|Feb23-13, 07:46 PM||#1|
Coloured light on complementary surface
I have observed the following phenomenon: When I shine coloured light (say red) on a surface that has its complementary colour (in this case green) it becomes significantly less saturated (even to the point of becoming completely desaturated, i.e. grey).
I had no opportunity to try this out in darkness, but my intuition is that a perfectly green surface completely absorbs red light and it would appear as if there was no light at all, i.e. remain black.
This seems in conflict with what I have learnt about light so far.
There are two systems, a subtractive and an additive. The subtractive is when the sum of all colours gives black (for example with paints) because it subtracts certain wavelengths from being reflected. The additive is when all colours give white, because they add up to the full spectrum (which is white, e.g. light).
Shining a coloured light onto a coloured surface seems to fall under the subtractive category.
But is it not the case that when you look at the wavelength histogram of a coloured light it is not a column on one very specific wavelength but rather a bell shape around it? If this is true, subtraction could work only with two identical curves. But with red and green the curves are different and subtracting them would create leftovers on the outside edges of the curves which would produce a colour.
Where am I wrong?
Could someone please explain and/or proof this to me.
References and further reading is appreciated. I do not fear math.
|Feb23-13, 09:56 PM||#2|
A colored surface absorbs some wavelengths and scatters others - it may also reflect and transmit different wavelengths differently.
The red surface scatters red wavelengths and absorbs others ... so if you shine non-red (i.e. green) light on it, a perfect red would scatter none and appear black. The less perfect the red, the more of the other colors also get scattered, so you get shades of grey ... if the surface just scatters red strongly, and other wavelengths to different amounts, then the color will appear to whatever mix is available to be scattered by the amounts scattered, which will be different for each wavelength and you'll get a slight colored tinge to the grey.
In practice - pigments tend to absorb a range of wavelengths and scatter the rest ... i.e. in the case of a primary like magenta, the pigment is absorbing (but not usually 100%) a wide range in green - scattering red+blue from white light. If you supply only green light though, you should get a clear greyish appearance.
Red pigment just has a lot of magenta in it and a bit of other colors to redden it.
Note - the scattered/absorbed colors are not bell-shaped curves around the dominant color.
Google for "absorption spectrum".
|Feb24-13, 05:47 AM||#3|
Hey Simon Bridge!
Thank you for your quick response.
I reasoned under the wrong assumption that the absorption spectrum was bell-shaped around the dominant colour. That is very good to know!
A further question arises for me.
Assume I shine red light on a green surface in a room where there is some white light as well. If the green surface completely absorbs red light, then that surface would still appear green because of the other white light - is that correct?
|Feb24-13, 06:35 AM||#4|
Coloured light on complementary surface
A white surface would look red - probably light red.
A surface that absorbs 100% of the red part of the spectrum would be cyan colored in white light.
It'll look like a dark cyan in the mixture.
You can try the effect out under a sodium street light if you are lucky enough to find some.
Take a white-light torch with you - or just use a red filter and two torches.
You can vary the ratio of the colored and white light on different colored bits of card.
There are two other effects to consider when trying to figure what something will look like in terms of color.
There is specular color, and how color vision actually works.
I've been concentrating on diffuse color since that is where you question lies.
Color vision has all sorts of artifacts in it that make an analysis difficult.
|Mar9-13, 02:50 AM||#5|
Thank you for your detailed response.
I was looking for isolated sodium lamps, but could find none so far that would be in an otherwise light-free environment. I will definitely try this out with some coloured source of light if I find one.
In practice, I assume, absorption is imperfect and a surface that has the complimentary colour of the light that shines on it, would still reflect a little bit. This would make it appear grey (not black as absorption is imperfect), because our colour vision requires a certain intensity of light for us to see colour.
|colour theory, coloured light, light, wavelength of light|
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