Is there any relation between wavelength and brightness?

In summary: I used the "middle gray" of the smoke and compared it to the other colors of the rainbow to determine how many bits the digitizer was using for color resolution. In summary, The brightness of a color is determined by the sensitivity of the optical equipment used to measure it, as well as the colors and context surrounding it. Brightness is subjective and can vary depending on individual perception. Intensity is the amount of energy passing through a specified area in a specified amount of time and can be affected by frequency and the number of photons. However, brightness is not directly proportional to the number of photons. Instead, it is influenced by the way our eyes perceive color and the technology used to capture and display it.
  • #1
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?
 
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  • #2
tris_d said:
What makes, for example, blue color appear darker or lighter?
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.
How bright a color appears also depends on the colors around it and the context you are looking at it in.
 
  • #3
"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.

One photon of blue light has more energy than one photon of red light, but you can have a red light source which is more intense than a blue light source.

"Brightness" as we experience it with our eyes depends on how our eyes work. We can only detect a small slice of the electromagnetic spectrum, and we sense different frequencies of light in different ways.
 
  • #4
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?

Be aware that what you are categorizing as bright appears to be based on what you perceive with your eyes. Do you think this is a good detector?

Zz.
 
  • #5
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.

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?

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.

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.
 
  • #7
ZapperZ said:
I suggest you read this for a discussion of "brightness" when applied to optics.

http://www.rp-photonics.com/brightness.html

Zz.

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.
 
  • #8
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.

You could have done a quick search and come up with the same thing.

https://www.physicsforums.com/blog.php?b=3588 [Broken]

Zz.
 
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  • #9
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?

Turning some image to gray scale is just removing information. If we are talking digital photography, then the detector responds to certain frequency ranges in such a way to add a certain number of red, green or blue bits to your image. In some ways traditional photographic emulsion using chemical excitation is more faithful to the original scene.

Still, brightness is in the eye of the beholder, even if it is an electronic eye.

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.

Frequency and color are related, but what you perceive is not frequency. You have three different optical receptors in your retina. A combination of frequencies can cause the same sensation as one frequency. The colors of the rainbow are "honest", frequency-related colors. The colors of your TV screen are "playing tricks on your eyes". They are causing your receptors to fire, due to the amount of red, green or blue sent from each pixel.

I actually spent some time trying to figure out what certain brightnesses indicated in relation to some video shot on 9/11/01 of the World Trade Center Building #2. There was an outflow of brightly glowing molten material which I wanted to determine the temperature of.

What I found is that it's a very tricky problem. Many of the pixels are saturated, which means that the information is truncated. 100% saturated pixels only tell us the minimum intensity. They are silent as to how far above the minimum the original scene was.

This isn't one of the images I worked with, but it is an example:

molten-metal-pouring.jpg


Comparing the black body radiation of the surrounding flames (which were not saturated) gave me some means of calibration.

Second Law of Thermodynamics: heat don't flow uphill.

You might be interested in the theory of black-body radiation. It has a lot to do with frequency and "brightness". It is also instructive if you are interested in photography, and someone tells you the Kelvin temperature equivalence of your light source.

It can be a lot of fun to play with RGB (red, green, blue) combinations, and their anti-combinations (yellow, magenta and cyan).
 
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  • #10
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.

Let me use a real world example that I commonly deal with when using my CCD camera for astrophotography. My telescope focuses light down onto the CCD sensor. The charge on each pixel is built up during exposure by photons being absorbed by the pixels. Each pixel can hold a certain number of electrons before reaching "saturation". The CCD's charge amplifier reads the amount of charge during readout and converts that into a voltage signal. This voltage signal, being 16 bits, is capable of being represented by 65,535 different numbers. Some CCD pixels can hold, say 55,000 electrons, while others can hold around 80,000.

The actual number is generally different for different sensors and doesn't matter that much really. It only matters that these electrons are converted into a 16-bit signal which is then displayed as a certain color or grayscale pixel on your screen. Since we can't detect a difference of 1/65,535 in brightness with our eye, we have to "stretch" the values upon display, which only means that we make our high and low points, IE the value of each pixel that represents black or white, different. So I could bring my white point down to 5,000 if I have a very low light picture to make it possible to even see anything. If I just left the white point at 65,535 the whole picture would just look black.

So let's say a pixel is readout and the charge is converted into a signal that measures as 65,000. What does this tell us about the actual light itself? Can it tell us the wavelength of the light? No. It can only tell us how many electrons were built up in each pixel.

