Light, colour and reflection

[Function]

Hello everyone

 

Just asking myself this question over and over again:

Is every color we see the negative one from what it actually is?

 

Light, colours as we see it, is reflected light. If white light falls on an object, which in its turn sends out light in that colour that we accept is the colour of the object, then it must absorb the rest of the white light - the negative of the reflected colourful light.

 

E.g.

I have in front of me a computer mouse which appears black to me (let's assume it's pure black - not very realistic, but hey, we're reasoning theoretically here). Black (or 'no light') is being reflected from it. No light is reflected, so all light must be absorbed, making this mouse in fact white.

 

Is this actually the case, and are photo negatives the true image of our world as we don't know her?

 

Thanks. Thread open for discussion

 

-F.

 

P.S. Thread open for discussion, even discussions which will lead us to a deeper state of thinking (e.g. how do we know that we see colours all the same, that the colours that truly are, are not as they are seen by us (so actually another problem than the original problem stated here))

Edited September 7, 2014 by Function

[studiot]

Lizzie L put a lot of effort into this thread.

 

http://www.scienceforums.net/topic/82666-can-a-white-piece-of-paper-reflect-light/

[elfmotat]

The color of an object is defined as the particular way in which our brain interprets the frequency of light that reflects off it and reaches our eye. A blue object is defined as blue because the light reflecting from it is blue. Why would we define its color to mean the frequencies of light that it does absorb, i.e. the ones we can't see? That makes no sense to me.

[swansont]

You can remove the human perception from the problem, which muddies the problem (your eyes can be tricked), and analyze with an instrument. White light has all the colors of the spectrum present. Green light has wavelengths around 550 nm in it, not all of the other wavelengths.

[Sensei]

Light, colours as we see it, is reflected light. If white light falls on an object, which in its turn sends out light in that colour that we accept is the colour of the object, then it must absorb the rest of the white light - the negative of the reflected colourful light.

In most cases it's correct.

 

There is yet another explanation:

white light is falling on object, then reflected photons are reacting us, but other also reflected photons with different frequencies, or at different angles, are not reaching us (instead of absorption).

It's called Iridescence.

http://en.wikipedia.org/wiki/Iridescence

 

Also instead of absorption there can happen refraction.

 

E.g.

I have in front of me a computer mouse which appears black to me (let's assume it's pure black - not very realistic, but hey, we're reasoning theoretically here). Black (or 'no light') is being reflected from it. No light is reflected, so all light must be absorbed, making this mouse in fact white.

Majority of black body objects are emitting light at frequencies invisible to human eye. For instance infra red, or microwave. You can see them emitting light (especially when they're really hot) using IR camera.

This can be used to detect fire, before flame really appeared.

 

Photon with frequency f, has energy E=h*f

f = c/wavelength

and

wavelength = c/frequency

so

wavelength = h*c/E

 

My red photon laser has wavelength 650 nm,

green photon laser has wavelength 525 nm (24% more energy than red),

blue/violet photon laser has wavelength 405 nm (30% more energy than green, and 60% more than red).

So it's not just matter of "color", it's matter of energy.

Energy can be used to heat substances.

Red photons will heat substances to lower temperature, than green, than blue (if we send them in the same quantity) (assuming they will be all absorbed).

 

Photons with certain frequencies (energies) trigger (or not) chemical reactions.

Edited September 8, 2014 by Sensei

[swansont]

 

My red photon laser has wavelength 650 nm,

green photon laser has wavelength 525 nm (24% more energy than red),

blue/violet photon laser has wavelength 405 nm (30% more energy than green, and 60% more than red).

So it's not just matter of "color", it's matter of energy.

Energy can be used to heat substances.

Red photons will heat substances to lower temperature, than green, than blue (if we send them in the same quantity) (assuming they will be all absorbed).

 

 

Emphasis added: that's a key caveat. Light absorbed from a non-thermal source could heat an object to an arbitrary temperature, dependent only on the power. There's a higher power in blue light vs red light if the photon count is the same. But 1 W of absorbed light from each would have the same heating effect.

[studiot]

 

Light, colours as we see it, is reflected light.

 

Not always.

 

Just look at some traffic lights.

The colour you see is the colour radiated by the source.

Edited September 8, 2014 by studiot

[Sensei]

Emphasis added: that's a key caveat. Light absorbed from a non-thermal source could heat an object to an arbitrary temperature, dependent only on the power. There's a higher power in blue light vs red light if the photon count is the same. But 1 W of absorbed light from each would have the same heating effect.

Exactly.

 

Anyway it is visible from equations:

 

[latex]E = \frac{h*c}{\lambda} = \frac{6.62607*10^{-34} * 299792458}{650*10^{-9}} = 3.056*10^{-19} J[/latex]

 

[latex]E = \frac{h*c}{\lambda} = \frac{6.62607*10^{-34} * 299792458}{525*10^{-9}} = 3.784*10^{-19} J[/latex]

 

[latex]E = \frac{h*c}{\lambda} = \frac{6.62607*10^{-34} * 299792458}{405*10^{-9}} = 4.9*10^{-19} J[/latex]

 

quantity of red photons in [latex]1 W = 1 \frac{J}{s}[/latex] is [latex]\frac{1 J}{3.056*10^{-19} J} = 3.27*10^{18}[/latex] photons per second.

 

quantity of green photons in [latex]1 W = 1 \frac{J}{s}[/latex] is [latex]\frac{1 J}{3.784*10^{-19} J} = 2.64*10^{18}[/latex] photons per second.

 

quantity of blue photons in [latex]1 W = 1 \frac{J}{s}[/latex] is [latex]\frac{1 J}{4.9*10^{-19} J} = 2.04*10^{18}[/latex] photons per second.

Edited September 8, 2014 by Sensei