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tris_d
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What makes, for example, blue color appear darker or lighter? Is it just amount of photons (intensity) or is there any relation between brightness and wavelength? Or is there something else that comes into equation as well?
The sensitivity of the optical equipment being used to measure it - i.e. the human eye has evolved to respond strongly to yellows and greens so these seem brighter and more noticeable.tris_d said:What makes, for example, blue color appear darker or lighter?
tris_d said:What makes, for example, blue color appear darker or lighter? Is it just amount of photons (intensity) or is there any relation between brightness and wavelength? Or is there something else that comes into equation as well?
Hetware said:"Brightness" as you have described it is fairly subjective. We might consider "light blue" to be brighter than "dark blue", but "black light" (really dark blue) has more energy per photon then visible light. That's the quantum physical way to describe it.
Intensity is the amount of energy passing through a specified area in a specified amount of time.
In classical electromagnetism, the intensity of light is proportional to the square of the amplitude of the waves. In quantum mechanics it depends of the frequency of the light as well as the number of photons.
ZapperZ said:I suggest you read this for a discussion of "brightness" when applied to optics.
http://www.rp-photonics.com/brightness.html
Zz.
tris_d said:Thank you. If you have some more detailed articles among the same lines let me know. I want to know all there is to know about brightness, intensity and whatever else comes into that equation.
tris_d said:Ok, let me define brightness as captured light on some photo that we open in Photoshop, turn it to gray-scale image, and then those pixels will have some value from 0 to 100, where 0 is black, 100 is white, and in between are shades of gray.
So then having all the photons be the same wavelength, brightness of each pixel will be proportional to the amount of photons that impacted that particular pixel?
tris_d said:I can see how light intensity would be defined by the number of photons, but what does frequency have anything to do with it?
If frequency defines intensity too, would that mean that one blue photon with higher frequency could produce more bright pixel than the other blue photon with lower frequency?
I thought frequency defines color, that frequency is just another face of the wavelength and that they are always in constant relation.
tris_d said:Ok, let me define brightness as captured light on some photo that we open in Photoshop, turn it to gray-scale image, and then those pixels will have some value from 0 to 100, where 0 is black, 100 is white, and in between are shades of gray.
So then having all the photons be the same wavelength, brightness of each pixel will be proportional to the amount of photons that impacted that particular pixel?
If frequency defines intensity too, would that mean that one blue photon with higher frequency could produce more bright pixel than the other blue photon with lower frequency?
I thought frequency defines color, that frequency is just another face of the wavelength and that they are always in constant relation.
tris_d said:I thought frequency defines color, that frequency is just another face of the wavelength and that they are always in constant relation.
tris_d said:Oh man! This is not simple. -- Thank you all, I'm chewing on it.
Simon Bridge said:I prefer to distinguish "complex" and "difficult" ... the second is subjective though the two terms are often used as synonyms in common language. So: Unless you are working simple idealized problems, nothing is ever easy!
It would be much harder for the eye to respond to each possible wavelength with a unique signal ... I suspect the resulting process would be more complex. Breaking the photon into three parts and doing the reconstruction is a handy way to simplify the process and it works very well despite the odd fudge (like "magenta").
This is certainly more difficult to follow than "color comes from wavelength" but it is no more complex.
The color of light does depend on it's wavelength, just like we tell grade-schoolers ... however, the experience of color is subjective and depends on how the eye and brain work together. How we get so much agreement on which color is which is certainly complex. iirc studying it is a big part of work on the mind-body problem.
tris_d said:Ok, let me define brightness as captured light on some photo that we open in Photoshop, turn it to gray-scale image, and then those pixels will have some value from 0 to 100, where 0 is black, 100 is white, and in between are shades of gray.
So then having all the photons be the same wavelength, brightness of each pixel will be proportional to the amount of photons that impacted that particular pixel?
I can see how light intensity would be defined by the number of photons, but what does frequency have anything to do with it?
If frequency defines intensity too, would that mean that one blue photon with higher frequency could produce more bright pixel than the other blue photon with lower frequency?
I thought frequency defines color, that frequency is just another face of the wavelength and that they are always in constant relation.
