## Optics B/C

jkang
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### Re: Optics B/C

Here's an interesting question from my graduate course. Suppose electromagnetic radiation propagates though a neutral dielectric in a region of normal dispersion. The medium has a refractive index n. Let s be the signal that first arrives at point x within the medium. At what velocity does s propagate? Answer in terms of the speed of light in a vacuum, c.
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Tom_MS
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### Re: Optics B/C

jkang wrote:Here's an interesting question from my graduate course. Suppose electromagnetic radiation propagates though a neutral dielectric in a region of normal dispersion. The medium has a refractive index n. Let s be the signal that first arrives at point x within the medium. At what velocity does s propagate? Answer in terms of the speed of light in a vacuum, c.
`By definition of the index of refraciton, n=c/v, where v is the speed of the electromagnetic radiation in the dielectric medium. Therefore, wouldn't s, the speed of the signal, just be c/n?`

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### Re: Optics B/C

Tom_MS wrote:
`By definition of the index of refraciton, n=c/v, where v is the speed of the electromagnetic radiation in the dielectric medium. Therefore, wouldn't s, the speed of the signal, just be c/n?`
Surprisingly and yet unsurprisingly, this is false. I'd recommend thinking more about the physics of the actual situation described, rather than equations and definitions.

Edit: the answer isn't immediately obvious (until you know it, then it is) so I feel like this question deserves some hints. I'll give them in terms of questions to guide thinking.
1) What causes light to slow down in a dielectric medium in the first place?
2) Why does it matter if the medium is neutral?
3) What exactly is a signal?
4) What significance does being the first signal to arrive at x have?

The purpose for my questions is not necessarily to just answer my question, but to give deeper insight into the physics of the problem, and a way to approach physics and the sciences in general. I think Science Olympiad does a good job in introducing students to a wide variety of subjects, but due to its competitive nature can neglect understanding and deeper insight into the concepts. Hope this helps!
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Tom_MS
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### Re: Optics B/C

jkang wrote:
Tom_MS wrote:
`By definition of the index of refraciton, n=c/v, where v is the speed of the electromagnetic radiation in the dielectric medium. Therefore, wouldn't s, the speed of the signal, just be c/n?`
Surprisingly and yet unsurprisingly, this is false. I'd recommend thinking more about the physics of the actual situation described, rather than equations and definitions.

Edit: the answer isn't immediately obvious (until you know it, then it is) so I feel like this question deserves some hints. I'll give them in terms of questions to guide thinking.
1) What causes light to slow down in a dielectric medium in the first place?
2) Why does it matter if the medium is neutral?
3) What exactly is a signal?
4) What significance does being the first signal to arrive at x have?

The purpose for my questions is not necessarily to just answer my question, but to give deeper insight into the physics of the problem, and a way to approach physics and the sciences in general. I think Science Olympiad does a good job in introducing students to a wide variety of subjects, but due to its competitive nature can neglect understanding and deeper insight into the concepts. Hope this helps!
`The first signal S to arrive at that point x produces the smallest relaxation time of any frequency of the light. As frequency increases to this frequency, the relaxation time of the dielectric keeps decreasing and thus the signal propagates faster and faster. At frequencies higher than that of S, the neutral dielectric cannot respond effectively and thus the susceptibility of the medium is zero (n is 1). This means that the signal S at this cutoff point has a refractive index of approximately 1 and thus a speed of c.`

jkang
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### Re: Optics B/C

Tom_MS wrote:
`The first signal S to arrive at that point x produces the smallest relaxation time of any frequency of the light. As frequency increases to this frequency, the relaxation time of the dielectric keeps decreasing and thus the signal propagates faster and faster. At frequencies higher than that of S, the neutral dielectric cannot respond effectively and thus the susceptibility of the medium is zero (n is 1). This means that the signal S at this cutoff point has a refractive index of approximately 1 and thus a speed of c.`
The answer is correct but the reasoning is not. It's unlikely that this will be answered completely correctly anytime soon so I'll let you go.
```I'll first answer each of my "hint" questions.
1) What causes light to slow down in a dielectric medium in the first place?
Light slows down in a medium because excited ions and electrons in the medium emit radiation. These radiation waves superpose with the incident light, and the result acts like light traveling with the reduced velocity [i]v=c/n[/i].
2) Why does it matter if the medium is neutral?
If a medium is neutral, it can be assumed that there are no excited ions and electrons in the medium. Thus the superposition described in (1) physically can't happen immediately.
3) What exactly is a signal?
A signal is in this case a wave carrying information. Obviously by relativity, it has a maximum velocity of [i]c[/i].
4) What significance does being the first signal to arrive at [i]x[/i] have?
Until the first signal, aka information, arrives at [i]x[/i], the ions and electrons there have no idea that light is present in the first place. Without superluminal transfer of information, there is thus no way for these to be excited or emit radiation that could slow down the main light wave. The first signal acts to deliver this information, exciting the electrons and ions.

So in answering these four questions, we arrive at the physics of the problem. The first signal tells the medium that light is there, and the medium in response emits radiation that slows the main signal down. Until the first signal arrives, the medium has no such information - it only makes sense for the first signal to be propagating at [i]c[/i], regardless of whatever [i]n[/i] could be. Thus we find that for light propagating through a dielectric medium, there is a "precursor" to the main signal that travels through the medium, and is what causes the medium to respond to the main signal and slow it down. In reality though, there are actually two precursors. If you guys are curious about these, they are known as the Sommerfeld and Brillouin precursors and you can read about them [url=https://en.wikipedia.org/wiki/Precursor_(physics)]here[/url].```
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Tom_MS
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### Re: Optics B/C

jkang wrote: The answer is correct but the reasoning is not. It's unlikely that this will be answered completely correctly anytime soon so I'll let you go.
```I'll first answer each of my "hint" questions.
1) What causes light to slow down in a dielectric medium in the first place?
Light slows down in a medium because excited ions and electrons in the medium emit radiation. These radiation waves superpose with the incident light, and the result acts like light traveling with the reduced velocity [i]v=c/n[/i].
2) Why does it matter if the medium is neutral?
If a medium is neutral, it can be assumed that there are no excited ions and electrons in the medium. Thus the superposition described in (1) physically can't happen immediately.
3) What exactly is a signal?
A signal is in this case a wave carrying information. Obviously by relativity, it has a maximum velocity of [i]c[/i].
4) What significance does being the first signal to arrive at [i]x[/i] have?
Until the first signal, aka information, arrives at [i]x[/i], the ions and electrons there have no idea that light is present in the first place. Without superluminal transfer of information, there is thus no way for these to be excited or emit radiation that could slow down the main light wave. The first signal acts to deliver this information, exciting the electrons and ions.