So how do we get nice color pictures? We use filters and we take multiple exposures. OR we use a single exposure, but every single pixel has it's own filter in front of it of either red, blue, or green. This is known as a Bayer Array. So, if your software doesn't know you are using a color CCD sensor with a bayer array, it will simply display your image as a grayscale monochrome image. I personally use a monochrome CCD, which means that my sensor doesn't have a bayer array. Instead I use a separate filter wheel with RGB or other filters to get my different color frames and then combine them into a color image.
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?

Nope. ALL photons, of ANY frequency with enough energy to excite an electron will be able to contribute to the pixel's final value. This is why filters are important. We reject the wavelengths that we DON'T want to see. An unfiltered CCD typically has a range of wavelengths that it responds to, with light of around 1,000 nm being the lowest energy capable of exciting an electron, to 300 nm being the highest energy light that it can respond to. Higher energy light is usually absorbed in in the small features of the CCD pixels before it can reach the photosensitive layer.
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?

One more thing on frequency. While CCD's have a range of wavelengths they respond too, they do not respond to all of these wavelengths equally well. See page 5 of the following link: http://www.stargazing.net/david/QSI/KAI-2020LongSpec.pdf

The graph in the lower left of page 5 represents the Quantum Efficiency of the chip. The QE is the percent of light that reaches the CCD that will end up being converted to photoelectrons. The graph is labeled from 0.0-0.6, with 0.6 being 60% efficiency. As you can see, the graph peaks in the 450 nm range, which is the blue-green region, for a monochrome CCD. The colored lines represent the 3 different color filters of a bayer filter that come with the color chips. So even light of the best wavelength for this particular CCD is only being read with a 56-57% efficiency, and much of the spectrum is far worse.

Not that this CCD is bad, this is actually a very good CCD efficiency for the price range. For comparison, your average QE for photographic film is well under 10%. The human eye is far harder to quantify a QE for, as vision is not just a mechanical process of detecting light and turning it into a value, but an extremely complicated process involving multiple receptors, timing of these receptors firing, and dozens if not hundreds of other things. Personally I would venture a guess and say that the QE of normal color vision in the daytime is generally less than 1%. But don't quote me on that.

I thought frequency defines color, that frequency is just another face of the wavelength and that they are always in constant relation.

You are correct.
 
  • #11
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.

Frequency and wavelength are related by
c = fλ where c is the wave speed.

A monochromatic wave will excite the three colour receptors in the eye and produce three signal values. The brain interprets this and assigns the combination a name which we call the colour. We are aware of vastly more colours than the ones corresponding to spectrum. The eye is easily fooled into believing that the result of excitation with several frequencies is the same as a spectral colour. Otherwise colour TV and printing wouldn't work.
 
  • #12
Oh man! This is not simple. -- Thank you all, I'm chewing on it.
 
  • #13
tris_d said:
Oh man! This is not simple. -- Thank you all, I'm chewing on it.

Unless you are working simple idealized problems, nothing is ever simple!
 
  • #14
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.
 
  • #15
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.

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"?
 
  • #16
It is logically possible that we are giving the same label to different conscious experiences of light with the same physical properties - if I had to bet I'd say it is certain. How would I find out if my experience of what we both label "blue" is the same as, or similar to, yours? Does it actually matter? ... where differences are important is where they mess with communication eg. color-blindness. There are whole industries (eg fashion) with a basis in the subjective experience of colors - what color "goes with" which and so on. Our aesthetic sense is important for our reproductive chances even.

The range of phenomena is a whole area of study in itself and probably won't be resolved without a decent model for how consciousness works.

In physics, the word "color" ("colour") has a specialized use ... like "work" and "force".
It is usually used as a shorthand for the type of light - particularly when it comes from, or close to, the visible spectrum (we don't normally talk about the color of a gamma ray). Photons are properly characterized by their energy or momentum - which are related to a characteristic frequency. As you advance through your course you'll find yourself using "color" less and less in physics ... then you learn about quarks.
 
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  • #17
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.

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"?
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.

The idea of brightness can also have a more quantitative meaning. That is, how much energy is passing through some (conceptual) surface area in a specified amount of time. That's basically "power", or energy per time.

It boils down to how "hot" the source is. Except some sources can cheat.

