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24

A result known as Birkhoff's theorem forbids spherical electromagnetic radiation. The statement of the theorem is that any spherically symmetric vacuum solution to Maxwell's equations must be static. It is rather simple to prove. In a spherically symmetric solution $\mathbf E$ and $\mathbf B$ must be radial. Make an Ansatz, $$\mathbf E = E_0 \exp(i(\mathbf ...


18

You have to distinguish, which interactions take place, when electromagnetic radiation passes through a solid and interacts with it. There is a nice plot on Wikipedia, showing the dielectric response of solids for different wavelengths/frequencies. Basically, as the frequency gets higher, the wavelength becomes shorter, and the molecules or atoms are no ...


17

Very reasonable question. I will try to answer it in an intuitive way. If you have a scattering medium, photons are reflected in random directions; but when you have a refractive medium, something else happens. The photon is not absorbed and re-emitted: instead, the photon interacts with the electrons in the medium, and since these electrons are somewhat ...


13

As many have said, the inverse square law applies to point-sources. These are idealized light sources which are sufficiently small compared to the rest of the geometry that their size is of no importance. If a light source is larger, it is typically modeled as a collection of idealized light sources, potentially using integration. The exact definition of ...


12

The inverse square law applies to point sources. For extended sources becomes accurate at distances that are large compared to the size of the source. At large distances the source looks like a point. What "large" means depend on the application. In the case of light fixtures, the Illuminating Engineering Society and other organizations have made ...


11

Light is composed out of a large ensemble of photons, and photons are quantum mechanical elementary particles. Matter is composed out of atoms and molecules , which have small dimensions and are in the quantum mechanical range. The quantum mechanical "size of interaction region" is given by the Heisenberg uncertainty relation. Even though a photon is an ...


6

The inverse square law applies to point sources. A real emergency light is not a point source, and therefore the law appears to not apply at close distances, because any real point is at a varying distance from different parts of the emergency light.


4

Your question is based on the assumption that a photon is a fundamental object i.e. that photons are something we can point to and say here is photon 1, here is photon 2, and so on. The trouble is that quantum field theory particles are somewhat elusive objects. This is particularly so for particles like photons that are their own antiparticles because such ...


4

Classical electromagnetism is perfectly compatible with special relativity. In classical E&M, light is an electromagnetic wave and there is generally no useful formulation in terms of particles. The most widely used technique to combine quantum mechanics with special relativity is relativistic quantum field theory. The relativistic QFT that ...


3

The inverse square law says that the intensity of incident light falls off in proportion to the inverse of the square of the distance from the light source. The important word here is "the distance" — the inverse square law implicitly assumes that all parts of the light source are at the same distance from the measurement point, or at least ...


3

Photons are boson, so it follows the Bose-Einstein statistics which is only true if the particles are truly indistinguishable. If you can distinguish between two photons, then it will follow the classical Boltzmann statistics which is not what happen in experiments. That means photons with same properties are the same. Even in your situation with photon ...


3

It is a matter of definition of "same". Classically one can define "same" condition of particles by labels stuck on them. Light classically is a wave, and same needs a new definition. We apply the everyday definition by identifying the light beam with the source. The light leaving the sun is the same light arriving on earth. The light reflected from the ...


3

Saying "the light wave splits" is not an accurate description of what we understand about the theory of light. That's why you haven't seen it discussed. The shape of the electromagnetic field (mode) fills all of space, subject to boundaries (to include containers, obsticles, etc. That is, the mode has a shape determined by the boundaries.). Some of the ...


3

In quantum mechanical domain these type of question does not have meaning. Every single photon is associated with a wave and vice versa. But to talk whether an electromagnetic wave contains a single photon or not is an ambiguous statement. When people say an electromagnetic wave necessarily contains many photons it only means that a incident beam of ...


3

This is a common misconception about what boundary conditions do and how they do it (for example here). You discussed two types of boundary conditions, Neumann and Dirichlet. In Neumann boundary conditions, we impose that the derivative of the variable normal to the boundary is specified, generally to be zero. With Dirichlet, we impose the value that the ...


