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14

The main reason is due to the fact that the prism refracts light in such a way that the "blue" part is more spread than "red" part. So that overall the energy hitting the thermometer is greater in the infrared and red part than on the blue part of the spectrum. Edit: I have just seen your edit. You're right. There you can see the details. Here's a quote ...


6

There are several different things that need to be explained / explored here. First - the speed of light in vacuum is independent of frequency / wavelength. The same is not necessarily true for light in any medium other than vacuum: this is why we can see rainbows! Second - not all objects emit "white" light. The emission spectrum of a star depends, among ...


6

It depends on your definition of a "color". There is a crayon in my box of coloring pencils that is "white" - you could call that a single color. But if I look at the spectral components of a light source that my eye perceives as white, there could be many different compositions, all of which look "white" to me. It might be "all the colors of the rainbow" ...


6

I wrote an earlier answer to the question: "Is true black possible?" which addresses some of the points you mention. It includes this image of "the blackest substance known": In particular, most "black" objects have a small amount of reflectivity, from which you can deduce clues about their shape. But the sample of really black material which I show is SO ...


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 ...


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 ...


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 ...


4

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 ...


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 ...


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 ...


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

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 ...


2

Even if a perfectly black object reflects no visible light, we can still distinguish its shape from a non-black background. Most real-life black objects are just much darker than the surroundings, not perfectly black.


2

The leading answer to this (closed) question gives some good reasons why IR vision did not develop widely across the animal kingdom. To paraphrase, sensing even the near-IR spectrum would require a different type of sensor compared to more or less regular chromophores, and there would be limited evolutionary pay-off for detection. The advantage gained over ...


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.


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 + ...


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

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 ...


2

"Black Gold: Plasmonic Colloidosomes with Broadband Absorption Self-Assembled from Monodispersed Gold Nanospheres by Using a Reverse Emulsion System" describes a similar phenomena of red gold nanoparticles assembling into black 3D structures, except they had to go through more steps than just drying them.


2

The object is red because it absorbs all other wavelengths more than it absorbs red, which gets reflected. The blue and green parts of the spectrum get converted into heat. The process in solids is more complex than in atomic physics because more than a single atom is involved in it, which allows the conversion of light (photons) into phonons (quantized ...


2

In most situations we are familiar with when it comes to light in everyday life, the interaction with light and matter is linear with respect to the electromagnetic field. This means that if you have a material that scatters 50% of a beam of light, then a 2mW beam will scatter 1mW or a 100mW will scatter 50mW etc… In these normal everyday situations, ...


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


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 ...


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. ...


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

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 ...


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 ...


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 ...


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 ...


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).



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