Yes, it's complicated. Can you tell us more about your working domain so that we can better address your rather abstract question?

Not that abstraction is a bad thing. Physics and mathematics wouldn't exist without it.
 
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  • #18
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").
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.
I am aware how much people are attracted to the idea of giving explanations in terms of photons but this is a great example where it is not appropriate and the explanation just falls on its face.
Stick to waves, wavelength, power and all the other classical ideas where they are appropriate here. They are moire than adequate for this sort of discussion. Avoid Photons until you have a proper idea of what they are considered to be by Physicists.
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.
 
  • #19
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.
OK - I should signal better when I'm not being technically correct.

Stick to waves, wavelength, power and all the other classical ideas where they are appropriate here.
I actually didn;t need to refer to a particular model for the point I was trying to make.

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.
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?---------------

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.
 
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  • #20
http://www.cv.nrao.edu/course/astr534/Brightness.html

This above is a link to internet article that talks exactly about the things I want to know, it's just that some parts of it do not seem to quite fit with what I can read everywhere else. Here are some statements that do not seem to compare:

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?

Wikipedia says:
http://en.wikipedia.org/wiki/Apparent_brightness
- Note that brightness varies with distance; an extremely bright object may appear quite dim, if it is far away. Brightness varies inversely with the square of the distance.

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"?
--//--

6.) The specific intensity or brightness is an intrinsic property of a source, while the flux density of a source also depends on the distance between the source and the observer.

How can intensity and brightness be intrinsic property of a source? Is intensity and flux not one and the same thing? -- I'd say intensity is a property of emitted light rather than a property of a light source, and that brightness is a property of an image, rather than property of either emitted light or light source itself.
 
  • #21
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?


---------------

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.

I did sound a bit more grumpy than I meant to be - sorry.
But you have taken my point about not introducing more technical terms than necesssary - especially when it's a bit tenuous. 'Splitting photons' is not what it sounds like and it is nothing to do with the way three separate analysis filters work. If you must introduce the photon at this stage, you could just say that each sensor detects photons with a different range of energies. But I don't see how that improves on the term 'wavelength response'.
 
  • #22
tris_d said:
Is intensity and flux not one and the same thing?

No, they are not. Look at two lightbulbs, side by side. They have equal intensities. Turn one of them off and the remaining one has the same intensity as before but the flux reaching you has dropped to half. Move the two bulbs (both on) together or apart (not too far, or the geometry may change) and the flux from them is the same. However, if you were to superimpose the two (having two similar filaments in the same frosted envelope) then the intensity would double but the flux would be the same as having the two side by side.

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.
 
  • #23
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.

My ideas? Take on board new ideas? Huh?! What in the world are you talking about? Did you confuse me with someone else?
Do you think all of the other contributors are idiots by putting things in the way they are doing?

What are you talking about? Is this twilight zone or something?
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.

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?
 
  • #24
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?

I already explained this. The brightness of the Sun per unit of angular diameter does not change. Only the overall intensity, or flux does since the Sun get's smaller as you recede from it.

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?

No real light source is a true 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?

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.

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.

The overall flux from the source DOES fall off with the inverse square. Just not the brightness of the source. It's confusing terminology.


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"?

I think it just means you have to measure the flux from the whole image of the source, which can be quite large if it isn't a point source.
 
  • #25
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.

It sure is confusing. Thanks. Some more clarification please.

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?

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.

How flux relates to brightness?
 
  • #26
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?

No confusion. Just read your responses to the last few replies and then read the contents of them again. They are not wrong.
 
  • #27
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.

#12
Oh man! This is not simple. -- Thank you all, I'm chewing on it.
Something wrong with this post?

#15
And if we swap eyes, maybe you would see it's purple what you previously called green.
Something wrong with this post?

#20
This above is a link to internet article that talks exactly about the things I want to know..
Something wrong with this post?



By the way, can you answer this:

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?
 
  • #28
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 RADIANT 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.
 
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  • #29
Drakkith said:
Let's look at an example...

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?
 
  • #30
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?

If you mean Radiant Flux, that would be Watts. Hopefully I haven't butchered optical terminology too badly in my example lol.
 
  • #31
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.

Great. I think you could/should make an internet article out of this stuff. It's fairly simple, but not very obvious if you don't consider there is a lens in between and that focusing is involved, to which I was completely oblivious previously.
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!

