Question

When we see white light, is it a constructive interference of red light, green light and blue light? How could a light bulb make so many different wavelengths of wave?

Answer 1

Hi! Fun, fun question! I'll answer quickly though. My knowledge is only elementary in this area. So hopefully somebody else can fill an any gaps that I leave you with.

First, I'm not sure about the first question. I'm not sure if you consider interference too much when you have very different wavelengths. But I don't see why you wouldn't. Someone else could help out here.

In your eye, your receptors will connect to the brain, and the brain will realize that ALL the receptors, each of which can detect certain bands of light well, are being hit with their detectable bands. So the eye will have some pick up on reddish color, others on greenish, until all the receptors have put together what they're getting. And the brain says, "white."

Why so many wavelengths of a light bulb? That question is easier, answering vaguely, I think. When atoms are excited, they'll have electrons absorbing energy and then releasing. In a light bulb, they absorb the electrical energy. Maybe the energy that comes from the high temperature, too. When they release energy, they create an electromagnetic wave.
The more energy released is the greater energy of the wave.
The greater energy of the wave, the greater frequency.

So the color of light depends on the frequency. The frequency depends on the energy, which depends on the material as to how much energy the electrons can absorb and release.

Hopefully someone else here can check that for you, though.

Answer 2

You're asking a question about something called Blackbody Radiation. When an object is heated it gives off a spectrum of light from the very short ultraviolet out to very long radiation. Theoretically, such a body gives off light with wavelengths from infinitely short to infinitely long. Max Planck was the gent that came up with the equation that describes the spectrum of light a blackbody gives off. The color you actually see is the result of averaging of the intensity of each wavelength in the visible spectrum . Here's a graph of the typical spectra of two incandescent bulbs. The vertical axis is energy density and the horizontal axis is frequency. Note the filaments on incandescent bulbs glow at about 2900-3000K.

Answer 3

Here's a plot of the Sun's blackbody radiation compared to an incandescent bulb.

Answer 4

I see. Thank you very much, @theEric and @PsiSquared.
PsiSquared, how about interference ?
Questions:
1. When Blackbody produces lots of varying wavelength radiations, is there any interference between them?
2. If there is no interference, can we see the blend of two/more colors? In other words, if there is no interference, can we still see white light?
Thank you!

Answer 5

Yes there is interference between different wavelengths, but with white only there are only a few very special cases where fringes are visible. Those special circumstances are the cases when the path length difference is very short, such as with the case of Newton's rings. Normally, the path length differences are large enough that the fringe spacing and the fringe contrast (visibility is the technical term) is too small for the eye to resolve. If you do an internet search for white light interferometers you'll some other cases where interference fringes can be observed with white light.

As for the colors we see, colors are detected by cone receptors in the eye. There are three types of cone receptors: cones that detect short wavelengths; cones that detect medium wavelengths; and cones that detect long wavelengths. The brain mixes the response of each of three cones to produce a color at a given point or area. This similar to how your brain, via the eye, interprets colors from your computer screen, a screen which is made up of red, blue, and green pixels.

Answer 6

There are no interferences between lights of different wavelengths: since the wavelengths are different, the waves cannot be coherent and cannot interfere.

'White' perception is how your brain interprets the signals received on your retina by the activation of your three cone receptors.

Answer 7

Well.. what? Yes or No= =?

Answer 8

Different color wavelengths can absolutely interfere. Such interference is the basis of temporal coherence. It just happens that with light sources with a broad range of frequencies, the coherence length is very short. We interfere lasers all the time, and there is no truly monochromatic laser, since all lasers have ∆ν>0. If different wavelengths couldn't interfere, we wouldn't have white light interferometers.

Answer 9

White light has a coherence length on the order of a micron, which means once it's traveled beyond a micron, it will no produce fringes with sufficient visibility, which is why there are only a handful of white light interferometer types.

Answer 10

White light can produce interference patterns, but these patterns are only the superposition of all the interference patterns produced by the individual wavelengths.
It does not mean that, within the white light, lights of different wavelengths can interfere.

