New answers tagged

1

I make f.lux, and I've chosen not to ship an overlay app for Android because I think it is a poor approach. It's hard to watch people say that these approaches are the "same" in any way, but I will try to bring numbers to the discussion, so maybe I will convince you. Measuring "melanopic response per lux" answers the question: at the same visual brightness ...


1

Disclosure: I got to see your question because one of the replays refers to my article on blue light filters (glarminy.com). There is one important though, at first glance, subtle difference between approach 1 (f.lux) and 2 (Twilight). It is that approach 2 reduces contrast considerably more than approach 1. I'll get back to this point. First on blue ...


0

Well, consider some wave fronts emerging from the light source. Now, as the distance from the source increases, so will the curve of the wave fronts increase. As we go farther and farther from the source, the curve of the wave fronts will become nearly straight. Light rays are considered to be falling normally on these curves and as with increased distance ...


1

This means that the techniques reduces the amount of light the organism under the microscope is exposed to - it is efficient in its use of light. I haven't read the paper in depth to understand why this might be so.


0

Your cristaline is not a lense made of homogeneous glass, it is made of cells. These are mostly transparent, but still there is a difference at their borders, and theses are aligned at large scale, forming a structure. This structure makes a slight variation of the optical properties, but it does diffract a bit of the incomming light, and at huge intensity ...


1

TL; DR Short answer is the energy from the photon causes the electron to jump. Conservation of energy dictates that the photon would lose some energy, and it would be from the particle matter that it would act this way. Long Answer This is called the photoelectric effect. Basically this is caused by an energy transfer from a photon, acting as a particle ...


2

Electromagnetic (EM) radiation between 400nm and 700nm in wavelength is the same thing as light. There is no difference. Neither is there a distinct difference between light, ultraviolet, x-rays, gamma rays, infrared, microwaves, or radio waves. Those names are all just human convention to specify EM radiation in certain frequency ranges. And regardless ...


1

Relative to the normal $N_s$ of the plane, $\Phi$ goes from $-S$ (reaching the sky horizon) to $+\pi/2$ (when it is parallel to the plane. Further, it would be on the back side of the plane, which must not count).


1

To answer your question Do these approaches produce the same effect/outcome when it comes to reducing the blue light emission of the display?, it is important to keep in mind the "normalization" hypothesis you make (assume the brightness of the display is set to a fixed level in all possible scenarios), which says that the emission spectrum of the ...


0

Light is at first a wave, and like any wave the simplest description of his propagation is the Huygens-Fresnel principle : at any point a new wave is re-emitted in every direction. When you sum the contributions, in free space the only location that does not vanish trough interferences is the next location in straight line direction. But if there is a hole ...


7

There is the question you asked, and the question you might have asked... and they are slightly different. I will answer both (I had answered the second before you clarified that you meant to ask the first... seems a shame to delete the second answer) First - the question you asked (about electronically controlling the display properties). There are (at ...


2

My attempt of answer will be based on my non-specialized knowledge of programming and computers workings, because the links you provide don't seem to give out the details on the workings of the code. I'm a physicist with a non-academic conceptual and practical understanding of computers. Whatever layers your software might have, the display will react ...


3

The following diagram may be helpful: If you have an incident ray that is polarized with the E field up and down (in the plane containing the incident ray and the normal to the surface), then when that ray is refracted, it contains a component of electric field that is perpendicular to the refracted ray (and still in the same plane). The reflection is ...


3

If the wall is made of a perfectly reflective material (such as a mirror) no, you won't see the dot. However, most walls are covered with paint or made of a diffusive material : when the laser beam hits the wall, its light gets diffused in all directions. Thus you are able to see the laser spot on the wall. To summarize : reflection depends on the ...


0

Essentially, the smaller the aperture gets the more it starts acting like a point source. As mentioned by dominecf, a point source creates circular wavefronts therefore resulting in a "bending" of light. A quantum mechanical explanation of the phenomenon has to do with the Heisenberg uncertainty principle which relates the uncertainties in position and ...


-1

In the case of refraction we have to know that electromagnetic waves travel at a different velocity in different mediums, for example they propagates faster in air that they do in water. So, if we have a ray of light which is emitted from below the surface of water it will accelerate when it reaches the surface of the water. Now, If the ray hits the surface ...


3

A properly collimated laser beam is called a Gaussian Beam whose transverse magnetic and electric field amplitude profiles are given by the Gaussian function. The Gaussian beam is a transverse electromagnetic (TEM) mode. The mathematical expression for the electric field amplitude is a solution to the paraxial Helmholtz equation: The width of such laser ...


