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7

No. There is no void left by the lack of an aether. The very notion of aether should serve as a warning as to how catastrophically analogical reasoning can fail. "Water waves are in water, sound waves are in air, therefore there must be something in which light propagates." This is flawed logic, and decades of physics were arguably hindered by adhering to ...


0

There are many possible cases of total transmission at an interface (assuming that the media are lossless), and below is a list of the underlying physics: if the media are impedance matched any the given incidence angle. Brewster's angle can be seen as a special case. Usually, at normal incidence, this condition would require the materials to have the same ...


6

As most people know, "let there be light" is a famous biblical quote, from Genesis. Now, on to the teacher's shirt. Those equations on his back are Maxwell's equations. "Let there be light" is a joke, because Maxwell's equations describe electromagnetic fields, and light is a form of electromagnetic radiation, so the equations can be used to describe ...


2

Briefly, the formula $E=mc^2$ applies only particles at rest in an inertial frame of reference. Since there is no rest frame for a photon, no inertial reference frame in which a photon is at rest, one cannot apply the formula $E=mc^2$ to a photon. In more detail, the four-momentum of a particle has components $(\frac{E}{c}, \vec p)$ The four-momentum for ...


1

The correct general equation is $E^2=m^2c^4+p^2c^2$, since $m=0$, $E=pc$. A photon's momentum is $p=\hbar k$, where k is the wavenumber ($k=\frac{2\pi}{\lambda}$). This would be consistent with $E=h\nu$ where $\nu$ is the frequency.


4

A "sharp" tip typically has a finite curvature; there will be a very small part of the "tip" that is therefore angled at such a way that light will be reflected off it. The sharper the tip, the smaller the radius of curvature, and the smaller the "twinkle" or glint. The second effect is diffraction: Light that passes an object will be diffracted. For ...


0

You are asking about the number of photons fired from a device such as a laser at a unit time. We cannot say what the precise number of photons would be at any interval due to a version of the Heisenberg uncertainty principle: $$\Delta E \Delta t \geq \frac{\hbar}{2}$$ It basically states we cannot suppress our uncertainty about energy fluctuations in time, ...


0

If atoms have well defined energy levels and those differences correspond to the frequencies of light that can be absorbed, how is it that opaque objects absorb all or most visible light frequencies get absorbed Photons in almost all frequencies hitting an object are absorbed in different ways (absorbed, reflected, refracted, scattered, ...


2

If atoms have well define energy levels and the differences correspond to the frequencies of light that can be absorbed, This is correct how is it that opaque objects absorb all or most visible light frequencies get absorbed and you basically don't have any visible light coming out on the other end? The crux of the matter is the word "objects". ...


0

Tadeas Bilkas answer let me think about the sence of all and all time citing the quantum mechanics. I write his answer in terms of common mechanics and get the same result: You have an emitter of balls which radiates just one single ball but in a spherical area. You place a lot of baskets some meters apart (with same distance) from the emitter. Mathematics ...


2

In case you "run out of photons", you must switch to probabilistic description of quantum mechanics. Let's consider an extreme case: You have an emitter of spherical waves which radiates just one single photon. You place a lot of detectors some meters apart (with same distance) from the emitter. QM says that the photon propagates as a probabilistic wave to ...


2

The classical electromagnetic field given mathematically by Maxwell's equations can be proven to emerge from a confluence of individual photons, which photons are described by the Quantum Mechanical form of Maxwell's equations. Thus the classical wave is made up by zillions of photons with energy $h\nu$, where $\nu$ is the frequency of the classical wave. ...


-1

At very low energy, the difference between photons and E&M waves becomes clearer: After every spin-flip in a hydrogen atom, "something" with E=6 µeV and spin=h/(2*pi) is released. Is that "something" a photon? Nobody can detect a photon with such a low energy. But if you have a radio receiver at 1420 MHz, you get a clear signal. The electric component of ...


1

"Running out" of photons simply means that your wavefront is absorbed or scattered in a different direction or something like that. Either way, the original wave is "consumed", so you loose intensity or photons, depending on which picture you like better. For the case of a single photon source: One photon can only interact with one electron. However, there ...


1

I think you have invented the zone plate, a kind of specialized flat circular diffraction grating that acts as a lens. It consists of a set of concentric transparent zones of decreasing widths. The width you have derived is the diameter of the first zone. Subsequent zones interfere constructively by allowing paths differing by integer multiples of the ...


