Thread: Questions

Hi. I'll gather in this topic questions about infomation "self studied" so they are likely to be... novice. Be patient

Let's begin

1) I read a lot about dispersion but I still can't answer a vey basic question : why does a prism separate colors while a normal glass (e.g. my window) doesn't? What is the main difference between the two?

2) About black body radiation. Apparently, when heated, a black body's glow color changes from red to yellow to white to blue. Is this sequence only for heated black body or for any heated body? (cause one example he gives is lava). I was also wondering why the green isn't present.

That's for now. Thanks everybody.

Originally Posted by Stranger

1) I read a lot about dispersion but I still can't answer a vey basic question : why does a prism separate colors while a normal glass (e.g. my window) doesn't? What is the main difference between the two?

Because in your window the edges of the glass are parallel. Refraction occurs when light moves into a medium where its speed is different. The sine of the angle of refraction is equal to the sine of the angle of incidence times the ratio of velocities. The catch is that the speed of light in media is frequency-dependent, with higher-frequency light having higher index of refraction.

In a prism or a plane of glass light is refracted twice: Once upon entering and once upon leaving. In a plane glass this results in a shifting of the beam (I've tried to illustrate that here), while in a prism, the geometry of the situation means that the beam is further bent (illustration here).

So in the window, the blue light is shifted maybe 10 nm further than the red light, but since the eye has a resolution about two orders of magnitude poorer, you don't see that. Furthermore, the light moves in the same direction after passing through the glass as it did before passing, irrespective of wavelength.

In the prism, OTOH, the light is bent twice in the same direction, meaning that light of different wavelenghts moves in different directions after passing through the prism.

2) About black body radiation. Apparently, when heated, a black body's glow color changes from red to yellow to white to blue. Is this sequence only for heated black body or for any heated body?

[...]

I was also wondering why the green isn't present.

Any heated body. A 'black body' is one that absorbs all light incident on it. Any heated body emits thermal radiation in a Planck distribution.

A body that absorbs only a fraction of all light incident upon it emits only a fraction of the light a black body would emit, but it doesn't (IIRC) change the distribution.

The sequence you mention is simply the sequence you get if you go from lowest to highest energy per photon in the visible spectrum. Red light is the least energetic, then yellow, green, and blue.

The Planck distribution has a single peak at a temperature-dependent energy, so as the temperature climbs, the peak moves from infrared through red, yellow, green, and blue into UV.

The reason you don't see green is that green is roughly in the middle of the electromagnetic spectrum, and the peak has a finite width, which means that when the peak is at green, the entire visible spectrum is represented more or less equally, and our eye interprets that as white.

Ah, that must be why I never noticed green stars before The link about Planck is great. No math equations (which I automatically ignore). It's not that I'm bad in math, but I'm only concerned with the physical meaning.

And this prism stuff, I had asked before in a another site but didn't understand what he meant.

Thanks. See you later

Originally Posted by JS

A body that absorbs only a fraction of all light incident upon it emits only a fraction of the light a black body would emit, but it doesn't (IIRC) change the distribution.

Surely you are mistaken, else the facility for colour vision would be redundant.

Originally Posted by Ophiolite

Originally Posted by JS

A body that absorbs only a fraction of all light incident upon it emits only a fraction of the light a black body would emit, but it doesn't (IIRC) change the distribution.

Surely you are mistaken, else the facility for colour vision would be redundant.

Sorry Ophiolite, everything JS says look good to me and I don't understand your objection.

I am probably being hugely simplistic in my understanding - where hugely simplistic is a euphimism for wrong.
If an object appears red this is because it preferentially absorbs other wavelengths, but emits red. When you explain how fouled up my thinking is, please be gentle.

Perhaps the difficulty here was that JS was restricting his consideration to black body radiation only, while in your objection you were thinking about the absorbsion and emission of light in general.

The black body radiation is based on temperature only which means it is based on the absorbsion and emission of energy stored only as molecular motion (i.e. heat). An object emitting black body radiation in the red visible light is what we call "red hot".

The other kind of absorbtion and emission of light is a complex thing involving both chemistry and physics. There is the preferential scattering, interference, transmission, reflection and emission of different wavelengths involved. The process of florescence is the image that I have always had in my head as the reason for a materials color, but now I suspect that scattering may be a more significant factor.

