# Tag Info

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Refractive index of a medium depends upon the refractive index of the surroundings (when you consider the light ray passing from the surrounding into the medium), optical density, wavelength of the light and temperature. Yeah it does depend on the the wavelength of light, because: As wavelength of light decreases, the velocity of light decreases. Now, we ...

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Yes it depends on wave length it is inversely proportional to it. Let $u$ denote the refractive index, then: $$u = c/v$$ where $c$ in the speed of light in vacuum and $v$ is the speed in a medium. And: $$v=fl$$ where $f$ denotes the frequency and $l$ the wavelength. So we get: $$\frac{U_1}{U_2} = \frac{l_2}{l_1}$$

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Two comments: First, you see a mirror surface on a lake at shallow angles, true, but only because there is light in the atmosphere ( $n_1$ in your equations). For your problem, assume there is no such light source, so only light emanating from the water ($n_2$) is of concern. Next, essentially you are correct that the limiting ray angle exiting the water ...

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Actually, total internal reflection is impossible in spherical raindrops; an internal reflection occurs, but it is not total. And the rainbow is not just an arc, it is a full disk. The disks are sized differently for each color, and are brightest at the very edge, which is what makes the colored arcs appear. For the primary rainbow, these disks are centered ...

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Here's the logic (well a particular rendition): Recall that $n$ is defined as the ratio of the speed of light $c$ in vacuum to the speed of light $v$ in the given medium; \begin{align} n = \frac{c}{v} \end{align} Note that in a linear medium, Maxwell's equations are exactly the same as in vacuum, except $\mu_0$ and $\epsilon_0$ are replaced by $\mu$ and ...

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Light changes direction when passing through a prism because the density of air is different to the density of glass. Therefore the speed changes, when something (i.e. glass or a prism) is optically dense it is harder for light to travel through it, thus making it's speed decrease. When the waves meet the prism they slow down, so if they meet it at an angle ...

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Another interesting way to see this, if you live or work where you can see a mountain range, is to train a telescope to some object on top of the mountain. Not only will you see the quivverring image you speak of, if your telescope mounts are stable and you don't adjust them, you will see that the object will move up and down in the telescope from day to day ...

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A one way mirror isn't really a one way mirror. It lets the same amount of light through in both directions. It works because one side of the mirror is light while the other is dark. Suppose you're on the light side looking at the mirror, and suppose that the light side is 99 times as bright as the dark side. Finally assume the mirror lets through half ...

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I found a solution to my problem using the vertical displacement: $$$$\tag{2} h_d = d_1 tan(\theta) + \frac{d_2 tan(\theta)}{n} + d_3 tan(\theta)$$$$ solved for $\theta$ this becomes $$$$\tag{3} \theta = \arctan \left( \frac{h_d}{d_1 + d_3 + \frac{d_2}{n}} \right)$$$$

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Sure, that is what the wavy motion over a hot car is, and also the cause of mirror-like mirages on deserts and hot roads. Stars also twinkle due to the same effect. Not sure about street lights per se, but I'm sure there are combinations where air refraction of light would show up there also.

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There is certainly an interaction there between the optical medium and the photon. Actually, there are two photons in the interaction: the incoming one is absorbed by an electron in the material so that the latter fantastically fleetingly rises to an excited state. A fantastically short time later, another, outgoing, photon is emitted and the electron ...

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I'm not sure I understand your question correctly, but let me know if the below helps. Let \begin{array}$n^2(\omega) &= 1 - \frac{\omega_p^2}{\omega^2} - i\epsilon\\ &=-\Omega-i\epsilon\end{array} where$\Omega:=-\left(1 - \frac{\omega_p^2}{\omega^2}\right)>0$by assumption. Then, expanding$n(\omega)$for small$\epsilon$we find (using ... 0 It's a simplification. Typically, you use boundary conditions to relate the fields outside and inside the dielectric that say that the normal component of the electric field$\mathbf{E}$has a discontinuity so that the auxiliary field$\mathbf{D}$is continuous. There is also another condition that says that the tangential component of$\mathbf{E}$is ... 1 Yes. Different colors of visible light have different speeds in a particular medium. The index of refraction of a medium is defined as the ratio of$c$, the speed of light in vacuum, to$vthe speed of light in the medium; \begin{align} n = \frac{c}{v} \end{align} Therefore, as you essentially point out, if the index depends on wavelength, then so does ... 0 A teacher I very highly regard explained it at a high school level to me as follows: (I was already taught something about interference of waves) You have a light-wave incident on the medium. The light-wave is just an oscillating electric field (plus magnetic field, but we know that electric effects are dominant in such situations). The oscillating electric ... 0 The explanation is very simple! The reason light changes direction ("bends") when traveling through glass, is because light travels slower in glass than in air. If now, you also want to know why light travels slower in glass than air, it is because the density of glass is higher than air and the electromagnetic fields of the glass molecules interfere more, ... 0 Apparent depth of the object is given by the following equation, $$AD=\frac{RD}{n_{ab}}$$ WhereAD$is apparent depth,$RD$is real depth, and$n_{ab}\$ is the refractive index of medium b (denser medium) w.r.t medium a (rarer medium). As real depth and refractive index is going to be constant always (for the given pair of medium), apparent depth will be ...

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Keeping in mind that a surface cannot have a refractive index, but rather the bulk material which has that surface, what shape is the material you're measuring, and do you get to modify it? If, for example, it's a planar solid (front and back faces parallel), point the laser at an angle and measure the lateral displacement of the output beam. Given that ...

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Let me answer this one with some drawings:

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It depends on what you mean by the energy of light. A light wave has an associated energy flux, that is how much energy passes through a surface at right angles to the light ray. This energy flux is a constant and doesn't change when a light wave passes through materials of different refractive index. However, since the velocity of light is reduced in a ...

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The energy of a photon (which I presume you are referring to) is given by: $$E=hf$$ Since the frequency does not change when light transfers from medium to medium, and Planck's constant is... constant, it follows that the energy of a photon remains the same.

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The medium (the water or whatever else) has a preferred frame, that is its frame of rest. In that medium, the speed of light is not the same for all observers and it doesn't have to be because there is a special restframe now. In vacuum there is no such preferred restframe, that being the root of the principle of relativity.

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No. The speed of light is always c. And only that is constant to every observer. You use only c in Lorentz transformation. In some medium light would appear to slow down because it would jiggle electrons of atoms in that medium which would generate electromagnetic waves themselves. The combined result is a wave traveling slower in a medium. You can see it ...

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Yes, you are correct. The sun is still/already below the horizon at the time of apparent sunrise/sunset, and each of these effects will extend the length of each day by exactly the same amount because the situation is completely symmetrical. Well caught!

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It depends on the position of the sun. A rainbow does not exist at a particular location in the sky. Its relative position depends on the position of the observer and the sun. All raindrops refract sunlight in the same way, but only the light from some raindrops reach the observer's eye. This light is what constitutes the rainbow for that observer. The bow ...

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The best is probably to give you an insightful link. You will find there an applet which illustrates the following, which is your intuition: Coming from the sun, light rays hit the droplet and enter it with refraction air to water They reflect internally in the droplet They come to your eye, following a second refraction water to air Since rays from the ...

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