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$F=ma$ Don't do that! You can't mix Newtonian mechanics and special relativity, let alone Newtonian mechanics and general relativity. Gravitation is fundamentally very different between Newtonian mechanics and general relativity. In general relativity, gravitation is a result of geometry. It is not quite a force. Mass-energy tells space-time how to ...

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The graph below shows the given data as red dots, with the angle in air as the x-axis, and the angle in water as the y-axis. The blue line is a least squares fit of the data to a line, and the purple line is a least squares fit to a parabola. The data clearly doesn't quite fit a line, but it does fit very well to a parabola. The parabola shown, calculated ...

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Refractive index describes the speed of propagation of light in a medium. So to restate your question: why is the speed of light slower in some media than in others? The wave equation tells us that speed of propagation depends on two factors: one is an inertial term, while the other is an elastic term. Let's look at a simple case of a string. The ...

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This is a good question that I struggled with myself for some time. I believe that the correct answer is the following. The imaginary part of the index of refraction, i.e. $\kappa$, quantifies the dissipation of light through a medium. However, if one wants to quantify the dissipation due to nonretarded electric fields alone, the quantity that quantifies ...

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Electromagnetic radiation in a medium propagates according to the law $$\mathbf E,\mathbf B \propto e^{\imath(\pm k_xx-\omega t)}$$ where $$k_x^2 = \frac{n^2\omega^2}{c^2}\;.$$ The refractive index $n$ can also be complex, in which case its imaginary part describes the absorption of the EM wave in the medium. But the oscillating part is in any case $$... -1 This is basically about dispersion: EM waves with different frequencies travel at different speed in a medium because of interaction. Usually, as in a standard textbook experiment wherein a light beam is bent by a prism, higher frequency waves bends more. The more it bends, the more slowly it travels. So, lower frequency waves usually go faster. However, ... 5 It is so possible that it actually happens on Earth, where I suppose we are both breathing. I think you should read the mirage page of Wikipedia (where I learned all I am saying here) as this can actually happen on our planet. It is called a superior mirage, and happen when you have temperature inversion, with a ground colder than the air above. The ... 2 After a lot more searching, I have found the answer to my question! :D Below is a summary of the information I found. There is no specific webpage I can link to because I relied on sources who quoted other sources which no longer exist, but maybe this information can be useful to someone else someday. Most of what I learned comes from Professor Lou ... 4 Of the choices given, I would favor explanation #2. It doesn't require quantum physics; modeling atoms as ball-and-spring systems works pretty well. In his famous textbook for undergraduates Griffiths does this, and if you have some math training that would be a fine place to head for the details. I think #5/#6 are also, arguably, correct if you treat the ... -1 In fiber optic measurement the effective refractive index is calculated as follow: RI eff=(L opt* RI opt)/L eff where: L eff is the cable or physical distance between two know even on the OTDR L opt is the optical distance between two known events 3 It's because most materials have (many) natural resonances. I assume you are alright with the phase velocity being different in different media, that is I assume you are alright with something of the form$$ \frac{ \omega }{ k } = \frac{c}{n} = \frac{ c}{\sqrt{ \mu \epsilon }} \sim \frac{c}{\sqrt \epsilon}  where $k$ is the wavenumber of a wave, ...

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Just blindly multiplying the overall answer by some factor isn't the way to go about it. I have an alternative proposal which may work well as a zeroth-order approximation at least. You already have the expression for the 2-layer case, and if I observe correctly, you are only concerned with the reflected part, not the transmitted part. So, a smaller ...

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