What Is the Nature of Black in the Visible Spectrum VIBGYOR?

[shihab-kol]

In the visible spectrum VIBGYOR, there is no black colour.So, what do we percieve as 'black' ?

Another of my queries is that when dispersion takes place there is a change in wavelength but not so in case of frequency. But they are related inversely. So, why does this happen?

 

[anorlunda]

In the visible spectrum VIBGYOR, there is no black colour.So, what do we percieve as 'black' ?

Black is the lack of light. White is the superposition of all visible colors.

Just think of a black-and-white picture. The dark parts are black and the bright ones are white, medium intensities are grey.

 

[shihab-kol]

Black is the lack of light.

What do you mean by that? How is light absent? Is it something related to absorption?

 

[sophiecentaur]

Bright / dark / black relate to intensity and not to wavelength or frequency (or chrominance).

 

[anorlunda]

Go into a sealed room and turn off the lights. What color do you see?

Is it something related to absorption?

Yes, that's one way. A surface that reflects no light to your eye is perceived as black.

 

[shihab-kol]

Go into a sealed room and turn off the lights. What color do you see?
Yes, that's one way. A surface that reflects no light to your eye is perceived as black.

So, something absorbing all light is perceived to be black?

 

[sophiecentaur]

So, something absorbing all light is perceived to be black?

Yes

 

[shihab-kol]

Bright / dark / black relate to intensity and not to wavelength or frequency (or chrominance).

The frequency part of the question does not relate to black.

Yes

Thanks for the clarification .
And about the second part ?

 

[nasu]

In your second part I would guess you misuse the term "dispersion". Dispersion means that the speed of the wave through a medium depends on frequency. You also can write velocity of wave as a function of wavelength. But dispersion does not mean a change in either frequency or wavelength. The speed in the medium is different than in the air and so the wavelength will change while propagating through that medium. The frequency does not depend on the propagation speed, is a characteristic of the source of the wave. But this happens even in a medium without dispersion.

 

[shihab-kol]

In your second part I would guess you misuse the term "dispersion". Dispersion means that the speed of the wave through a medium depends on frequency. You also can write velocity of wave as a function of wavelength. But dispersion does not mean a change in either frequency or wavelength. The speed in the medium is different than in the air and so the wavelength will change while propagating through that medium. The frequency does not depend on the propagation speed, is a characteristic of the source of the wave. But this happens even in a medium without dispersion.

Yes.
I tried to mean refraction.
Thanks

 

[sophiecentaur]

Frequency and intensity are independent. Intensity does have an effect on perceived color, though. You can see a yellow area and a brown area in the same scene. They can both have the same chrominance co ordinates. Our color perception falls off as light level drops and our vision becomes monochrome at very low levels. It's a bit of a grey area. ( pun intended)

 

[pixel]

Another of my queries is that when dispersion takes place there is a change in wavelength but not so in case of frequency. But they are related inversely. So, why does this happen?

It's true frequency and wavelength are inversely related, but in refraction the wavelength and the speed of the wave change, the frequency doesn't change.

 

[Drakkith]

So, something absorbing all light is perceived to be black?

Well, perfect absorbers do not exist. Typically an object with a "black" color absorbs a much larger percentage of incoming light across most of the visible spectrum compared to surrounding objects, though it never absorbs all of it. I am currently surrounded by about a dozen different objects that I would call "black" and each one actually looks slightly different than the others.

 

[russ_watters]

So, something absorbing all light is perceived to be black?

Or not having them created to begin with. Consider a white movies screen.

 

[shihab-kol]

Why does the frequency not change?

 

[jbriggs444]

Why does the frequency not change?

Why does the frequency not change when a wave enters a medium like a prism and thereby speeds up or slows down?

Because every time a wave crest arrives at the boundary the wave crest passes through and proceeds past the boundary. The frequency at which crests arrive is equal to the frequency at which crests proceed.

 

[sophiecentaur]

Black is not a particularly quantitative thing. It is not 'less than 10 photons arriving per nanosecond'.
The doorway into a bar can appear black at noon but, with the same lighting inside, it can be the brightest object in the street on a moonless night.

 

[XZ923]

Well, perfect absorbers do not exist. Typically an object with a "black" color absorbs a much larger percentage of incoming light across most of the visible spectrum compared to surrounding objects, though it never absorbs all of it. I am currently surrounded by about a dozen different objects that I would call "black" and each one actually looks slightly different than the others.

