# Coloured light on complementary surface

 P: 5 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.
Homework
HW Helper
Thanks
P: 12,961
 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).
Well yes it would ... the shade of grey would depend on the type of surface.

A colored surface absorbs some wavelengths and scatters others - it may also reflect and transmit different wavelengths differently.

Loosely:
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.