For the macroscopic (bigger than quantum) effects, you are better to stick with the wave concept of light (electromagnetic radiation). In that realm, "brightness" can either mean what a little girl says when she says a yellow balloon is "brighter" than a blue balloon. That is subjective, and of little value to the physicist. Though it may be of extraordinary value to the cognitive neurophysiologist. Let's call it qualitative.tris_d said:And if we swap eyes, maybe you would see it's purple what you previously called green. Well maybe not to that extent, but would it be theoretically possible that we see different colors and just use the same names to describe them? Is that what you mean when you say "agreement on which color is which"?
How do you intend to "Break a photon into three parts"? Before you use the term 'Photon' you should understand what it represents. It is the smallest amount of energy which can be carried by a given frequency of EM. It is not 'made up' of other photons.Simon Bridge said:Breaking the photon into three parts and doing the reconstruction is a handy way to simplify the process and it works very well despite the odd fudge (like "magenta").
OK - I should signal better when I'm not being technically correct.sophiecentaur said:How do you intend to "Break a photon into three parts"? Before you use the term 'Photon' you should understand what it represents.
I actually didn;t need to refer to a particular model for the point I was trying to make.Stick to waves, wavelength, power and all the other classical ideas where they are appropriate here.
Well called - I was too focussed on the point I was trying to make and slipped up elsewhere.The way that our colour vision (three colour analysis) works is pretty well established and 'personal' interpretations can seriously damage the understanding of newcomers to the subject.
Simon Bridge said:OK - I should signal better when I'm not being technically correct.
I actually didn;t need to refer to a particular model for the point I was trying to make.
Well called - I was too focussed on the point I was trying to make and slipped up elsewhere.
I should have talked about the receptor's response to the incoming light (simpler to have the three "signals" that a continuous frequency response) and left it at that. I was trying to convey a sense of the increased simplicity of this method and it does not help to do this if I use a complex and technical-sounding language.
Perhaps you can show me how I could have made the same points better?
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Aside: parametric down conversion of photons is often described by physicists as "splitting a photon in half".
Also see Hübel H. et al. Direct generation of photon triplets using cascaded photon-pair sources Nature 466, 601–603 (29 July 2010)
... for more on how physicists understand "photon splitting". It is just not the process that happens in the eye.
tris_d said:Is intensity and flux not one and the same thing?
sophiecentaur said:You may have noticed that people are getting a bit fed up with your responses. You seem to be desperate to show that you are not wrong in your ideas, rather than willing to take on board new ones.
Do you think all of the other contributors are idiots by putting things in the way they are doing?
It could just possibly be you who could do something about this to resolve it.
"Don't understand" and "won't understand" are separated by a fine line.
tris_d said:1.) The number of photons falling on the film per unit area per unit time per unit solid angle does not depend on the distance between the source and the observer.
Are they talking about intensity? Should not number of photons per unit area per unit time drop off with the square of the distance?
--//--
2.) Thus we distinguish between the brightness of the Sun, which does not depend on distance, and the apparent flux, which does.
Now they say flux depends on the distance, but is flux not the number of photons per unit area per unit time that they just previously said does not depend on the distance?
3.) Brightness is independent of distance. Brightness is the same at the source and at the detector.
I guess this is true if the light source is not point source?
4.) If a source is unresolved, meaning that it is much smaller in angular size than the point-source response of the eye or telescope observing it, its flux density can be measured but its spectral brightness cannot.
What in the world did they just say here?
So when I try to put everything together my interpretation is this: If light source is "resolved", that is when its focused projection has angular size greater than point source, then brightness does NOT fall off with the distance. But when the light source gets so far away that its focused projection covers no more than one pixel on the image, then it becomes "point source" or "unresolved", and then the inverse square law starts to apply in such way that the brightness DOES drop off with the square of any further distance from that point on.
5.) If a source is much larger than the point-source response, its spectral brightness at any position on the source can be measured directly, but its flux density must be calculated by integrating the observed spectral brightnesses over the source solid angle.
What is "spectral brightness" and how is it different to just "brightness"?
Drakkith said:The overall flux from the source DOES fall off with the inverse square. Just not the brightness of the source. It's confusing terminology.
Stars are so small in telescopes that their true size cannot be seen and the appear to be point sources, so we cannot measure how bright they are per unit of angular diameter, only the flux from the whole star.
tris_d said:My ideas? Take on board new ideas? Huh?! What in the world are you talking about? Did you confuse me with someone else?
