a cone around the source so no light would come at me unless through extreme diffraction. The only way for me to see the light in this situation would be for the light to reflect off the walls, although i understand that i could just turn the flashlight towards me and see the source.

http://www.warren-wilson.edu/~physics/PhysPhotOfWeek/20050401LaserRefraction/LaserRefractn.jpg

Description: This is a photo made by student Alicia Safdie showing laser light entering a tank of water. An aquarium is filled partway with water - the water level is about 1/4 from the top of the image. Laser light enters the air above the water from the upper left and strikes the water. Most of the laser light then enters the water, but the light is bent as it enters the water. Normally, the laser beam is not visible, but some milk was dispersed in the water making the water slightly cloudy, which makes the laser beam visible. Smoke from a smoldering match, blown into the air above the water level, makes the laser beam visible in the air. To obtain the picture without much light for the camera, auto exposure was used with the camera well-supported on the table. The exposure time was 1 second.

Image and description credits: http://www.warren-wilson.edu/~physics/PhysPhotOfWeek/20050401LaserRefraction/

Here, the laser might be visible because of scattering of rays from the smoke particles above the water surface and scattering of rays from the milk molecules inside the water.

Similarly, I expect light path to be visible because of light scattered towards your eye by the dust particles coming in the light path.

 

[RyanXXVI]

Thank you all, I understand this concept rather well now.