3

At first it does seem unnatural for something to be a field only, no particles actually moving. The velocity of sound is so slow, that you can actually see the propagation of sound visually. Therefore we can visualize it directly. The wavelength of a water wave is directly visible with the eye. Electromagnetism does not allow that. First, you cannot see ...


3

Yeah, @EddyKhemiri, as @almagest wrote it isn't just visible light, but rather that all electromagnetic waves moving through a vacuum travel at $c$. Also- just a pet peeve, but remember that this is the speed in a vacuum; it can be slower in different mediums.


3

Well spotted. The average intensity of the whole fringe system must be $1+4=5\;\mu$W. However there will be places where the intensity is a maximum, $1+4+2\sqrt{1\times4} = 9\;\mu$W and places where the intensity is a minimum, $1+4-2\sqrt{1\times4} = 1\;\mu$W. Your chosen position is one which is nearer a maximum than a minimum. Later Here is a ...


3

Photons don't have an associated wave function. You either use the (fully classical) Maxwell's equations for the electromagnetic field (without photons) or Quantum Electrodynamics (which doesn't work with wave functions at all). For more details, see What equation describes the wavefunction of a single photon?.


3

You cannot simplify the effects of EM radiation on biological systems to simply $E=hf$ because different materials absorb or transmit different frequencies preferentially. $E=hf$ tells us the energy per photon, but it doesn't tell us how much is absorbed by any particular type of cell. It also doesn't tell us the intensity of the radiation (energy per ...


2

Before knowing about X-rays, you should understand - What is Electromagnetic Radiation(EM)? Answer: Electromagnetic radiation is a energy released by electromagnetic processes and light, ultraviolet rays, infrared light etc are all examples of electromagnetic radiation.It consists of electromagnetic waves. Electromagnetic waves are produced whenever ...


2

Not all the light shining on the top surface of your glass block can pass straight through the block. Inside the block, total internal reflection occurs at the side faces. The light which was "supposed" to exit through the sides is refracted to the inside. This causes the shadow you observe and a slighly brighter middle section, which is hard to see. In ...


2

Most glass contains iron oxide as an impurity which gives the glass a slightly green hue. When you're looking straight through a pane of glass you don't notice it because the glass is so thin. But when you look down the side it becomes apparent because of the thickness.


2

I think the answer is simply: "Yes". What you should keep in mind is energy conservation: As long as there are no sources, the total energy of the electromagnetic field is conserved. But then what, in free space, would cause the initial change in the electric or magnetic field to get the oscillations going? A source, which is possibly localized ...


2

Frequency of the emitted light depends on the nature of the source. Eg: Incandescent -- depends on the temperature of the material Discharge Tubes -- depends on the characteristic spectrum of the gas


2

I think this is an interesting question. Unfortunately, many hasty sketches of the history of physics, as they are taught, tend to draw somewhat biased conclusions for the sole purpose of avoiding delving into these types of questions (some people consider it to be a waste of time apparently). As far as I can tell, the classical scattering theory at the ...


2

One can use an easy to understand picture. Suppose, you have an antenna composed from 6 rods along the axes of a cartesian coordinate system. At one moment electrons together get accelerated outwards in all 6 rods. Using a hand rule one could see that the magnetic fields around the rods cancel each other out exactly. So you are free to make an experiment ...


2

The quote from the reference says it all: (I added caps) "The minimum test distance IN PHOTOMETRY of these sources is called the 'minimum inverse-square distance.'" The minimum distance is therefore a photometry issue, in other words, a measurement problem. The essence of the measurement problem is how far away you have to be before you can approximate the ...


2

Cort and Ilmari have given good answers about the practical issue: the inverse square law is for point sources, and so a non-point source (like an emergency light) will only appear to have the same properties at some minimum distance that depends on the geometry of the real source. However, it seems nobody has mentioned a different "minimum distance" that ...


1

To be clear, Maxwell's equations are known as "Lorentz-invariant" equations, which means that they take the same form in every Lorentz-transformed frame of reference. Special relativity actually came about from studying Maxwell's (classical) equations without charges or currents. Then we get: $$\nabla \cdot \mathbf{E}=0$$ $$\nabla \cdot \mathbf{B}=0$$ ...



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