# Tag Info

1

Radiant flux does add linearly - two lamps with radiant flux P will sum to a total flux of 2P and match the single lamp with 2P. With lamps, we should all be careful to separate electrical power from radiant power. Different lamps will have a different conversion rate of electrical power to radiant power (especially in the visible wavelengths - many lamps ...

0

The professor is basically correct, though they left out some details. Here is where the 'larger' part comes in: For the same unit projected area (say, the area of your pupils projected onto the light source), an incandescent bulb must produce a lot more flux per unit projected area in the steradian direction being measured to achieve the same illuminance on ...

1

I don't quite agree with your numbers. In order of magnitude terms, if the planet receives 1400 W m$^{-2}$ and reflects 30% (which is implicit in your question), then if the average photon is say 600 nm (questionable given spectral type of Proxima Cen), there are $\sim 4\times 10^{21}$ photons per second per square metre incident on the planet. The area this ...

6

Other answers are saying it's frequency alone. That is not true. Frequency does, however, play the biggest role in perception of color. Especially when neglecting edge cases like humans hardly perceiving any color in low-light conditions due to cones being less light-sensitive than rods. Rods only support monochrome vision (brightness only; no color). ...

2

The answer is energy. Now for photons, E=h*f, so this will give you the answer as frequency. Frequency is proportional to energy. Now you are confused, because most people would say wavelength. It is true, for vacuum, where frequency and wavelength could both be proportional to energy. But in a medium, frequency is the only one that is proportional to ...

3

There is this similar question on a sister site which is well answered IMO. Actually It is not the wavelength or frequency that determines light absorption- it is the energy of the photon that matters. The energy of incident light should match the excitation energy of the chromophore. The medium itself can also affect light absorption by electronically ...

18

Frequency. As you mentioned, wavelength changes in different mediums, but frequency doesn’t. If you look at a red ball underwater, it still looks red, even though the wavelength of the light is quite different.

-4

The equation of velocity of light is $v=f λ$. If the source of light (frequency) is same then color of light depends on its velocity & wavelength. In optics color depends on bending of light e.g dispersion of light.

3

Wavelength or Frequency Wavelength ($\lambda$) and frequency $f$ are related by the speed of light ( $c$ ) : $$\lambda = \frac c f$$ So either gives exactly the same amount of information. I'll use wavelength for the remainder of the answer to be consistent. Velocity The speed of light in vacuum has no effect on color. It's a constant. The speed of ...

4

You're right to be confused, light is a special case because its speed (in a vacuum) is always the same value. In the equation $$\nu=\frac{c}{\lambda}, \tag{1}$$ $c$ is a constant. This means that for light (not for general waves), as soon as you know the frequency you immediately know the wavelength, and viceversa. "Colour" on the other hand is a little ...

4

Unfortunately, there's no easy explanation for this. It involves some basic atomic physics, so you will need a picture of how electromagnetic fields interact with matter. Consequently, some of this may be difficult to understand, but hopefully you can ask about (or read up) whatever you don't understand. Semoi's answer is correct, but a lot of details have ...

3

If an analogy helps: Picture walking towards a plowed field, which you meet at an angle to the edge of the field. The furrows run parallel to the edge of the field at which they meet. The furrows make it harder to walk over the furrows at an angle, and you compensate by turning into the field. However, longer legs mitigate this effect so that you don't ...

1

This question is still yet fully defined in Physics, because they require the analysis of light to be a duality; in which they are understood as both: 1) a “particle”, called a photon. 2) a “massless wave”, measured by it’s frequency. I believe the theoretical issue lies between: A) ”Newtonian Physics” (rules governing our understanding of ...

21

Snell's law tells us that the angle of refraction depends on the index of refraction, $n_1 \sin{\alpha_1} = n_2 \sin{\alpha_2}$. However, the question remains, why $n_{\text{blue}} > n_{\text{red}}$. In order to address this, we need a model for the refractive index. The refractive index $n$ of a material is related to the atomic transitions of the ...

12

To explain this, we need to use the Fermat's principle of least time. According to the principle of least time, a light ray always takes the path which is the quickest. We also need to know another fact that the speed of all different colours of light in vacumm is the same, but in any other medium like glass, water, etc. their speed is different. And it has ...

3

So your question is what is the microscopic reason for material dispersion in optics, meaning $n=n(\lambda)$ ("dispersion relation"), where $n$ is the refractive index of the medium and $\lambda$ the wavelength. Because by snells law $\frac{\sin \theta_{2}}{\sin \theta _{1}}=\frac{n_{1}}{n_{2}}$ (see FakeMod's answer) this means a different refraction for ...