What are you talking about? Is this twilight zone or something?
Understand what? Resolve what? What are you talking about? What is it you are upset about? Is that link I posted yours? Is this some joke? What is your problem?
sophiecentaur said:No confusion. Just read your responses to the last few replies and then read the contents of them again. They are not wrong.
Drakkith said:Let's look at an example...
tris_d said:Thank you Drakkith, that's great! Let me chew on it.
Just one thing in the mean time, what are the units of light flux?
Drakkith said:Let's look at an example. Let's say that the Sun puts 1,000 photons per second onto a sensor of 100 pixels at the focal point of a telescope here on Earth. Then we take this sensor and move it twice as far away from the Sun as it was. The Sun now puts 250 photons per second onto the sensor. BUT, in both cases, each pixel that receives light receives the same amount of photons per second. The reason that there are 1/4 as many photons hitting the sensor is that the image formed at the focal plain is half the size as before in both the X and Y directions. So the surface area of this image at the focal plane is 1/4 what it used to be and only 1/4 as many pixels are even hit by light from the Sun. So it was originally 1,000 photons over 100 pixels is 10 photons/second/pixel. Now its 250 photons over 25 pixels, which is still 10 photons/second/pixel! This also means that the TOTAL amount of photons falling onto the aperture of the telescope has fallen from 1,000 to 250, so as you can see in both the focused and unfocused case the RADIATIVE FLUX, the photons per second, has decreased to 1/4 just by doubling the distance.
Now, what about brightness? I am not familiar enough to figure out which of the many units (See here) to use, so I will have to explain it my way again instead. Let's say that brightness is the number of photons coming from an angular section of the sky, as that seems to be the only way it makes sense.
Let's say I measured the number of photons per second coming from an area of the Sun that is 15 arcminutes x 15 arcminutes. So the area would be 225 arcminutes. Since the Sun was putting a total 1,000 photons per second onto the sensor, and 15x15 arcminutes is 1/4 the size of the Sun from the Earth (the Sun is 30 arcminutes across), the number of photons per second from this area is 250.
Now, I move the telescope twice as far away. How many photons per second to I now get from this same 15x15 arcminutes? Well, if the Sun has had it's dimensions halved, it is now 15 arcminutes across instead of 30. Which means that the area is 1/4 what it was, which means the whole Sun now fits in this 15x15 area! And if the Sun was putting a TOTAL of 250 photons per second onto our sensor earlier, then it must still be doing the same thing now since we are at the same distance. So even though I've moved twice as far away and the total light from the Sun has decreased to 1/4 what it was, I still have the same amount of light coming from the same angular area of sky. (Note that I've simplified the explanation by using the area of a square, not a circle. However the result is the same.) So the BRIGHTNESS, which I mean as the number of photons per area of sky, is exactly the same. Note that this also happens if we move CLOSER to the Sun. At half the distance to the Sun the light is quadrupled, but the image of the Sun is now 4 times as large! So 4,000 photons, over 400 pixels is still 10 photons/second/pixel!
But what about far away stars? Here we run into an issue. My telescope focuses the light down to a point called an airy disc. Let's say I'm measuring 500 nm light. With an aperture of 250 mm and a focal length of 1,000 mm my telescope will focus 500 nm light down to a spot that is 4.88 microns in diameter. But, what if my star image is even smaller than that? Like, much smaller? Well, in that case we treat the star as a "point source". At this point we cannot measure the brightness of the star, only the total FLUX. If we know the size of the star and it's distance we could calculate the brightness, however we cannot measure it.
tris_d said:Light intensity is number of photons per unit area per unit time?
Light flux is number of photons per unit area per unit time, or what?
Drakkith said:Let's look at an example. . . . .
sophiecentaur said:I am having great difficulty in seeing why you keep bringing photons, pixels and peculiarities of imaging systems into this particular question. I realize the bottom line for a practical astronomer is looking at resolvable, bright enough images but is all this fundamental to the actual question?
You will have read my reservations about the use of photons on account of the variable energy. Watts were good enough for all the original calculations on this stuff.
(You know I have a general aversion to explanations of things that bring in Photons when their actual nature is not specified initially. It is a potentially risky process and the raw results are suspect. Surely it isn't any harder to consider light as a continuum for basic optics than ignoring the fact that a massive object consists of atoms, in the context of mechanics)