I'm just curious, isn't a black hole a perfect absorber beyond the event horizon? Further, would it be accurate to state that the only known existence of "perfectly black" that we know of at this point is inside a black hole's event horizon as perceived by someone outside it? Or is this negated by the existence of Hawking radiation which should be visible in the IR range?

I'm sorry if these are fairly stupid questions; I'm fascinated by the concept of black holes but don't know much about them.

 

[jbriggs444]

I'm just curious, isn't a black hole a perfect absorber beyond the event horizon?

Wiki can answer this. If one considers a black hole as if it were a "black body"...

"A black hole of one solar mass (M☉) has a temperature of only 60 nanokelvins (60 billionths of a kelvin); in fact, such a black hole would absorb far more cosmic microwave background radiation than it emits. A black hole of 4.5×10²² kg (about the mass of the Moon, or about 13 µm across) would be in equilibrium at 2.7 K, absorbing as much radiation as it emits. Yet smaller primordial black holes would emit more than they absorb and thereby lose mass.^[11]"​

 

[pixel]

Why does the frequency not change?

Another example: a light string attached to a heavier one. If you start a wave train in the lighter string, as each crest meets the heavier string it will become a crest there, so the frequencies are the same in each string. But for equal tension in each string, the wave in the heavier string is slower and the wavelength there is smaller than in the lighter string.

 

[Drakkith]

I'm just curious, isn't a black hole a perfect absorber beyond the event horizon?

I'm talking more about physical objects you could encounter in real life. Not crazy spacetime geodesics.

 

[sophiecentaur]

Why does the frequency not change?

How could it? What would drive the parts of one medium at a different frequency from the parts in the previous medium? You need Phase Continuity at an interface and that demands no frequency change.

 

[shihab-kol]

But shouldn't the wave lose energy along the way? And frequency is a measure of energy.In a wave the energy is transferred from one place to another but along the some dissipation should occur.
So, why?

 

[jbriggs444]

But shouldn't the wave lose energy along the way? And frequency is a measure of energy.In a wave the energy is transferred from one place to another but along the some dissipation should occur.
So, why?

Why should a wave lose energy along the way?

In the case of electomagnetic waves quantized as photons, the energy in a photon is determined by its frequency and vice versa. But photons do not "dissipate" (disregarding cosmological red shift).

In the case of more mundane waves (e.g. sound waves, surface waves on a lake or transverse waves on a taut string) there is no association between frequency and wave energy. The energy is associated with amplitude, not frequency.

 

[shihab-kol]

Why should a wave lose energy along the way?

In the case of electomagnetic waves quantized as photons, the energy in a photon is determined by its frequency and vice versa. But photons do not "dissipate" (disregarding cosmological red shift).

Ah, I am being taught the classical nature of waves where they do dissipate energy (like sound), so the quantization of energy in a photon did not occur to me.
Thanks.

 

[shihab-kol]

Well, another question.
I draw lines of black ink consecutively on a paper.When viewed from a distance they appear to be one solid thick black line.But when viewed from a near position one can make out the separate lines.Why is this so?

 

[A.T.]

I draw lines of black ink consecutively on a paper.When viewed from a distance they appear to be one solid thick black line.But when viewed from a near position one can make out the separate lines.Why is this so?

https://en.wikipedia.org/wiki/Angular_resolution

 

[sophiecentaur]

Well, another question.
I draw lines of black ink consecutively on a paper.When viewed from a distance they appear to be one solid thick black line.But when viewed from a near position one can make out the separate lines.Why is this so?

Diffraction.

 

[shihab-kol]

Diffraction.

How?

 

[shihab-kol]

I read in a brief history of time that the Earth is a geodesic and assumed it to be some sort of a sphere. But Drakkith says that black holes are also geodesics .
What exactly is a geodesic?

 

[davenn]

How?

AT's angular resolution was the better response ... did you read the link he gave ?

I read in a brief history of time that the Earth is a geodesic and assumed it to be some sort of a sphere. But Drakkith says that black holes are also geodesics .
What exactly is a geodesic?

Have you googled you question yet ?
I know you will get multiple hits for answers

 

[shihab-kol]

Have you googled you question yet ?
I know you will get multiple hits for answers

I got a meaning.
The shortest possible line between two points on a curved surface.
Uh, but how about a black hole?
I don't understand.(Of course I get the'crazy' part!)