2

Your understanding is right, but you have to also realize that there is a rich variety of experimentally verified relativistic effects where the constant $c$ is occurring and this constant would suddenly have to be changed. Namely all the particles suffice the relation $$E = mc^2 = \sqrt{m_0 c^4 + p^2 c^2}$$ This relation is used e.g. to compute the mass ...


7

It's the water itself that forms the lens. Lenses work via refraction. The refractive index of water is about 1.333, which is different from the refractive index of air (about 1.0), so rays of light bend at the junction of the air and the water.


3

Actually you can argue that circular polarization is the "more natural" basis for a single photon. The photon carries one unit $\hbar$ of angular momentum, and circularly polarized light carries real angular momentum (an opportunity for me to mention one of my favorite experiments ever, using photon polarization to drive a pendulum). A single photon that's ...


0

There is also a geometric limitation for seeing that far. I have Q&A'ed it on mathematics.SE. If standing on level ground, Legolas would have been able to see only 4.8km far due to the planet's curvature (assuming that Middle Earth is on a planet resembling ours). To see that far, he would have to have climbed a hill or tree of about 50m height.


0

The diagram is simplified, because refraction actually occurs at the front surface of the cornea and at both surfaces of the lens. (Also the eyeball is not a sphere.) However, taking the cornea and lens as a single compound lens, that lens does have an "optical centre", such that rays passing through that point are not deviated. The author has chosen to ...


0

I believe both rays drawn in that diagram go thru the center of the lens aperture, so they in fact are not refracted. That is, both the top and the bottom of the circular object are at equal radiuses (distances) from the optic axis. Your confusion may be with the rays drawn from a single object point when finding the size of the focussed image, in which ...


0

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 ...


-1

There is a philosophical question of what happens to your shadow, after it would travel faster than the speed of light. This is easily realized with the thought experiment of an object is moving around a point source of light. Given that the radial velocity of a rotating body is $\omega = vr$, then the shadow projected at a distance $r$ can then be ...


2

My theory would be that your eyelids, like windscreen wiper, wipe away some of the exes water on your eyeball. Due to adhesion between the water and the eyelids the surface of this water is slightly curved near the eyelids and if the eyelids are close enough near each other this surface basically acts as a concave lens, which I tried to demonstrate in the ...


2

Well, here's my first answer on Stack Exchange! All light, not just visible light which we see, consists of many different electromagnetic waves. These are all part of the electromagnetic spectrum. The "color" of the light we see is defined by the wave's wavelength. http://science.csustan.edu/chem1112_4/beers%20law/beers%20law081.htm Here is a ...


1

It's quite common for ac-powered electrical devices which are designed without moving parts to "hum," usually with an audible frequency of 60 Hz, 120 Hz, 180 Hz, or other 60 Hz harmonics. (In other parts of the world, where the power distribution system frequency is 50 Hz, you'd obviously get harmonics of that frequency instead.) For instance, when I'm ...


2

Is the shaking immediate upon turning the bulb on or after a while? This could mean heating causes Is this the only bulb doing this or is it same with other (of same type)? This could mean bulb towards end of its life or bulbs of this type do that due to construction of filaments Is this the only bulb doing this or is it same with any other? This could mean ...


2

This really depends on the design of the bulb in question. High-power or hard use bulbs have filaments under some tension and strong supports. Other (mainly decorative) bulbs have very wispy support wires and straight, loose filaments. In this case, the wire can be very sensitive to mechanical vibration through the floor. The bulb and lamp ( a bulb will be ...


2

Assuming that the bulb is powered by some variation of alternating current, and that there is a static magnetic field present in the environment, the vibration could be caused by the interaction of the fixed external magnetic field, and the alternating magnetic field caused by the AC flow through the filament. Bulb filaments are often in the form of a ...


1

It's an optical illusion. The trees at all distances will move in a direction opposite your motion, but the nearer trees will appear to move faster than the ones farther away. If your eyes get used to the speed of the nearer trees, it may look like the farther ones are moving in the opposite direction.


11

Your shadow, surely, is not just the apparent darkening of a 2D region of a diffuse surface you happen to be standing in front of; it is the entire volume of space that your presence is preventing light from reaching, the extrusion of your silhouette from the light source out to infinity (for a point light source, or possibly to a point a finite distance ...


17

Note: This is basically item 3 in jkel's answer. If you move at an appreciable fraction of the speed of light, then your shadow can appear to be "trailing" you, although it will always be "attached" to your feet if you're on flat ground. Suppose a person is moving in the direction shown below and that there are plane waves coming in from an angle. The ...