So in answering these four questions, we arrive at the physics of the problem. The first signal tells the medium that light is there, and the medium in response emits radiation that slows the main signal down. Until the first signal arrives, the medium has no such information - it only makes sense for the first signal to be propagating at [i]c[/i], regardless of whatever [i]n[/i] could be. Thus we find that for light propagating through a dielectric medium, there is a "precursor" to the main signal that travels through the medium, and is what causes the medium to respond to the main signal and slow it down. In reality though, there are actually two precursors. If you guys are curious about these, they are known as the Sommerfeld and Brillouin precursors and you can read about them [url=https://en.wikipedia.org/wiki/Precursor_(physics)]here[/url].```
Okay thanks. Now I have one. The image below shows the absorption coefficient of water as a function of photon wavelength. Now, imagine sunlight hitting the surface of a lake. Determine the ratio between intensities of red light of wavelength 670 nm and blue light of wavelength 475 nm 2 meters below the surface.

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### Re: Optics B/C

Are you guys all seeing the image? If not, it should be the second image here: https://en.wikipedia.org/wiki/Electroma ... n_by_water

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### Re: Optics B/C

jkang wrote:
Tom_MS wrote:
`The first signal S to arrive at that point x produces the smallest relaxation time of any frequency of the light. As frequency increases to this frequency, the relaxation time of the dielectric keeps decreasing and thus the signal propagates faster and faster. At frequencies higher than that of S, the neutral dielectric cannot respond effectively and thus the susceptibility of the medium is zero (n is 1). This means that the signal S at this cutoff point has a refractive index of approximately 1 and thus a speed of c.`
The answer is correct but the reasoning is not. It's unlikely that this will be answered completely correctly anytime soon so I'll let you go.
```I'll first answer each of my "hint" questions.
1) What causes light to slow down in a dielectric medium in the first place?
Light slows down in a medium because excited ions and electrons in the medium emit radiation. These radiation waves superpose with the incident light, and the result acts like light traveling with the reduced velocity [i]v=c/n[/i].
2) Why does it matter if the medium is neutral?
If a medium is neutral, it can be assumed that there are no excited ions and electrons in the medium. Thus the superposition described in (1) physically can't happen immediately.
3) What exactly is a signal?
A signal is in this case a wave carrying information. Obviously by relativity, it has a maximum velocity of [i]c[/i].
4) What significance does being the first signal to arrive at [i]x[/i] have?
Until the first signal, aka information, arrives at [i]x[/i], the ions and electrons there have no idea that light is present in the first place. Without superluminal transfer of information, there is thus no way for these to be excited or emit radiation that could slow down the main light wave. The first signal acts to deliver this information, exciting the electrons and ions.

So in answering these four questions, we arrive at the physics of the problem. The first signal tells the medium that light is there, and the medium in response emits radiation that slows the main signal down. Until the first signal arrives, the medium has no such information - it only makes sense for the first signal to be propagating at [i]c[/i], regardless of whatever [i]n[/i] could be. Thus we find that for light propagating through a dielectric medium, there is a "precursor" to the main signal that travels through the medium, and is what causes the medium to respond to the main signal and slow it down. In reality though, there are actually two precursors. If you guys are curious about these, they are known as the Sommerfeld and Brillouin precursors and you can read about them [url=https://en.wikipedia.org/wiki/Precursor_(physics)]here[/url].```
Is that only for pulsed light (impulses), or is it true for monochromatic continuous wave light as well? You're original question did not specify. An impulse (or pulse) has many frequencies in it.

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