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When white light falls on an object the body absorbs all colors of light and reflects some colors of light which we see the object to be.For eg.A red ball reflects red color and absorbs the rest colors.So,if I throw green light on a red ball the ball will absorb that green light.Now,will the part on which the light falls must appear black as it absorbs all the light and reflects no light.But this does not happen in reality(I have experimented using laser light).Why?

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    $\begingroup$ Colours aren't discrete entities. An object is never "blue" or "red" instead an object will reflect different portions of the EM spectrum. $\endgroup$ – Kenshin Jun 27 '15 at 12:33
  • $\begingroup$ @Mew-Image the situation that the ball is commanded that it will reflect only red light.Now,green light falls on it.So the ball will not reflect it.Does the portion of the ball on which the light falls appear black?It should but doesn't(seen experimentally by me). $\endgroup$ – SHM Jun 27 '15 at 12:37
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    $\begingroup$ as I said, there are no discrete colours in physics, there is no wavelength that equals red. The red cone receptor in our eyes responds to a whole range of wavelengths of light, not a single wavelength. $\endgroup$ – Kenshin Jun 27 '15 at 12:53
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    $\begingroup$ you will likely see the blue looking dark red in colour. $\endgroup$ – Kenshin Jun 27 '15 at 13:09
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    $\begingroup$ If the ball was perfectly red, and if the only light falling on it was perfectly green, then the ball would indeed appear black. But, if you see a red ball, that means there already is some red light reflecting off of it from somewhere. Shining a green light on the ball won't make the red light go away. $\endgroup$ – Solomon Slow Jun 27 '15 at 17:19
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One half of the problem is light. Light is an oscillating electromagnetic field. The frequency of oscillation determines the color. Higher frequencies have shorter wavelengths and are blue. Lower frequencies have longer wavelengths are red.

Light is a mix of a range of wavelengths. Sunlight contains a wide range of wavelengths, including some that are too long or short for us to see. A laser contains a very narrow range.

How light interacts with matter is the other half of the problem. Electromagnetic fields exert forces on electrons. Electrons in atoms or molecules are constrained in how they can move.

Metals have electrons that can move freely. This makes them good conductors. Freely moving electrons reflect light. Light makes electrons vibrate and absorb the light. Vibrating free electrons emit light and stop vibrating. Thus metals are shiny.

But even in metals, electrons are constrained. For example in copper, electrons can vibrate more readily at low frequencies. Copper reflects red light better than blue. It reflects better still in the infrared and even worse in the ultraviolet. Here is a graph from a random website, http://www.minoanatlantis.com/Minoan_Mirror_Web.php

enter image description here

So if you shine a red laser on copper, it reflects almost all of it. If you shine a green laser on copper, it would reflect less.

Other materials have different ways of interacting with light. But at each wavelength, they may reflect some and absorb some. They reflect more at some wavelengths and less at others. Even black materials reflect some light.

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When we say a red ball absorbs other colors, it doesn't mean it absorbs them perfectly. It can reflect non-red light, just less strongly than it reflects red. Beyond that, color vision is extremely complex. Your brain does amazing things trying to compensate for the color of light in a scene.

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I believe the point Mew is trying to make is that 'red' is a range of wavelength. Although your laser has a very narrow range of red wavelength, I suspect the range of wavelength reflection in the green pigment of the ball is much broader and may even have a range of red in it. An interesting, although perhaps unrealizable experiment would be to find a pigment that exactly measures red with the same wavelength and bandwidth as your laser. I predict that would indeed appear black.

You also can't discount the possibility that a pigment (material) can absorb light at one frequency and emit it at another such as in the case of fluorescence. A ball may show brilliant green in the sunshine due to fluorescence by UV light, but when exposed to only green light of the same green wavelength show as black.

I drive a red car and recall one time I had parked in a lot during the day. That night when I returned to drive home I could not find my car. The lot was illuminated wit sodium vapor lamps which give off a fairly narrow wavelength in the yellow-orange color. I finally realized the black car I was standing in front of was my car. It was the first time I experienced such an impressive effect that I had to touch the car to prove to myself someone had not repainted it.

The moral is - get the right light and the right pigment and you can realize the effect.

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