The florescence explaination of the the color of an opaque object is due what kind of electron energy jumps the atoms of the material have in the visible range. This means that a red object both absorbs and emits well in red. Other frequencies in the visible range will be absorbed only partially by the atoms. So that a green photon will interact with an electron of these atoms causing the electron to absorb the energy equivalent of a red photon leaving a photon that is so reduced in energy that it is probably not in the visible range anymore. In fact it will probably be in the infra-red range where it can be absorbed by the whole molecule as heat motion. However the only photons emitted by these atoms in the visible range as the electrons return to their ground state will be red photons.

The color of a transparent material will be more like what you say, in that it is transparent (transmission as opposed to scattering, absorbsion and interference) in a certain color range.

Originally Posted by mitchellmckain

Perhaps the difficulty here was that JS was restricting his consideration to black body radiation only, while in your objection you were thinking about the absorbsion and emission of light in general.

What you said later in your post made sense Mitchell, however, JS said this:
A body that absorbs only a fraction of all light incident upon it emits only a fraction of the light a black body would emit, but it doesn't (IIRC) change the distribution.
Here he appears, to my untutored eye, to be saying that he is not now talking about black bodies and black body radiation. He has therefore descended, as it were, into the mundane world of everyday objects where the distribution of emitted lightwave frequencies will definitely be effected.
I do understand that in the case of perfetc black bodies what he stated would be true. I do still fail to understand, however, how it can be correct, as he has stated it.

Originally Posted by Ophiolite

If an object appears red this is because it preferentially absorbs other wavelengths, but emits red.

Then if we shine a green object with red light, it should appear black? :?

Cause, if I understand what you say, it absorbed wavelenghts in the white light and emited the green. Fine, if the light only have red wavelenghts, then the object should appear black cause all light is absorbed

Please note the use of the word preferentially.

So? It'll apear darker? I'm not sure I understand.

Originally Posted by Stranger

Originally Posted by Ophiolite

If an object appears red this is because it preferentially absorbs other wavelengths, but emits red.

Then if we shine a green object with red light, it should appear black? :?

Cause, if I understand what you say, it absorbed wavelenghts in the white light and emited the green. Fine, if the light only have red wavelenghts, then the object should appear black cause all light is absorbed

This in fact depends on the material and which of the processes are producing the red color. If it realy looks red under a pure green light then the material must be florescing. If the main reason for the red color is a preferencial scattering and/or reflection of red light then the material would appear black under the pure green light.

P.S. Just because a light looks green doesn't mean it is a source of pure green light.

P.S.S. Apparently I switched red and green in this answer to Stranger probably because the original discussion concerned a red object. Hopefully my point is understandable regardless.

hi there, another beginner question:

3) When talking about Plank's interpretation of black body radiation, the article says:

"This has the effect of reducing the number of possible modes with a given energy at high frequencies in the cavity"

Ok, what is a "mode"? I searched many articles but I still don't know what it actually is... Is it simply a wave?

Thanks everybody.

It seems to me it could mean several things. I would need to see the context.

Sorry I didn't reply earlier,

The "mode" is mentioned in wikipedia when talking about ultraviolet catastrophe and hyperphysics when interpreting blackbody radiation.

http://en.wikipedia.org/wiki/Ultraviolet_catastrophe
http://hyperphysics.phy-astr.gsu.edu/hbase/mod6.html

I searched in wikipedia, found an article "normal mode" and, if I understand well, then "mode" means, in this context, wave.

You misunderstand, I don't need a definition of mode. I need to see the whole paragraph or page of the text in which the original statement you discuss was found.

But in regard to the ultraviolet catastrophe, the words wavelength or frequency is usually used instead of the word mode these days.