4

Just as Jon Custer wrote in his comment, even a perfectly collimated laser beam with a planar wavefront will diverge. The way it happens is determined by the Huygens principle, and depends on the beam profile: When the light intensity is abruptly cut by a sharp flat obstacle, the light will indeed diffract in almost all angles. A razor blade cutting a laser ...


1

It does not. Neglecting huge gravitational fields (e.g. black holes), which distort even the traveling path of light, a light wave propagates in a straight line. The "wave" part is expressed in the electric and magnetic field of the light beam/pulse, but these two fields oscillate in the plane transverse to the propagation direction. Lastly, the ...


0

Actually it is interesting to calculate the radiation frequency for an electron in low orbit about a neutron star. The orbital frequency depends only on the density of the star or planet (radius doesn't matter!). Wikipedia tells me a neutron star has a density of 10^17 times as great as earth. The frequency goes as the square root of the density. The orbital ...


2

Is a neutron star's residual light released similar to an exited atom the difference is gravity hold in the electrons instead of protons? No. Atomic energies are of order of keV at most, the electrons are bound in energy levels about the atom. There will only be photons produced if an electron is kicked to a higher energy level and then decays back ...


5

Photons mediate the electromagnetic force. Atoms are not necessary for photons to exist. You just need charged particles (electrons, protons, etc) to interact with each other from a distance. There are many ways for a photon to be created and destroyed. Depending upon its wavelength, as it propagates in free space , it could "disappear" and a pair ...


-1

intensity is defined as power/area, here, more energy that reaches your eyes, more light you observe, By symmetry if we consider a isotropic (symmetrical) point source with total power being emitted P, intesity at any d(A) {small area} at the surface of imaginary sphere will be P/4π(r)^2 so, Intesity falls by second power of radial distance


1

This is just an effect of the optics used to take the images. The size of the aperture causes the light to be bend: Wikipedia: Airy Disk. Now, as the aperture of most lenses is not circular, the edges cause some visible rays in the image, as is depicted here. The left column shows an assumed shape of the aperture and the right column the resulting image of a ...


2

There is light all around the lamp, and all around you (except of course for the points that light cannot reach because of walls or whatever). You just only perceive the light that comes into your eyes. The light that comes directly from the lamp is generally more intense than that reflected by the objects around you, so that you tend to ignore the latter. ...


1

This stuff is confusing, but you have it about right. Irradiance is $\Phi / A$ if the flux is constant over the area, which usually means that $A$ is small compared to $d^2$. Note that the total flux within the cone is the same wherever you choose to measure it. The irradiance at the surface $dA$ is $\Phi / dA$, and the irradiance at the surface $A_0$ is ...


2

If the 2 objects are at different locations, then the green light is hitting different group of cones in your eye than the red light. You should thank to the lens in your eye which is sending the light rays to different areas on your retina, depending on the direction from which the light is coming. Without the lens, all light would hit all cones at the same ...


1

We won't be able to observe the light beam because photons don't interact with each other -- there is no way for light to bounce off the other photon so we won't be able to see it! Even if we imagine a particle travelling parallel to our path, both at the speed of light, if a photon were to bounce off of the particle, it could never reach our position ...


1

Neither image displays diffraction. They both illustrate shadowing. The second photo was apparently shot on a clear sunny day. All of the rays come from the sun. Some are blocked, those that aren't go straight down to the floor. The second was probably taken on a hazy day. Rays originate from the haze; they come from all directions, and make their ...


2

The trick is that because the "slit" is infinitely wide, you shouldn't work in the far-field approximation (Fraunhofer difraction integral which leads to fourier optics), but with distances computed to the square order (Fresnel diffraction integral). The results are appropriately named Fresnel integrals (this time this is a name of a special analytical ...


0

When trying to understand light, there are two components to consider: the amplitude, and the polarization, and you can assign both these properties to all points in space. This is very difficult to visualize, so most of these images trying to visualize light are bound to be inaccurate. The first image you show in your post is accurate in showing how the ...


0

Any electromagnetic radiation - and in special case light as a small and visible for us part of EM radiation - is composed of photons. This is right for the process of emission as well as for the absorption of EM radiation. Any photon has a electric E and a magnetic B field component, both perpendicular to each other and to the direction of propagation v ...