21

Here is how you can "run faster than" (or at least, get away from) your shadow: you jump at sunset (I just realized 15 minutes after posting that this is the point that @jkej's answer made as possibility #2) Your shadow will detach from your feet, and it will "run away" from you. In the frame of reference of the shadow, you are running away from it. ...


18

Depending on exactly what you mean by away from your shadow, I can think of a number of methods: Position yourself in the shadow of some object larger than you. This would result in your shadow disappearing altogether; the ideal solution in my opinion, but perhaps considered cheating by the likes of Mr Elephant. Position yourself in such a way that your ...


4

This is borderline philosophical stuff, but... If you want to put some distance between you and your shadow, you have to fly. Actually, you have to put some distance between and the opaque surface right below you, so swimming in a tank would do the trick as well. This will disconnect you from your shadow. If you have a jetpack, or if you are in an ...


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 ...


4

Let's try to validate and quantify the conjecture first raised by Carl Witthoft in a comment to the question, which is basically that the sky only appears less blue in the second picture because a lot more light is scattering off of the windows towards your camera. If this is true, we ought to be able to see it. The first thing to do is convert the pictures ...


4

There is a favorite question on graduate schools' qualifying exams in physics, Why the sky is blue? The answer is the Rayleigh scattering. Shorter wavelength photons are more likely to change the direction in the atmosphere which is why the bluer, shorter-wavelength light is overrepresented in the light coming from random directions of the sky. I ...


1

Due to the high speed of light, the light pollution would clear, at least in a practical sense, almost instantaneously from the area once all lights were shut off. One's eyes might need up to about 30 minutes to fully adjust to the dark, so it would take about that long for most of the stars to appear to "come out" to the naked eye.


0

The lowest frequency limit is provided by the size of the universe. If we could make an antenna that size, the frequency would be "very close" to zero. The highest EM frequency limit is provided by the smallest antenna we could make. I believe that would be the size of a hydrogen atom, giving us, $$F_u = \frac{2.997x10^8}{6.28x5.29x$10^{-11}} = ...


0

Listen from the Feynman: I want to emphasize that light comes in this form-particles. It is very important to know that light behaves like particles, especially for those of you who have gone to school, where you were probably told something about light behaving like waves. I'm telling you the way it does behave-like particles. You might say ...


1

There is a good explanation of this in Matter and Interactions vol II by Sherwood and Chabay. I no longer have the text; I will try to summarize its explanation as I remember it. The electrons in a substance are analogous to charged masses on springs. The electrons in insulators are relatively tightly bound; those in conductors are loosely bound or unbound. ...


2

The classic experiment to demonstrate this is the double-slit experiment. Take an opaque material, and cut in it two small slits. Shine on these slits some coherent light, such as that from a laser. On the side of the slits opposite the light source, place a light detector. Firstly, you will observe diffraction and interference: This demonstrates that ...


6

When it's traveling through space, it's a wave. When it hits a wall, or a photo-sensitive chemical strip or something similar, it's a particle. No, this is wrong. It's not sometimes a particle and sometimes a wave. It's always a particle and always a wave. Here is an example of an experiment whose results can't be explained by a pure wave model or a ...


3

An overview in layman's terms: First, it is important to note that not any electric field will induce current in a conductor, because other than the fact the intensity of the field defines the speed of each charge (bigger difference of potential), the oscillation frequency of the $\mathbf{E}$ also plays a very important role, if the frequency is too high, ...


8

The reasoning has to be the other way around: Light acts on the metal and makes the electrons move. This, however, results in an energy loss, as the electrons feel a resistance and thus the radiation loses energy. This can be formulated more precisely with counteracting electric fields. That's why all good conductors are opaque. In insulators this can not ...


2

There is a small error in your math by the way - the factor of $4$ should be a factor of $1$ - check the denominator of the quadratic equation. Why is this so unreasonable? At a glance, it looks like the momentum of the sail is a bit larger than what it started out with. If I make the reasonable assumptions that: The sail is non-relativistic ...


0

The only person I know, who spoke about the structure of the photon, is de-Broglie, listen to him: For both matter and radiations, light in particular, it is necessary to introduce the corpuscle concept and the wave concept at the same time. In other words the existence of corpuscles accompanied by waves has to be assumed in all cases. However, ...


1

Since $n=\sqrt{\varepsilon_r\mu_r}$ (the relative permeability $\mu_r$ being almost always $1$), and $v=\frac{c}{n}$, you can also write $I = \frac{nc\varepsilon_0}{2} E_0^2$. (We used the decomposition $\varepsilon=\varepsilon_r\varepsilon_0$.) So, the intensity depends linearly on the refractive index.



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