Yes, I understand, this is why I gave you the links

Here is the pararaph

Planck had postulated that electromagnetic energy did not follow the classical description, but could only oscillate or be emitted in discrete packets of energy proportional to the frequency (as given by Planck's law). This has the effect of reducing the number of possible modes with a given energy at high frequencies in the cavity described above, and thus the average energy at those frequencies by application of the equipartition theorem. The radiated power eventually goes to zero at infinite frequencies, and the total predicted power is finite.

and

"Blackbody radiation" or "cavity radiation" refers to an object or system which absorbs all radiation incident upon it and re-radiates energy which is characteristic of this radiating system only, not dependent upon the type of radiation which is incident upon it. The radiated energy can be considered to be produced by standing wave or resonant modes of the cavity which is radiating.
The amount of radiation emitted in a given frequency range should be proportional to the number of modes in that range. The best of classical physics suggested that all modes had an equal chance of being produced, and that the number of modes went up proportional to the square of the frequency. But the predicted continual increase in radiated energy with frequency (dubbed the "ultraviolet catastrophe") did not happen. Nature knew better.

Originally Posted by Stranger

hi there, another beginner question:

3) When talking about Plank's interpretation of black body radiation, the article says:

"This has the effect of reducing the number of possible modes with a given energy at high frequencies in the cavity"

Ok, what is a "mode"? I searched many articles but I still don't know what it actually is... Is it simply a wave?

Thanks everybody.

Ok yes as I said before, these days we would say that it reduces the number of frequencies with a given energy in the cavity. So Plank is using the word "modes" as a synonym for frequencies. In a standing wave modes and frequencies are essentially the same thing.

The point is that when you use the classically derived Rayleigh law for the relative density of different frequencies of radiation in a cavity (or radiated from a blackbody), and then use integration to add up the energy contributed from all possible frequencies then you get infinite energy. Plank came up with a formula for this same density, which does not give infinite energy and then explained how this formula could be derived. Plank showed that if waves of a certain frequency could only have energies which are multiples of Planck's constant h = 6.626e-34 Joule-seconds times that frequency, then you could derive his energy density.

The statement in question is equivalent, for if the number of energies for a given frequency is limited to multiples of h times the frequency then the number of frequencies (or "modes") for a given energy is also limited by the same requirement. That is, given an energy E the frequencies are restricted to E/h, E/(2 h), E/(3 h), ....

Ok, I understand, the no. of frequencies decreases because it's limited to E/nh.

mmm. ok, but what about this

Probability of occupying modes:
Classical --> equal for all modes.
Planck --> quantized modes: require hf energy to excite upper modes, less probable.

and

This has the effect of reducing the number of possible modes with a given energy at high frequencies in the cavity described above

Why high frequencies? I mean, the no. frequencies is reduced, but why is high frequency less probable?

Originally Posted by Stranger

Why high frequencies? I mean, the no. frequencies is reduced, but why is high frequency less probable?

The answer is right in front of you. Look at the sequence I wrote down

E/h, E/(2 h), E/(3 h), ....

The first one is the highest frequency. This is an infinite sequence so there are an infinite number of frequencies but the farther you go in the sequence the lower the frequency. So picking any cutoff frequency, like E/(100 h) for example, there is a finite number of frequencies higher than that cutoff (99 in our example) but an infinite number of frequencies lower that cutoff frequency.

uh huh, I think I understand.

In the curve between radiated intensity and frequency,

1) It begins low because low frequency gives low radiated intensity.

2) It ends low because the no. of high frequencies is reduced.

3) The middle is high because it has an average frequency but the no. is not as reduced.

Is this right?

Originally Posted by JS

Originally Posted by Stranger

1) I read a lot about dispersion but I still can't answer a vey basic question : why does a prism separate colors while a normal glass (e.g. my window) doesn't? What is the main difference between the two?

Because in your window the edges of the glass are parallel. Refraction occurs when light moves into a medium where its speed is different. The sine of the angle of refraction is equal to the sine of the angle of incidence times the ratio of velocities. The catch is that the speed of light in media is frequency-dependent, with higher-frequency light having higher index of refraction.

JS is right. if you took regular window glass and made a prism out of it, you would get the same result as a regular prism. prism glass isnt made any diffrently besides the shape.

hello everybody,

I had a question about particles: what happens to a free electron after it spends all its energy (e.g. in photoelectric effect, the photoelectron uses some of its E. (binding E.) to get away from +ve nucleus and spends the rest ionizing surrounding atoms). Then what happens? is a particle with no energy possible?


Thanks

The first question can be simply answered.The window frame has two faces parallel ,when light enters it, without any refraction it goes out. But in case if a prism ,it has three sides and each angle of 45drgree. so it refracts the white ray and splits it in its constituent colours.