Would you agree that instead of defining brightness as a measure of "number of photons per area in the sky" would be better to say it is a measure of "number of photons per pixel", so that brightness be a property of an image rather than property of the light itself?
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.

Are most of the stars in our galaxy point light sources? Are most of the other galaxies point light sources? -- When you say "500 nm light" you refer to angular size of "airy disc"? I guess the size of airy disc depends on magnification, so how do you know it's 500nm and not 200nm, that is how do you know you focused it properly (if this question makes sense)? -- How do we measure the flux? You say we can not measure the brightness in case of point light sources, but would not image itself of such point light source be a measure of its brightness?
 
  • #32
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?

If you insist on using photons in your arguments the everything changes with frequency and that just doesn't help anyone. That's the whole point of using Watts - it works for any combination of wavelengths. Your attempted definition will not work for a mixture of frequencies.
Why bring photons into this at all? All this stuff was done and dusted before anyone came up with E=hf
 
  • #33
Drakkith said:
Let's look at an example. . . . .

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)
 
  • #34
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)

I'm interested in brightness, and I believe we established brightness is a property of an image rather than property of the light source or light itself. Therefore, since the image is digital object rather than analog, since it is a collection of discrete pixels, I think we also need to quantize the light so we can talk about the relation between the image and the light, and then we could perhaps define brightness as the number of photons per pixel.

That's where my original question came from. I was wondering if there is anything else beside the number of photons per pixel, like wavelength, that would define the brightness of a pixel.
 
  • #35
Why wouldn't it be fundamental to the question? I showed that before and after the optical system the results are the same. I am merely putting an optical system in because I am far more familiar with the workings of optics, sensors, and photons than I am waves and energy per area and such. I know we can't treat light as little particles traveling through space, but unless I've gravely misunderstood something I think my explanation works out the same either way. Please correct me if I'm wrong on something.
 
<h2>1. What is the relationship between wavelength and brightness?</h2><p>The relationship between wavelength and brightness is known as Wien's displacement law. It states that the peak wavelength of radiation emitted by an object is inversely proportional to its temperature. This means that as the wavelength decreases, the brightness increases.</p><h2>2. Why does the brightness increase as the wavelength decreases?</h2><p>This is because shorter wavelengths have higher energy levels, which means they carry more energy. As a result, when an object emits radiation at shorter wavelengths, it appears brighter to us.</p><h2>3. Is there a specific wavelength that is the brightest?</h2><p>Yes, the peak wavelength of an object is the brightest. This is known as the peak emission wavelength and it can be calculated using Wien's displacement law.</p><h2>4. Can the relationship between wavelength and brightness be seen in everyday life?</h2><p>Yes, the relationship between wavelength and brightness can be seen in everyday life. For example, the sun appears brightest at shorter wavelengths, such as blue and violet, while objects like fire emit radiation at longer wavelengths, appearing less bright to us.</p><h2>5. How does the relationship between wavelength and brightness relate to the color of light?</h2><p>The color of light is determined by its wavelength. As mentioned before, shorter wavelengths appear bluer and brighter, while longer wavelengths appear redder and less bright. This is why objects with higher temperatures, such as the sun, emit bluer and brighter light, while cooler objects emit redder and less bright light.</p>

1. What is the relationship between wavelength and brightness?

The relationship between wavelength and brightness is known as Wien's displacement law. It states that the peak wavelength of radiation emitted by an object is inversely proportional to its temperature. This means that as the wavelength decreases, the brightness increases.

2. Why does the brightness increase as the wavelength decreases?

This is because shorter wavelengths have higher energy levels, which means they carry more energy. As a result, when an object emits radiation at shorter wavelengths, it appears brighter to us.

3. Is there a specific wavelength that is the brightest?

Yes, the peak wavelength of an object is the brightest. This is known as the peak emission wavelength and it can be calculated using Wien's displacement law.

4. Can the relationship between wavelength and brightness be seen in everyday life?

Yes, the relationship between wavelength and brightness can be seen in everyday life. For example, the sun appears brightest at shorter wavelengths, such as blue and violet, while objects like fire emit radiation at longer wavelengths, appearing less bright to us.

5. How does the relationship between wavelength and brightness relate to the color of light?

The color of light is determined by its wavelength. As mentioned before, shorter wavelengths appear bluer and brighter, while longer wavelengths appear redder and less bright. This is why objects with higher temperatures, such as the sun, emit bluer and brighter light, while cooler objects emit redder and less bright light.

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