# Tag Info

1

I experienced the same problem: too often, the emission value of glass is misquoted in different sources on the internet, as I found to my own dismay. My own calculations found about an average reflectivity R of 0.2, transmittance T of 0.4 and absorption of 0.4 for glass of about 2mm thick. (I calculated those values using absorption spectrum graphs of ...

0

Linearly polarized wave can be represented as a superposition of a right circular wave and a left circular wave. We can use device like Fresnel's composite prism in which we use subtances such that refractive index for right circular wave is different than the refractive index for left circular wave. now,if we send plane polarized light though the prism ...

1

First of all ,what is unpolarized light? As Feynman said, ''Light is unpolarized if you cannot tell whether it is polarized or not.'' light from ordinary sources are unpolarized because our detectors only can detect the mixture of waves polarized in different directions.(not individual waves). waves emitted by any one molecule may be linearly polarized ...

0

Well, you use the amplitude $|\mathbf E|$ of the electric field to find the total amount of energy carried by the electromagnetic wave; the root-mean-square magnitude of the Poynting vector for an EM plane wave is $$\left< S \right> = \frac{\epsilon_0 c}2 |\mathbf E|^2 .$$ You seem to be after a way to specify the degree of elliptical polarization, ...

2

Using a process called interference, we can find wavelength, because the way that waves interfere is reliant of wavelength. Interference is based off of two key principles of waves: they are made up of peaks and troughs. When troughs overlap, they go lower. When peaks overlap, thy go higher. When a peak meets a trough, they cancel. Of course, the positions ...

0

Yes light can change propagation direction when scattering off of a grating into pre-defined directions that depend on the angle of incidence and the grating period (as well as the refractive medium the grating resides in). This is just Bragg's Law. The change in direction (or momentum) is determined by the above parameters. You can think of the change in ...

0

It all depends on the amount of radiation to which one is exposed due to the day-to-day things used. Radium releases alpha particles which have very low penetrating power. Hence we are not in danger of those radiations.

1

Yes, it all depends on the refractive index of glass at the wavelengths you are interested in. In the IR (infra red) I think ZnS is a material of choice for lenses and windows - but I know KBr can be used - Potassium Bromide - you have to be careful with KBr as it dissolves in water.... Above the UV it becomes difficult to use windows as all materials are ...

11

One has to make clear that the watches we are using now are no longer using radium , because of radiation danger awareness. Radium dials are watch, clock and other instrument dials painted with radioluminescent paint containing radium-226. The 1900s (decade) were the peak of radium dial production, as radiation poisoning was then unknown; subsequently, ...

1

By far the most common isotope of Radium is 226Ra, which decays by emitting an alpha particle. Alpha particles have almost no penetrating ability, and in general externally- occurring alpha particles are absorbed by the outer layers of skin which are naturally sloughed off, so no permanent damage occurs. If you swallow it, that's a whole other (very sad) ...

-5

Humans can tolerate a certain amount of radiation. The watch contributes less radiation to our bodies than the soil. Radium emits x-rays, so yes, they can excite an electron.

0

Light does become polarized in a magnetic field. The magnetic field of a black hole was detected due to the polarization of light. Check this article: http://www.iflscience.com/space/black-holes-powerful-magnetic-field-observed-first-time

2

The radiation produced by a mobile phone does have the same heating effect (dielectric heating) as a microwave oven. However, as you suspected, the difference is power. Power is a measure of energy per unit time which, for a given frequency, equates to the number of photons per unit time (not 'power per photon', as your title says). As listed here, ...

5

If we assume you are a sphere in space, at the same distance from the sun as Earth, then we can calculate the heat absorbed - and we can calculate how hot you need to be so heat in = heat out (assuming uniform surface temperature, and radiative heat transfer only). For this, we need the Stefan-Boltzmann expression for total emission at a given temperature: ...

2

Theoretically yes, the laser principle does not consume any material. There is a light source that excites the electrons in the material to higher levels, they deexcite to some intermediate one, here the avalanche of photons appears producing the laser light and leaving the electrons in the ground state. And you can repeat the process without a loss.

2

Q1: For photons of energies much less gamma rays, the quantum mechanical photon-photon interaction is negligible. This is consistent with the classical electrodynamic description where the principle of superposition holds (electromagnetic waves pass through each other unchanged, as well as through electric/magnetic fields). Q2: in reality, charge is defined ...

1

theoretically if its components never wore out then yes. however in practice things do wear out eventually and so no it could not be done in the same way that a perpetual motion machine can work in theory but not in practice.

1

Since your title starts with "Theoretically", I will give the theoretical answer: no. Intensity does not need to decrease. If you send polarized light into a slab of transparent material at the Brewster angle, then there will be NO reflection, and ALL the light will be transmitted. Theoretically, this means that you might get 100% transmission. In practice, ...