0

In the context of qed, photons are slowed down in the medium because they are absorbed and re-emitted by electrons. May I be forgiven for repeating what I recently wrote in an answer on reflection by a mirror? (The reason is the same although the situation is different. Although a single photon can only be absorbed and emitted by a single electron, it ...

0

Too long for a comment. In contrast to @Annav answer, i strongly assert that there is absolutely no quantum effect in this problem (by the way, the rigorous theory of QED in dielectric media has still some problems). I think that the OP has all the key ideas for the problems of propagation speed, except that it is barely related with damping or with ...

-1

This should be a comment, In mainstream physics, particles of light are called photons, and are axiomatically defined in the standard model of particle physics. Classical electromagnetic waves are a superposition of a huge number of photons, all travelling with velocity c because they are zero mass. In a transparent medium, which is the one you are ...

-2

"different color" is a feeling in your brain. Red and blue is different feeling, the root of the different feeling is some different property of the photon that can result in different feelings. In the case of human eye, the property that make the difference of feeling is the frequency/energy of the photon. Photons with different energy stimulate light ...

1

At an abstract high level, it's because the boundary conditions are such that the interface between media is a timelike hypersurface. That's what breaks the symmetry between space and time. If the medium's material properties (e.g. index of fraction) were uniform throughout space but then suddenly changed to new spatially uniform values, then the interface ...

2

I think the spirit of your question is "If I somehow manage to shake a charge at the right frequency, will it emit visible light?". In this case the answer is yes: this is exactly what free electron laser does. But instead of having an object holding and shaking the charge, the charge itself is passed through an alternating-polarity magnet and thus ...

6

It's the same reason that light follows a straight path in a constant medium and follows Snell's laws when changing medium: there's destructive interference between all the other paths. You can actually get light to reflect at another angle if you remove parts of the mirror. If you do it right, the parts of the wavefunction going in the new direction remain, ...

4

Let me add a few things. A photon is an elementary particle, and as long as it propagates, it is in a superposition of states, meaning it is in a superposition of frequencies, and does not have a well defined frequency. You cannot know its frequency until you interact with it or absorb it. As a quantum mechanical entity photons can be in superposition ...

3

By shaking your wand, you would produce electromagnetic waves with the same frequency as your wand-shaking. The waves thus produced would be extremely weak ultra, ultra, ultra-low frequency radio waves which your eyes are not sensitive to. So you would still be in the dark.

2

If an observer's distance from a light source exceeds the distance required for light to travel to the observer, and we have the advanced technology required to see something that far away, and also know the exact location in space to look (although data transfer speeds are bottle-necked by light-speed, but let's imagine that bottle-neck doesn't exist...), ...

55

Some areas of physics are counter-intuitive. For them, your everyday experience is a poor guide to how the universe really works. This is one of those areas. Photons have no mass. They all have the same speed. Yet they have energy and momentum, and it isn't the same for all photons. If you are used to $p = mv$, this doesn't make sense. The explanation is ...

16

They differ in their energy. Special relativity states that $E=\sqrt{m^2c^4 + p^2c^2}$. For a massive particle, there is a one on one relation between its energy and speed. In the limit $m \rightarrow 0$ this is no longer the case. All massless particles move at light speed, but their energy/momentum can vary.

14

The only difference between the two is the energy they have. $$E=\frac{hc}{\lambda}$$ As you can see from the equation above, different energies means different wavelengths. Different wavelengths means different colors. It is important to know that even though photons are always massless and always move with the speed of light, that does not mean that ...

32

Although a single photon can only be absorbed and emitted by a single electron, it leaves that electron in exactly its original state. There is no record, and no way of knowing, which electron absorbed and emitted the photon. According to quantum theory, to calculate the result when any electron could have absorbed and emitted the photon, we must form a ...

-1

The cosmic microwave background is the farthest back we can see due to before that the entire universe was composed of opaque plasma, if someone somehow went far enough to observe the birth of the universe they would not see anything, until the universe was cold enough to allow light to pass through, that is the cosmic microwave background (it is in ...

8

You are confused because you try to imagine mirror reflection as absorption and reemission. Though on this site, this might be helpful for certain reasons, in this case you need to understand that reflection is elastic scattering. Elastic scattering is a form of particle scattering in scattering theory, nuclear physics and particle physics. In this ...

23

The actual working-out on the level of single photons is quite tricky. I can only give you this hint: There is a coherent interaction of the photon with a lot of electrons in the silver layer. Silver or aluminium coatings work, because these materials have free electrons (which is also why the are excellent conductors) If there was (mainly) absorption+...

