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1

A "ray" in geometric optics is a locus of continuous propagation of light. Think of it as mapping where the energy is going in space. In principle there are an arbitrarily large number of them, but we draw a manageable number for visualization purposes. The various [letter]-rays were so named when people didn't know what they were beyond being things that ...


2

Individual photons are not considered rays. Because of the wave and particle nature of photons, they are much more complicated than what they are generally thought of: a projectile of light. In fact, they do not have an exact measurable position, but do travel in straight line trajectories. What we consider rays are lines perpendicular to the wave front of ...


1

Short answer: use your technique, but use scale factors of 0.690 for L, and 0.348 for M (instead of 0.542 and 0.575), and you will reproduce the luminosity function. Long answer: You're on the right track. I tried to find 'official' scale values for the LMS curves online to combine into the luminosity function, but couldn't quickly get them. You are ...


0

Something else that I don't think has been mentioned - most solar panels are limited by the lowest power generated by any cell in the array. So, to boost the power by reflection, you would have to ensure that the reflected light covered the entire panel to gain anything from it. This of course reinforces what has been suggested - adding power by adding ...


2

The colour you see in the sky on cloudy nights is due to the reflection of city lights off the clouds. In rural areas, a cloudy night is, as you expected, significantly darker. However, the massive amount of light given off in urban areas reflects back to Earth when there is cloud cover. And so, you see a red-orange hue, similar to the overall colour ...


2

The ray theory of light is equivalent to the Eikonal Equation, which in turn is essentially a slowly varying envelope approximation to Maxwell's equations. If we write the electric and magnetic field vectors as $\mathbf{E}\left(\mathbf{r}\right) = \mathbf{e}\left(\mathbf{r}\right) e^{i\,\varphi\left(\mathbf{r}\right)}$, $\mathbf{H}\left(\mathbf{r}\right) = ...


2

The accepted airglow answer might be technically true, but it does not answer the question! The existence of an additional and very faint source of green light in the atmosphere does not explain the absence of the green light in the sunset sky gradient. I wasn't satisfied with other answers either. The only satisfactory answer I could find is this one. ...


2

When a wave travels through a rope, the rope goes up and down, the position of all the 'rope-particles' changes, they oscillate and this makes up the wave. With light, it is indeed the electromagnetic field oscillating, but you shouldn't think of the arrows that represent that field in your first picture of light as 'extending into the rest of the space'. ...


1

What would happen to light passing through a narrow space between the event horizons of two equal-mass black holes? Would it deviate or follow a straight path? Like iharob and JohnnyMo1 said, the light goes straight. But something else happens to it. See this screenshot of Irwin Shapiro's seminal paper: See where he said the speed of light depends on ...


1

I'm going to assume you mean that the light travels on the precise center line between the holes, as iharob did. This sort of symmetry question is very common in physics. Here's a similar question in classical electrodynamics. "If I place a positive charge at the center of a perfect equilateral triangle of equal negative charges, will it move?" Let's say it ...


2

I think the answer is very simple if you ask another question, you are implying that both black holes generate the same gravity and that the light passes exactly through the middle between them, so the question is If the light deviates, where will it deviate to? Since given the conditions it's not possible to give an answer to this question, it means ...


0

White paper works well as a diffusor. If you can make the box a couple centimeters thick, then you can place sensors at least 10cm apart and they will easily detect the laser's light. For signal processing purposes you always want to modulate the laser with a couple kHz and use narrowband AC amplifiers to detect the signal of your photodiodes. This will ...


0

How about one or more layers of the plastic diffusers used in ceiling fluorescent fixtures? Go to a Big Box Hardware Store and you'll likely find acrylic sheets embossed with various diffusive patterns.


0

You might try something like a ring of plexiglass in front of the white layer, which would use internal reflection to carry the light to sensors.


5

For many materials the change in refractive index over the range of visible wavelengths isn't huge, so it's not a bad approximation to take a single value. The range of visible wavelengths is from about 400nm to 700nm, so the middle wavelength is 550nm. As it happens, the sodium D lines are not far from this, at 589nm, and since they are bright and easy to ...


1

Parallel rays reflecting on a concave mirror do intersect at one point, the focus, if the mirror is a parabola (in 2d plane geometry) or paraboloid (in 3d space geometry).


0

Perhaps it is working as a grease spot photometer as well. The display back light may just be much brighter than the diffuse light coming through the logo. http://www.phy6.org/outreach/edu/greaspot.htm As mentioned in the other answer it is possible that the LCD is transmitting less light when it is operating and so blocking out the back light (and the ...