2

When light hits a barrier, even transparent ones, some light is reflected and some is refracted. This is often described by the transmission coefficient for that material, and at that wavelength. This can happen at the macroscopic barriers and at the smaller barriers between crystals or grains within a material. It is a simple property of waves which does ...

0

Your most recent edit asks, where does the current that gets induced in the secondary coil come from? The current is the motion of the charges, it is produced when you make the charges move. To make them move you must give them kinetic energy. Energy is conserved, but can be converted to different types. And so, I'd like to talk about where the energy ...

0

The most important concept relating Faraday cage hole size to cell phone signal attenuation is the idea of a cutoff frequency. For round holes, you would model them as cylindrical waveguides. For simplicity, we'll consider rectangular waveguides instead. To match the electric and magnetic field boundary conditions, it's necessary that the waves go to zero ...

0

Without the phase change energy conservation would not be satisfied. To see why this is true you can think of a simple Michelson interferometer; without one of the fields having a phase flip you could get constructive (or destructive) interference at both sides of the beam splitter which would result in twice (or none of) the energy which you sent into the ...

1

There is subtle point I would like to make; for the transmitter to send $200\cos(1000\pi t)$, the transmitter would need to exist for all time. If the transmission starts at $t = 0$ and ends at $t = t_f$, the ideal transmitter would transmit $$200\cos(1000\pi t)\left[u(t) - u(t - t_f)\right]$$ where $u(t)$ is the unit step function. However, this is ...

1

Not all light bulbs are thermal emitters. Fluorescent lights do not use incandescence, hence they would not emit an equal spectrum to an incandescent source with an identical maximal light frequency. But in general yes objects do concurrently emit a whole spectrum of waves based on their temperature, regardless of whether their light is visible to us.

0

Whenever someone investigates the interaction between photons and edges we interpret fringes behind an edge as manifestation of particles wave character. And at the same moment we always emphasize that this waves are not observable direct. So it's only one of the possible interpretations that from fringes with a wavelike intensity distribution behind an edge ...

1

Since both you and your teachers are stumped I will give some pointers. Pointer 1 - light is an electromagnetic wave. The energy flow is given by the Poynting vector. In vacuum (or air) this is $$\vec S = \vec E \times \vec H$$ Conveniently, for plane waves the time averaged Pointing vector is (see wiki ) $$\langle S \rangle = \frac12 \epsilon_0 c E^2$$ ...

0

Are you asking why describing an arbitrary polarization can be described as a combination of linear polarizations (not necessarily in phase), or what the physical mechanism is within the polarizing material? The answer to the first case is relatively simple: any polarizer will block the component of incoming light perpendicular to its axis and pass the ...

0

I'd like to explore the question in two parts: first considering only fundamental particles and secondly considering composite particles and other system. For fundamental particles their mass is a part of their basic identity. It is not variable, so the reaction $$X \rightarrow X + \gamma \,, \tag{*}$$ (where $X$ is any fundamental particle) in free space ...

0

There exist a wiki article on this, project elf. The scaled-down system the Navy eventually constructed, called Project ELF, began testing in 1982 and became operational in 1989. It consisted of two transmitter facilities, one at Clam Lake, Wisconsin and one at Republic, Michigan. with a total of 84 miles[ of above-ground transmission line antenna. The ...

1

Every point of a wave front may be considered the source of secondary wavelets that spread out in all directions with a speed equal to the speed of propagation of the waves.

3

The measurement of the velocity vector of a nearby, large galaxy is not a simple matter. If you just dropped a spectrograph slit down randomly somewhere in Andromeda then you could get a wide variety of answers, since it rotates with velocities of $\sim +/- 200$ km/s in different parts of its disk. To estimate the centre of mass redshift one must use ...

0

I think you should be more specific in the type of black hole, because the spinning (and spinning-charged) black holes can make light reach the photonsphere and not be swallowed by the black hole, but enter an orbit around it. Eventually debris that are also in orbit will be heated up and started to glow, and it's the first step for a quasar to be born.

7

It is not clear from the question, but let's assume we are in free space/vacuum. Then the Maxwell equations read: $$\nabla \cdot \mathbf E = 0\\ \nabla \cdot \mathbf B = 0\\ \nabla \times \mathbf E = -\frac{\partial \mathbf B}{\partial t}\\ \nabla \times \mathbf B = \mu_0 \varepsilon_0 \frac{\partial \mathbf E}{\partial t}$$ A plane wave can be written as ...

3

In one sense the problem is simple, in that any solution with complex numbers (say with phasors) can be literally translated into a real version (just equate the real and imaginary parts of each complex equation as two real equations). However many phasor type setups are designed to only look for solutions of a particular type, so it might be considered ...

4

While for vectors $\vec{B}$ and $\vec{C}$, the cross product $\vec{B}\times\vec{C}$ is indeed perpendicular to both of the vectors, it is simply not the case that the curl of a vector field is orthogonal to the vector field. Do not read too much into the cross product notation. In particular, you can add any constant vector field to $\vec{A}$ without ...