2

Relation between transmission and density You ask for a mathematical relation between the intensity $I$ of the transmitted light and the number density $n$ of gas molecules. It is $$I = I_0 \, e^{-n\sigma r},$$ where $I_0$ is the emitted intensity (i.e. before the beam enters the gas), $\sigma$ is the (wavelength-dependent) cross section of individual ...

1

Clear air is colorless in the visible spectrum. Yet consider what happens when the Sun changes its altitude from high in the sky to just over horizon: intensity of solar radiation decreases for an earthly observer. That's due to scattering (Rayleigh scattering, in particular, is relevant here) from a larger amount of atmospheric gases. Also, when you go ...

1

If the opening is greater than one wavelength in size you will have a diffraction pattern, a beam pattern like a sin(x)/x function, and will have a main lobe and side lobes--and so will not have uniform illumination. To get relatively uniform illumination you would need an opening less than 1/4 wavelength in size. The opening acts like an aperture or ...

0

A gravitational well doesn't have an edge, the gravitational potential created by the Sun is smooth and extends to the infinity, reaching 0 only assymptotically. If you move in this potrntial, the differences in the potential will cause the shift in the observed spectrum of the light, but the shift is gradual. This can be most easily explained as photons ...

0

Your question mixes together light/matter interactions and light/light interactions. The pressure you refer to is light/matter interactions. Sound propagation in your context would refer to light/light interactions. With that context set, I can see a few ways of answering the question, depending on which conditions are allowed. One way of interpreting the ...

0

A beam of light can be modulated to carry an audio (or video) signal (sometimes in digital format as in fiber optic communication).

0

There is a difference between creating pressure due to force and creating a disturbance in a medium with bulk properties such as pressure, temperature, etc. You have not really asked a well posed question and based on your comments to existing answers it seems like you just want to pull one magic rabbit after another out of your hat. Acoustics is not a "...

0

"Pressure" just means force per unit area. Changes in pressure don't propagate. What can propagate through the body of some material substance is changes in its internal pressure--the cumulative forces exerted by the atoms or molecules of the substance on each other. There is no body of material substance in your thought experiment. The photons of light do ...

1

It's actually important to recall that this is true only by definition, because it's part and parcel of what refraction means. But refraction is not the only thing that can happen when light passes between two mediums, even though it is probably by far the most common thing that happens in this situation. There are solutions of Maxwell's equations where ...

0

Hmmm: $$R_A = 1000\,{\rm cd/lx/m^2}$$ where a lux (illuminance) is a lumen per square meter means: $$R_A = 1000\,{\rm cd/(lm/m^2)/m^2} = 1000\,{\rm cd/lm}$$ and a lumen (luminous flux) is a candela-steradian: $$R_A = 1000\,{\rm cd/(cd\cdot sr)} = 1000\,{\rm sr}^{-1}$$ which probably means it is 1000/4$\pi$ times brighter that isotropic re-...

0

In highschool we are taught about Bohr's atom model and how electrons emit light/ photons when transitioning from a higher energy orbit to a lower energy orbit. First question Where does this phenomenon actually occur in real life? Is it referring to when the sun/ stars emit light? Is it referring to when objects are heated? Visible light is a ...

0

Photons are emitted when a charged object transitions from a higher energy state to lower. This is the fundamental principle. And all sources of light can be explained using this. It is also to be noted that what we see as light is only a small range of frequencies, plus only when many many photons are involved. Now in the case of an incandescent bulb, ...

0

The commonest example of electrons jumping energy levels and emitting light is the fluorescent lamp, first popularised as neon lights. The light from a given band is of a single frequency or color. Different atoms have different energy levels and emit different colored light. For example neon primarily emits red, sodium yellow, mercury red, blue and ...

0

0

Let the following figurewhere the screen is at $x=0$,the object at $x=a$ and the convex lense(with focus $f$) is at variable point $x=b$. Then for sharpe image on the screen we have : $$\frac{1}{f}=\frac{1}{v}-\frac{1}{u}$$ $$\frac{1}{f}=\frac{1}{b}-\frac{1}{b-a}$$ $$\frac{1}{f}=\frac{b-a-b}{(b-a)(b)}$$ $$\frac{1}{f}=\frac{-a}{b(b-a)}$$ $$b^2-ab+af=0$$ It ...

0

As the other answer already tells you, computer colours are indeed fake, they rely on the way human beings perceive colour. I want to answer your question about generating UV and IR frequencies, but first, a detour. Let us say that the circle of minimum area that a human eye can discern has area A. For two such circles adjacently placed, the human eye can ...

0

The computer screen RGB color mapping protocol takes advantage of metamerism, which is the physiological optics effect wherein stimulation of the different color receptors in the human retina produces a full range of perceived colors in the brain- even though the retina is being presented with a combination of ONLY red, green, and blue wavelengths. We can ...

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