And, I get the angular resolution part. Thanks,AT

 

[sophiecentaur]

Diffraction.

How?

Every image is formed from the contributions of light from all the different paths from the source object. The lengths of these paths are all slightly different so the separate waves interfere with each other, not adding up perfectly. This is Diffraction and it occurs Always. Diffraction is what causes the blurring of images*. A small aperture will produce more visible 'diffraction effects'. This can be explained by the fact that 1. the diffraction 'fringes' are finer and finer as the aperture is wider and more coarse from a smaller aperture and 2. Proportionally more light passes through the middle of a large aperture so the fringes are less visible.
*I expect some people will disagree with this blanket statement but the aberrations that are caused in glass lenses are still, effectively due to the failure of the various paths to add up perfectly (in phase). To my mind that is just a sub-set of diffraction.

 

[shihab-kol]

Every image is formed from the contributions of light from all the different paths from the source object. The lengths of these paths are all slightly different so the separate waves interfere with each other, not adding up perfectly. This is Diffraction and it occurs Always. Diffraction is what causes the blurring of images*. A small aperture will produce more visible 'diffraction effects'. This can be explained by the fact that 1. the diffraction 'fringes' are finer and finer as the aperture is wider and more coarse from a smaller aperture and 2. Proportionally more light passes through the middle of a large aperture so the fringes are less visible.
*I expect some people will disagree with this blanket statement but the aberrations that are caused in glass lenses are still, effectively due to the failure of the various paths to add up perfectly (in phase). To my mind that is just a sub-set of diffraction.

Ok.Got it!
Thanks!

 

[sophiecentaur]

Ok.Got it!
Thanks!

Ok.Got it!
Thanks!

My post can be summed up by the term in AT's post 'angular resolution'. I gave you overkill, possibly.

 

[pixel]

I expect some people will disagree with this blanket statement but the aberrations that are caused in glass lenses are still, effectively due to the failure of the various paths to add up perfectly (in phase). To my mind that is just a sub-set of diffraction.

I guess I'm one of those people, although maybe we're getting more into the realm of philosophy here.

In diffraction, we consider the different path lengths that light takes to reach a given point. With aberrations, the light is directed to different points, e.g., with spherical aberration the rays that goes through the outer parts of the lens meet closer to the lens than the rays that go through the center of the lens. I would not consider the inability of all these rays to meet at the same point as diffraction. And usually diffraction is what provides the ultimate limitation on imaging, after all the aberrations are minimized, hence the term "diffraction limited" optics.

 

[sophiecentaur]

I guess I'm one of those people,

Fair enough. But I cannot see your distinction between how a lens produces an image and the way that, for example, a Zone plate will also produces an image. No one would claim that a zone plate doesn't work by diffraction - it produces points on a plane where the light from a particular point on an object adds constructively (in phase) but nowhere else from that source point. Precisely how a lens operates. You have chosen to look at the operation of a lens in terms of Rays but it can also be treated in terms of what it does to a spherical wave front, emanating from one point on an object. The lens delays each part of the wave front so that the the path length is equal at only one point in the image plane. The only difference between the two is that the zone plate works by selecting parts of the emanating wave front to produce constructive interference at certain parts of the image plane and the lens adjusts the phases by suitable delays through different thicknesses of glass.
I appreciate that the terms 'diffraction' is commonly used in a restricted sense but it's a fairly arbitrary classification - in the same way that the patterns from two slits tends to be referred to as an interference pattern - largely because it can be predicted fairly well by treating the slits as point sources. But it's still a diffraction pattern which can be calculated approximately using a simple summation of terms instead of using the full integration.

 

[shihab-kol]

I know it is a lot of questions but since the heading contains both light and colours here goes.
Blue + red = magenta
But just combining rays would not give me magenta.At least that's what I think.
So,what are the criteria?
I think an object which gives out magenta will absorb every color except blue and red.
But is this all?
And yeah,geodesics

 

[shihab-kol]

Oh, come on !
You all going to leave me hanging??!?

 

[russ_watters]

I know it is a lot of questions but since the heading contains both light and colours here goes.
Blue + red = magenta
But just combining rays would not give me magenta.

Why not? That's how a computer monitor does it.