0

Intensity is an objectively measurable attribute of light. It is the rate at which energy is delivered to a surface. Intensity is energy delivered per unit time per unit area. The intensity of light is a measurement of photon irradiance, which is the number of photons delivered per square meter per second. You can measure intensity with a photoelement, ...


0

First of all, polarization vectors ${\epsilon^\pm_i}^\mu$ can be shifted by gauge transformations such that a quantity proportional to the corresponding Minkowski momentum is added to it: $${\epsilon^\pm_i}^\mu \rightarrow ({\epsilon^\pm_i}^\mu+A_i p_i^\mu)$$ It is easy to show that the above equation is invariant under any such gauge transformations: ...


-2

Most likely is its always hotter than the outside. the frames absorb heat up into the 160 plus mark depends on where you live. 100 degrees usually about 150 degrees gained we call it solar gain.


1

This question is too broad. It involves ALL the objects in the universe which have a surface, i.e., everything. I'm going to avoid giving a lecture here. In some liquids and most gases the electronic structure of each individual atom or molecule is enough to describe their spectra. The "property" you are looking for in the case of solids is the band ...


0

Several answers here already talk in great detail about how electron orbitals affect if a photon will be absorbed or not, but this is not the whole story. The color from reflected radiation is indeed the only factor if the surface is completely flat and perfectly reflective, excluding the black-body radiation, but most surfaces are not. Take for example all ...


4

The atoms in the lattice can be thought of as coherent re-radiators of the incident photons. This is not unlike the scenario we have in a double slit experiment, where a Huygens construction of the wave front considers each point in the slit as a radiation source. So it might be "opinion" but I think that diffraction is an appropriate word to use.


0

An endoscope uses a bundle of multimode fibers. The number of fibers determines the resolution of the image transported through the fiber bundle. The angle of light isn't preserved in a single fiber. Usually - depending on the length and bending of a such a multimode fiber - angles/modes will be mixed in some way; therefore you need a bunch of fibers to ...


0

If the two sides of the prism are parallel, then the ray will exit the prism at the same angle it entered, just offset. If you need the offset, or the faces are not parallel, then you need to calculate the exact path the ray takes through the material. That requires knowing the index of refraction for the material. If you have that, then you can use ...


0

Here's a plot of atmospheric water vapour (fog) which is the green line. The horizontal line is wavelength and the vertical line is attenuation. Liquid water is red and ice is blue. From wikipedia: The absorption of electromagnetic radiation by water depends on the state of the water.The absorption in the gas phase occurs in three regions of the ...


0

Think of it this way: You are way up in the sky and the distance you see as a centimeter could be meters long since you see objects getting smaller as you go farther away. Assuming that an ordinary water wave travels with a velocity of 3 or 4 m/s at a maximum, it is not hard to imagine that you are seeing them as if they are standing on the ocean. ...


1

I imagine this effect has to do with the fact that velocity is relative. When you're on the shore, you gauge the velocity of the waves with respect to the shore. When you're in a plane, you're likely gauging the velocity with respect to the other wave crests, which are moving at the same velocity and so there is no apparent movement.


-1

Can electrons reflect light? Yes. Like CuriosuOne said, electrons are shiny. I kid ye not, google on electrons shiny. Metals are shiny because they have free electrons. Check out this question about the colour of metals, where Ali said a metal is are silvery because it "reflects all wavelengths specularly (more or less)". Also see this article by William ...


1

What you learned is correct. More simply, it's a consequence of the "time reversal symmetry" of most of fundamental physics. This symmetry is still present in general relativity. But, it's obscured by the standard system of coordinates. When you transform these coordinates into the Kruskal coordinate system, you not only have a black hole, you also have ...


-1

I learned in school (very long ago) that the path followed by light is independent of its orientation on this path. There's 2 things going on there. One is that our knowledge has changed since you went to school (or more correctly when the textbook author went to school), and the other is that these statements tend to hold true only at human timescales ...


1

White light is a mixture of all wavelengths in visible spectrum. The blue glass has a property to absorb all colors except blue.Hence only blue is transmitted and thus the light seems to be blue.Similarly that yellow light might not be pure ad must be containing some amount of red light,which gets transmitted singlehandedly.I hope this helps!


0

Along w/ the info in the other answers, keep in mind that "white" is not a fixed item. In very dim conditions, your cones don't work, and the rods in your retina only report intensity, not color, so everything looks white/grey. In very bright conditions, your retina overloads, and all you sense is 'white.' Now, one can define "white light" analogous ...