Imagine you produce an interference pattern using white light. If you put a red filter on the first branch of the interferometer and a green one on the second branch, all interferences will disappear. The red light and the green light cannot interfere because they are not coherent.

Answer 11

They do have a coherence length, just as white light does, as I just demonstrated.

Answer 12

What do you mean by "they"?

Answer 13

There's a coherence length associated with viewing interference fringes between red and green light, and it's longer than that of white light.

Answer 14

If green is 550nm and red is 700nm, the gives a center wavelength of 625nm and a bandwidth of 150nm, yielding a coherence length on the order of 2.6μm.

Answer 15

Thank you so much guys. I was away and i am going to look at the replies. Thank you!

Answer 16

The fact that lights of different colours can or cannot interfere is debated between PsiSquared and myself. But if we go back to your first question:
"When we see white light, is it a constructive interference of red light, green light and blue light?"
The answer is clearly "no". White light is only the superposition of red, green and blue light that reconstructs the full spectrum of white light.

Answer 17

@Vincent-Lyon.Fr: Yeah I think so. Thank you
@PsiSquared:
What I understood is:
Our brain detects different wavelengths of waves separately and therefore we can still see white light even though if there were no interference between different wavelength.
Am I right?
Well. Actually, if there is interference between different wavelength of waves, I think there will be change in intensity. Am I right? But how could fringe be formed? I think fringe will be only when there is diffraction.
Thank you also!

Answer 18

@Vincent-Lyon.Fr: Well. I shouldn't use @ to specify that paragraph, i just want somebody to check my understanding. So if you can check my understanding as well it will be great.
Actually, I want to know the result of your debate. @PsiSquared,@Vincent-Lyon.Fr Thank you guys!
Well. I want to ask one more question though I don't want to close this post but I think you guys are experts so I will post one more question here and can you response for all my replies please? Thank you very much!
Another question:
In general speaking, intensity of light is how much power per unit area, right? If we have a laser light reflects from a mirror which will cause a specular reflection, because the laser light travel in cone shape, the area will not change as it travel further so it will have the same intensity when it entered our eyes if the mirror doesn't absorb any energy. Right? But what about the laser light is reflected from a rough surface?(diffuse reflection) Will it have the same intensity when it enter our eyes? Because specular reflection just reflect all light waves into an specified area but diffuse reflection is not. Diffuse reflection will reflect light waves into all directions so if we look at it at one of the directions what intensity will we get?
Thank you guys!

Answer 19

Wrong. Correction: @Vincent-Lyon.Fr: I shouldn't use @ to specify a person for that paragraph

Answer 20

A fringe is formed when when one light field deconstructively interferes with another. The net result is that the intensity at the point is less than when the two fields constructively interfere. What the eye detects is light intensity (the light's electric field squared, i.e. I∝E^2). Diffraction is not required for interference. All that is required for interference is ether a path length difference between beams of light or a timing difference (i.e., frequency) between beams of light. We generally model interference in terms of a coherent beam (light from a single coherent source which also implies the light is monochromatic) but even incoherent light can interfere.

Light reflected from a diffusing surface will tend to be less intense than light reflected from a specular surface. What intensity you get from a diffusing surface will depend on the profile of the incident light and the the diffusing properties of that surface.

I've attached an image demonstrating the interference of three wavelengths of light. The top image shows each individual wave overlaid, while the bottom image shows the mathematical result, i.e. the interference, of those three waves.

Answer 21

Here's an image of 5 wavelengths interfering. As with the picture above, the first image is of the 5 individual wavelengths overlaid (as projected on a screen--I should added this comment in my post above), while the second is the mathematical addition of those waves. Note that the first posted image and this image show the intensities of each wave, which as I indicated above, is given by I∝E^2.

Answer 22

Ok. I think we had enough discussion about this topic.
Thank you guys!
@PsiSquared: Your answer is as detail as before and it seems that you are very good at that. Did you study them in university?
I will close this and ask another question. Thank you guys!