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91

You have created a rather poor pinhole camera (camera obscura). You can see an "image" of the sky, a green space (trees) and even a reddish brown blur that is your driveway. This is not diffraction or refraction - it's geometrical (classical) optics. Because the hole is pretty big, you see a very blurry image. But basically, the light from the sky falling ...


31

like even when light gets on the moon why does the space appears dark from the moon? For the same reason it appears dark from the Earth (when flying at an altitude of 80,000 feet or so): Image credit: View from the SR-71 Blackbird. The fact is, we can't 'see space' from the Earth's surface during the day because the atmosphere is 'in the way'- the ...


19

As you have probably noticed, the moon is tidally locked with the earth so that we always see the same side. You can look up in the sky and watch sunlight move across the moon's face. From the surface of the moon this change in illumination would look just like the day/night cycle on Earth ... except that it's roughly a month long. Until the advent of radar ...


15

Lorentz came with a nice model for light matter interaction that describes dispersion quite effectively. If we assume that an electron oscillates around some equilibrium position and is driven by an external electric field $\mathbf{E}$ (i.e., light), its movement can be described by the equation $$ ...


8

In non-relativistic systems both $E\sim k$ and $E\sim k^2$ are possible. Quadratic dispersion relations occur if $\langle 0|[Q_i,Q_j]|\rangle\neq 0$ for some of the generators. This occurs in a ferromagnet because rotational invariance is broken and $J_z$ has an expectation value. In terms of effective lagrangians the difference between ferromagnets and ...


7

Snell's law is given by $n_1 \sin i =n_2\sin r $ where $n_1$ is the refractive index of the material the light is initially in and $n_2$ the refractive index of the material the light is going into. These are not constants for a given material and change (although very slightly) with the wavelength of light. Thus for different wavelengths of light the ratio ...


6

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 $$ ...


6

Two media can have equal indices of refraction. For example, you could pick the densities of two different gases so as to make their indices of refraction equal. You could do the same thing with different liquids containing properly chosen concentrations of solutes.


6

Suppose you have an infinite plane wave. To find the momentum of this wave you Fourier transform it. Because it's an infinite wave the Fourier transform is a delta function and the wave has a well defined single value for the momentum. Now take a wave packet i.e. the same infinite plane wave but now multipled by some envelope function. When you Fourier ...


5

An easy way to make this intuitively plausible is by remarking that the Schroedinger equation in the absence of a potential is as follows $${\partial\over\partial t}\Psi = \nabla^2\Psi$$ up to constants, which is the heat equation if we ignore the fact that the omitted constants are complex numbers rather than real and of the right sign. If you consider ...


5

What you want to do is change the wave equation into a Klein-Gordon equation: $$\frac {1}{c^2} \frac{\partial^2 \psi}{\partial t^2} - \nabla^2 \psi + \alpha^2 \psi = 0,$$ where $\alpha$ is a constant of appropriate dimension and usually (in quantum theory) given by $$\alpha=\frac {m c}{\hbar}.$$ Inserting an ansatz of the form $$\psi=e^{i(kx-\omega ...


5

The simple explanation given in Hewitt's Conceptual Physics is that atoms in condensed matter have a high-frequency resonance, and the index of refraction for most substances is strongest at the blue end of the spectrum because that's the high-freqency end, which is closest to the resonance. The following is my attempt to flesh this out with a little more ...


5

The term dispersion refers to the speed of light in a material having a dependence on frequency (or equivalently wavelength). The refraction angle's dependence on frequency is caused by the material dispersion, not the other way around. In all materials the refractive index will have dispersion but it's often the case that certain materials in certain ...


4

The whole point of Snell's law is about taking into account wavelength! Remember that the fundamental property of the light is its frequency. Wavelength, on the other hand, is not a fundamental property of a beam of light since the wavelength changes all the time as it passes through various media. The index of refraction of a medium is a dimensionless ...


3

Depending on the physics underlying the particular wave equation in question, the three most fundamental limitations on dispersion are causality, stability and holomorphicity. These are most readily converted to mathematical statements about the operators in a wave equation if the wave equation is linear. I'll confine the following mainly to optics; ...


3

In a linear wave equation, there is nothing to pull a pulse or envelope of running waves apart. But there is nothing to hold it together, either. A minor disturbance such as a small obstacle or some dispersion, will change the waveshape, or break it up, such as losing some of its energy to outward spherical waves from the obstacle. Two or more pulses in ...


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, ...


2

I start my answer with the the second question (similar situations). This is eerily similar to the free-particle dispersion relation (or the relativistic energy-momentum relation) in natural units ($c = 1$ and $\hbar = 1$), $$\omega^2 = k^2 + m_0^2$$ This system of units is commonly adopted in particle physics and quantum field theory. It can be clearly ...


2

First of all, this is NOT a 2D photonic crystal since you send light in the $z$-direction (you are using all 3 dimensions) and if the rod is not infinitely long. I'm not aware of an analytic formula, but your band diagram exhibits several features characteristic of metallo-dielectric photonic crystals. For example, there's a bandgap starting at zero ...


2

Yes. There's an even easier notation for dispersion, or standard deviation which is $$D(Q) = \sqrt{ \langle Q^2\rangle - \langle Q\rangle^2 }$$ Both those terms are the same. So $D(Q)$ is zero.


2

The dispersion relation gives you information regarding the relation between momentum of electrons, and energy of such electron. Heisenberg's uncertainty principle relates uncertainty in the position versus uncertainty in momentum, which is a very different issue. If you consider a single massive free particle, it also possesses a dispersion relation in the ...


2

$k$ is just a quantum number. $\hbar k$ gets its name "crystal impulse" from the fact, that the formula for a band structure without interaction (free electrons) coincides with the formula you get with the definition of classical impulse in terms of $k$, but it is NOT an actual impulse. For a free electron we have the energy dispersion: $$ \epsilon(k) = ...


2

I will hand wave here, looking at the problem a photon at a time. We know from the double slit experiment that even individual photons impinging on the double slit geometry display an interference pattern, characteristic of the frequency/energy of the photon and the geometry of the slits. One can think of a crystal as a very large number of three ...


2

$E = \hbar \omega$ It doesn't matter what the form of $\omega(\vec{k})$ is, whether it's linear or not, $E=\hbar\omega$. e.g. For massive particles, $E = \frac{\hbar^2}{2m} |\vec{k}|^2$, which is parabolic, not linear, and $\omega = \frac{\hbar}{2m} |\vec{k}|^2$.


2

The dispersion that leads to the rainbow effect generated by transparent media results from an intrinsic property of the medium being considered: the dependence of its refractive index $n$ on the wavelength of light $\lambda$ passing through it. In this sense, water in a glass is just as dispersive as water droplets in a rainbow. When different wavelengths ...


2

The wave mechanics dispersion relation you cite is for EM waves propagating in free space. In other media, the dispersion relation is not necessarily linear (it can be quadratic or have some more complex dependence). So in this context, there's nothing special about quantum mechanics. More generally, the dispersion relation tells us about the phase speed ...


1

That the phase speed can have a dependence on the wavelength/frequency of the wave. For instance, a whistler mode wave can have a cubic dispersion relation at low frequencies. In this limit, the higher(smaller) frequencies(wavelengths) propagate faster than the converse. It results in a sort of "spreading out" of the wave modes. This if often seen ...


1

First: I'd guess that by light dispersion you mean chromatic aberration. In general refocussing the light will not undo chromatic aberration, but it's possible to specifically design pairs of lenses to (mostly) eliminate it. These are called achromatic doublets. Second: yes, you need a bandpass filter. It's not possible to transmit everything below 600nm ...



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