1

The vector potential is transverse to the direction of motion. To see this, perform a gauge transformation into the Coulomb gauge, such that $\nabla\cdot\mathbf{A}=0$. Using a plane wave solution $\mathbf{A}_0\cos(\mathbf{k}\cdot\mathbf{r}-\omega t)$ we then find $\mathbf{k}\cdot\mathbf{A}_0=0$, i.e. the potential is perpendicular to the direction of motion. ...

1

Let us work in the Coulomb gauge, i.e. $\phi=0$ and $\nabla\cdot\mathbf{A}=0$. Then the electric and magnetic fields are defined as $$\mathbf{E}=-\dot{\mathbf{A}},\quad \mathbf{B}=\nabla\times\mathbf{A}$$ Now one solves the wave equation for $\mathbf{A}$. The constraint $\nabla\cdot \mathbf{A}=0$ tells us that $\mathbf{A}$ is transverse to the wave vector. ...

3

In electrodynamics, one uses complex fields only as a calculation trick, since for instance terms like $$\exp(i \omega t)$$ are usually more easy to handle as $$\cos(\omega t)$$ . The same trick can be applied for example to the problem of the driven harmonic oszillator: $$\ddot x + 2\gamma \dot x + \omega_0^2 x = A \cos(\Omega t)$$. Adding $i \left( ... 0 In general, $$\mathbf E = -(\nabla \phi + \frac{\partial \mathbf A}{\partial t})$$ For the source free case, and in the transverse gauge with appropriate boundary conditions on$\phi$, $$\mathbf E = -\frac{\partial \mathbf A}{\partial t}$$ Which can be checked by taking the curl of both sides$$\nabla \times \mathbf E = -\nabla \times \frac{\partial ... 1 I don't really see a difference between the microwave measurement and the definitions of the SI units. We have Unit of time: second: The second is the duration of 9 192 631 770 periods of the radiation corresponding to the transition between the two hyperfine levels of the ground state of the cesium 133 atom. Unit of length: meter: The ... 1 We know that light is massless so why does a black hole's gravity attract light? Because gravity doesn't just attract objects with mass. It alters the path of light too. Because gravity is caused by a concentration of energy which "conditions" the surrounding space, altering its metrical properties, whereupon we talk about spacetime curvature. But note that ... 1 The problem with this question (although your question is still a natural one for those thinking about light to ask) is that it mixes the ideal and the real. You describe an ideal situation with your mirrors, but then ask for what would happen in real life. No actual mirror has reflection coefficient of 1 (which would represent 100% reflectivity) and so any ... 1 A photon has a rest mass of nought (where the rest mass$m$is the Lorentz-invariant quantity in the four-momentum's Minkowski norm squared$E^2/c^2 - p^2 = m^2 c^2$). However, a lightfield of energy$E$gravitates and itself has a gravitational source equivalent to a mass$E/c^2\$. Also, a system of photons has a nonzero rest mass (see reference), as does ...

1

light is supposed to possess relativistic moving mass even though it does not possess any rest mass. m2c2=M2c2-M2v2 where m is rest mass and M is relativistic mass and v=c. this gives m = 0 , but M is not zero the value of M can be calculated from the experimental data on radiation pressure.

1

We say the electromagnetic wave is oscillating because something waves as the wave passes by. Light does propagate as per the above image, but it isn't the full story. For a bit more, have a look at the Wikipedia electromagnetic radiation article and note this: "Also, E and B far-fields in free space, which as wave solutions depend primarily on these two ...

2

In light propagation, oscillation does not mean any movement in space. It is the value of the electromagnetic field, at one given point in space, that oscillates. The picture that you quote does not represent the movement in space, but the electromagnetic field value as a function of time. Compare to waves in water: if you put a little boat on the water, ...

2

(This type of question has been asked by 4 users but in those questions they either gave an example of a wooden box or a room and they got answers that the light is absorbed by the wood or the walls of the room. But in my question its the case of mirrors.) In this case, the light would be absorbed by de "viewer". You would need some type of device ...

0

An electromagnetic wave is more like an oceanic swell wave than a circular spreading ripple in a pond. If you've ever been on a ship you'll know what I mean. A swell wave is maybe two metres in height, two hundred metres in wavelength, and five hundred metres wide. It's this big hump of water barrelling across the ocean at maybe ten knots, going in a ...

5

For light bulbs and other thermal emitters this is definitely true. Their emission follows the black body spectrum (if you neglect absorption due to the glass container). If you want to be picky: Any device, which is operated above 0 K (which applies to all devices) emit thermal radiation according to their temperature. This is not directly related with the ...

19

Basically the same reason as what Floris said, but this also has another important aspect: Visible light has a far too small wavelength to affect a compass. Not only does the field oscillate too quickly around an average of zero – even at any single “snapshot” in time of the electromagnetic wave, there would nowhere be a large region where ...

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