 

[Drakkith]

Fair enough. But I cannot see your distinction between how a lens produces an image and the way that, for example, a Zone plate will also produces an image. No one would claim that a zone plate doesn't work by diffraction - it produces points on a plane where the light from a particular point on an object adds constructively (in phase) but nowhere else from that source point. Precisely how a lens operates. You have chosen to look at the operation of a lens in terms of Rays but it can also be treated in terms of what it does to a spherical wave front, emanating from one point on an object.

Indeed. I believe one could say that diffraction is just wave interference as well. But we tend to define different terms to talk about specific situations when it suits us.

I know it is a lot of questions but since the heading contains both light and colours here goes.
Blue + red = magenta
But just combining rays would not give me magenta.At least that's what I think.
So,what are the criteria?
I think an object which gives out magenta will absorb every color except blue and red.
But is this all?

There is much more to color perception, but that is the basic idea, yes.

Oh, come on !
You all going to leave me hanging??!?

You posted your previous question right after most people who use the forum went to bed (your post was at 11:38 PM my time). Give them time to get their morning coffee at least!

 

[sophiecentaur]

But just combining rays would not give me magenta.At least that's what I think.

Combining "Rays" is additive mixing. which is how TV produces the colours. Red light plus Blue light will produce Magenta. But when you refer to an"object" that is describing the colour due to Subtractive mixing - the object absorbs all wavelengths except red and blue and will be described as Magenta.
You need to google Additive and subtractive colour mixing. They are the 'opposite' idea to each other. Subtractive mixing is a much more difficult way to produce wanted colours which is why ALL colour film and printing fails for even simple colours and Spot Colours have to be used if you really want colour fidelity.

 

[sophiecentaur]

we tend to define different terms to talk about specific situations when it suits us.

yes, that is common usage but the reality is that whenever there is more than an ideal point source, diffraction takes place. The 'quality' of the image is just a matter of how well the various waves manage to arrive at the right time in the right place.

 

[Drakkith]

yes, that is common usage but the reality is that whenever there is more than an ideal point source, diffraction takes place.

Diffraction always occurs though. Even with an ideal point source. A "diffraction limited" telescope would image a point-source as an airy disk, just as expected from diffraction laws.

The 'quality' of the image is just a matter of how well the various waves manage to arrive at the right time in the right place.

Certainly. Geometric (ray) optics cannot adequately describe the final image quality in many optical systems.

 

[shihab-kol]

You posted your previous question right after most people who use the forum went to bed (your post was at 11:38 PM my time). Give them time to get their morning coffee at least!

Oh, by my time it was pretty early!

 

[shihab-kol]

Combining "Rays" is additive mixing. which is how TV produces the colours. Red light plus Blue light will produce Magenta. But when you refer to an"object" that is describing the colour due to Subtractive mixing - the object absorbs all wavelengths except red and blue and will be described as Magenta.
You need to google Additive and subtractive colour mixing. They are the 'opposite' idea to each other. Subtractive mixing is a much more difficult way to produce wanted colours which is why ALL colour film and printing fails for even simple colours and Spot Colours have to be used if you really want colour fidelity.

So, when i emit only blu and red or blue and green they will mix just like that but when I reflect it , I absorb other colours and reflect only those two.
Ok, got it.
Thanks.

 

[shihab-kol]

It seems the members won't answer to my 'geodesic' queries .
Why? Because its not mentioned in the tags?
Ok, look again, its there!
Now, please reply!

 

[sophiecentaur]

Diffraction always occurs though. Even with an ideal point source.

Sorry, I didn't mean it that way. I meant that it's only a point source that creates a pattern (itself) that is diffraction fringe free.
Unless you are a glow worm or bioluminescent squid, you have colour due to subtractive mixing from white light

 

[russ_watters]

It seems the members won't answer to my 'geodesic' queries .
Why? Because its not mentioned in the tags?
Ok, look again, its there!
Now, please reply!

Can't you Google that?

 

[sophiecentaur]

It seems the members won't answer to my 'geodesic' queries .
Why? Because its not mentioned in the tags?
Ok, look again, its there!
Now, please reply!

That topic is nothing to do with the colour black. Start a new thread (when you have done some googling) with an appropriate title and you may get some answers. The right title will grab the right people.
(And try to make your comments a little less demanding than that post. You need good will from PF.)