2

There is a fairly general discussion of relativistic aberration on John Baez's Physics FAQ site, and a more mathematical treatment on Wikipedia. The formula telling how the original angle is changed for the moving observer is: $$ \cos\theta_O = \frac{\cos\theta_S - v/c}{1 - \cos\theta_S\,v/c} $$ I knocked up a quick graph in Excel to see what happens with ...


2

Light waves are emitted spherically, however electromagnetic waves nevertheless have polarization. The Maxwell equations that the electromagnetic field satisfies are $$\begin{array}{rlcrl} \nabla\cdot \vec E ~=& c\rho &~~& \nabla\times \vec E ~=& -\dot {\vec B}\\ \nabla\cdot \vec B ~=& 0 &~~& \nabla\times \vec B ~=& \vec J + ...


0

Substantial difference between light and sound is: light is transverse wave and can be polarised; sound wave in gas is longitudal wave and cannot be polarised.


0

I don't think there is a universally agreed phrase to describe this, but I think the closest is the principle of geometrical reversibility.


0

A travelling wave in the direction of the picture(or whatever direction) can have two polarizations, each perpendicular to its direction of propagation. Now, in the black body radiation derivation, we usually use a box as a black body and inside the box standing waves are formed. But standing waves are nothing more(mathematically as well as physically) ...


0

What you are observing is the apparent total brightness reported to you via the optical sensing portion of your brain. A lot of things go into that subjective feeling: how wide your iris opens, how many rods/cones in the retina are stimulated, and what the relative peak and minimum intensities are in your field of view. If you in fact used a calibrated ...


0

Think of the rays at the extremities of the cone of light. The angle of incidence on the surface of the ceiling from these rays and so the scattering angles from the surface of the ceiling will be greater that those rays which hit the ceiling at smaller angles (you concentrated cone). So there will be more light heading towards the walls with your wide ...


2

Photons are simply quanta (very small pieces) of energy - they have no physical size, and no physical 'structure'. This video is a good introduction to our modern understanding of what light is.


1

light is made of photons, not of protons or anything else. And as an electromagnetic wave, it's structure is often the structure of its interaction with the matter (typically made of dielectrics and conducters). Then, it all depends what you accept as "seeing": you can see it because it reach your captors (which are not exactly at the place of focus) you ...


6

What you describe is the phenomenon called optical dispersion. For example in most optical media the refractive index and therefore the velocity of light depend on frequency. The question is whether empty space shows the same effect. This has been studied as part of attempts to detect any granularity in spacetime due to quantum gravity effects - short ...


0

Energy cannot be stored in this way. Optical fibre will lose all the light energy in few seconds. Optical fibre do not have 100% efficiency. In case of 100% efficiency , No effect would be there due light energy in optical fibre outside.


0

I will address the "see" part. Why do we see black objects? Colors of objects are formed when the spectrum of that color is reflected. Example Green objects are green because they reflect the green spectrum of light, red objects are red because they reflect red spectrum of the visible light and white objects because they reflect all the visible spectrum ...


0

As long as there is a discontinuity in the index of refraction ( real or imaginary parts) then there will be a reflection from the "black" object. How your eye-brain perceives it is a different matter..


-1

Paradoxically enough, the speed of light in a vaccum is completely unchanged by any outside factors. It doesn't matter what speed the object is moving, the speed of light is always the same in any frames of reference.


0

A related question is, why do we see white? From a physics perspective you are correct, we should not see the object. We should see a hole, a complete absence, like looking into the face of a shadow rider or a dementor. However, from a biological and neuroscientific perspective, we autocorrect. The limited light is detected by the super sensitive rods ...


3

The local speed of light is always $c$. By this I mean that if any observer anywhere measures the speed of light they will get the value $c$. It doesn't matter if they are accelerating or not. However for an accelerating observer, even in flat spacetime, the speed of light at distant locations will not be $c$. The geometry of spacetime for an observer with ...


0

If we ask a bottom observer, he says that when the gravity field gets stronger, it takes the pulse of light a shorter time to travel from the bottom sensor to the top sensor than when the gravity field was weaker, because of the gravitational time dilation that slows down his clock and slows down the light, but slows down more his clock than the light. If ...


2

Yes, it's necessary for example on some telescopes to keep the image the same way up on a camera as the telescope tracks across the sky. VLT naysmyth focus There are a couple of optical designs, using either rotating prisms or a rotating set of mirrors. Look up field rotator. eg http://www.ing.iac.es/~eng/optics/documents/OPT-WHT-001.pdf



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