0

Electromagnetic waves are only an observed phenomenon. There is no travel. As an example we can choose a photon which is "traveling" from Sun to Earth (for simplicity we do not care about gravity issues here). That means that Sun is emitting a bit of energy (a momentum) which is received by Earth. So far everything is OK. But what is happening between ...


0

The speed of light it is medium-dependent, note $c/v= \mu$. But in this case, it is correct to say that a photon has a KE given by $E = hc/ \lambda$ in vacuum. At the interfaces between mediums, however, using Huygen's principle, $\lambda$1$/ \lambda$2$= v$1$/ v$2. So light has a definite speed in a given medium, with its energy (note it only has kinetic ...


0

Since, electro magnetic waves have electric and magnetic vector. Due to this EM waves show electric and magnetic field. An electric and magnetic field have no need a medium to show thier effect. Hence in the presence of electric and magnetic field vector which vibrate perpendeculer to each other and get pertervation EM waves travels in vacuum.


1

"It is known" that each atom has a characteristic atomic emission spectrum, as long as the atoms are isolated from one another. Emission spectra are usually observed in gases at low pressure. But when the atoms are compressed into solids or liquids, the close proximity of the atoms distorts the environment in which the emission takes place, and shifts the ...


2

Yes and no. Fusion inside the sun produces light - but the atom are moving so fast that their electrons are not attached - it is a plasma. As such, you would be hard pushed to find emission lines in the sunlight. You will see some absorption lines - the colder hydrogen and helium further out will absorb little bits of the radiation. What you are left with ...


0

You may be inquiring about a mechanism to separate sound vibrations which originated yesterday from sound vibrations which originated just an instant ago. If you stood inside the Capitol dome with sensitive enough apparatus, could you hear the residual vibrations of voices which spoke yesterday or a hundred years ago? The question is (1) whether vibrations ...


3

Time is a continous flux that goes from the past to the future; you can't stop it, you can only accelerate or decellerate it. From relative theories and from quantum mechanics we know that, in the most of case, the max speed of informations is the speed of light. So we can see events from the past: we can observe the light of a star that, now, is died. Only ...


0

A high level description. In the case of reflection at a metallic interface the electric field of the photon forces the electrons in the metal to oscillate. The oscillation means that the electrons are being forced at the same frequency as the photons. Thus as the electrons in the metal oscillate they begin to emit light in response. The frequency, ...


0

To make an object "invisible", one important thing you have to do is to stop it from blocking the light behind it from the observer's perspective. If you have an opaque object that emits/reflects no visible light, then what you will see is a black silhouette of the object. The only way it will be invisible then is if it's against a completely black ...


1

What if I say that the mirror doesn't flip left and right? You've heard it right the mirror doesn't do the flipping. As the above answers say the mirror shows what is right infront of it. It's you(we humans) who think it is flipping. Let me get this in detail Before we begin tell me , 'What makes you think that the mirror flips your left and right?' Or ...


0

To summarize: hotter => bluer, but more radiant => brighter. Something which is very hot, but invisible, would need to be small and have an energy output proportionate to the lower surface area. An energy source of a given number of watts is indeed easier to "hide" if it emits mainly at higher frequencies. This is what makes cobalt-60 dangerous when ...


0

1) It's true that the peak wavelength for a black body decreases with temperature. But let's say you want to know what temperature has what peak wavelength. Well, you can Google on "peak wavelength temperature calculator" and try for yourself. But I'll give you the short form. Since visible light is in the range of 400 to 700 nm, your body would have to be ...


13

Planck's Law gives us the intensity of black body radiation as a function of temperature: $$B(\lambda,T)=\frac{2hc^2}{\lambda^5}\cdot \frac{1}{e^{\frac{h c}{\lambda k_B T}}-1}$$ If we plot a normalized plot of this curve for different temperatures, you see the following: As you can see, it does look like the higher temperatures make the relative ...


37

You're right that as the temperature increases, shorter wavelengths receive a higher proportion of thermally radiated power, and longer wavelengths a smaller proportion, because of the shifting Boltzmann distribution of your molecules' kinetic energy, and therefore the shifting power spectrum of the light they emit. However, most of the objects you see ...


2

Infrared lasers are much more dangerous to the human eye compared to a visible laser of the same power, because infrared lasers do not trigger a blink reflex, which means the laser has much more time to damage your retina. Your other questions can be answered by reading about the many differing ways that visible and infrared light interact with matter via ...



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