# Tag Info

0

To add to the answers so far, here's a outline of a pretty crude calculation you could do. Let's assume the house is in space. That way, we won't have to worry about the other parts of the Earth equilibriating with the house. Now let's assume your house is a perfect blackbody (which may or may not be a good approximation depending on how your house is ...

0

Yes. All matter that interacts with light absorbs it to some degree. This is true whether you're discussing UV or infrared or visible light. For example, the earth wouldn't be nearly as warm as it is if it didn't absorb visible light.

1

Yes, it would, though not as quickly as if you were getting the full spectrum of sunlight. All frequencies of the light spectrum carry energy, so it becomes a question of how much of that energy is absorbed by the house. For example, if your house was completely black, all that visible light energy would be absorbed by the house and converted into heat. If ...

0

Given the context of the question, the fact that it seems to be about $E=mc^2$ specifically, and that the OP says he's having a hard time understanding it, i'm going to try and give a simple answer in plain english without a load more complicated formulae. I am no physicist, and although the concept may not be that easy, the formula is pretty simple, maybe ...

1

I will try to answer this question with my basic understanding of special relativity: Is matter condensed energy? It kind of is, but a better way to phrase it would be that everything that has energy, (behaves like it) has mass. Imagine you have a hollow box with the insides covered with perfect mirrors and you put it on a scale. If you shone a light ...

15

The famous equation $E = mc^2$ is actually just a special case of the relativistic equation for the total energy: $$E^2 = p^2c^2 + m^2c^4 \tag{1}$$ where $p$ is the relativistic momentum and $m$ is the (constant) rest mass: $$p = \frac{mv}{\sqrt{1 - v^2/c^2}}$$ For an object that isn't moving $p=0$ and equation (1) becomes: $$E = mc^2$$ which is ...

6

does this equation mean masses are just condensed energy? No, it means that mass is just another form of energy, just like heat, motion, electric attraction, etc. For example, the energy of a charged sphere is $$E=\frac{3}{5}\frac{Q^2}{R}$$ This equation doesn't mean that charge is just condensed energy; it means that charged objects have energy. ...

1

As all the other commentators have said, both diffraction and interference are manifestations of the same thing: the fact that waves superpose. The two words are slightly different, but it's clunky to work with two concepts when you only need one. You never do a "diffraction calculation" or an "interference calculation", you just do wave mechanics and ...

1

These high reputation members aren't confused about the difference between diffraction and interference. The two terms are synonymous in the physics community (we're not arguing on the basis of linguistics. I can agree the terms are linguistically different just because people sometimes prefer one over another in certain contexts). The way the physics ...

0

The electromagnetic radiation of gamma rays and X-rays is generated through accelerative processes involving atomic nuclei. The involved photons are the so called hard radiation and could destroy human tissue. The acceleration of electrons generates photons in the range of UV radiation, visible radiation and IR radiation. It is possible to send informations ...

1

Whether the amount of diffraction is 'negligible' depends on how you define this criterion. The first order minimum in the diffraction pattern from a single slit occurs where $\sin\theta = \lambda/d$ where $d$ is slit width, $\theta$ is diffraction angle and $\lambda$ is wavelength. If $d = \lambda$ the central lobe of the diffraction pattern will spread ...

0

The key thing is that the surface have facets. That is, it has to have smooth flat parts that can reflect light like a mirror. If the surface is just amorphous then the scattering will tend to be too disorganized to see the polarization. I have seen polarized light coming off quite surprising surfaces. A manhole cover for example. It had been polished fairly ...

0

In the classical theory of reflection (and refraction) of electromagnetic waves, there are equations which describe the reflection of light in two specific orientations. They are known as the Fresnel equations. However, the polarizations of light lie in a 2D vector space, so as long as you decompose any incoming wave of light into the two linearly ...

1

See Snell's Law. $$n_1sin(\theta_i)=n_2sin(\theta_o)$$ where $\theta_i$ is the angle the incoming ray makes with the perpendicular to the surface between the two substances, and $\theta_o$ is the angle the outgoing ray makes with the perpendicular to the surface between the two substances. This can actually be derived from Maxwell's equations, but that's ...

1

To be clear, Maxwell's equations are known as "Lorentz-invariant" equations, which means that they take the same form in every Lorentz-transformed frame of reference. Special relativity actually came about from studying Maxwell's (classical) equations without charges or currents. Then we get: $$\nabla \cdot \mathbf{E}=0$$ $$\nabla \cdot \mathbf{B}=0$$ ...

0

Yet another way. Consider a wave of a form $f(kx-\omega t)$ in the emitting frame, being observed as some other wave $f^{\prime} (k^{\prime} x^{\prime}-\omega^{\prime} t^{\prime})$ in a frame moving at $v$ along the x-axis. We actually don't care what $f^{\prime}$ is, only the argument of the function matters. So use the Lorentz transforms to change $x$ and ...

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Classical electromagnetism is perfectly compatible with special relativity. In classical E&M, light is an electromagnetic wave and there is generally no useful formulation in terms of particles. The most widely used technique to combine quantum mechanics with special relativity is relativistic quantum field theory. The relativistic QFT that ...

0

Each of those spectral regions involve electromagnetic waves (photons) at a range of frequencies, and it is that which determines how they are generated, how they interact with matter (i.e. Particles with charge, one by one or in various groupings and configurations like atoms, molecules, bulk material composed of different molecules, different temperatures, ...

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Neither origins nor emitters nor receivers are required: consider a photon with momentum $\vec{p}$: it has 4-veclocity $p_{\mu}=(||p||/c, \vec{p})$. To compute what another observer moving at $\vec{v}$ sees, do a Lorentz transform: $p'_{\mu} = \Lambda_{\mu}^{\ \nu}p_{\nu}$. From this you'll get the relativistic (time-dilated) Doppler effect and stellar ...

2

Cort and Ilmari have given good answers about the practical issue: the inverse square law is for point sources, and so a non-point source (like an emergency light) will only appear to have the same properties at some minimum distance that depends on the geometry of the real source. However, it seems nobody has mentioned a different "minimum distance" that ...

2

The quote from the reference says it all: (I added caps) "The minimum test distance IN PHOTOMETRY of these sources is called the 'minimum inverse-square distance.'" The minimum distance is therefore a photometry issue, in other words, a measurement problem. The essence of the measurement problem is how far away you have to be before you can approximate the ...

3

The inverse square law says that the intensity of incident light falls off in proportion to the inverse of the square of the distance from the light source. The important word here is "the distance" — the inverse square law implicitly assumes that all parts of the light source are at the same distance from the measurement point, or at least ...

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As many have said, the inverse square law applies to point-sources. These are idealized light sources which are sufficiently small compared to the rest of the geometry that their size is of no importance. If a light source is larger, it is typically modeled as a collection of idealized light sources, potentially using integration. The exact definition of ...

1

The first thing to note is that each of the slits produces a diffraction pattern the width of which is controlled by the width of the slit and the wavelength of the light. The amount of light travelling from a slit in a particular direction is controlled by the diffraction pattern due to a single slit. The light waves from each of the slits superpose ...

0

Wether you get destructive interference or constructive will depend on the position. What you get is something in the image. Where the diagonal lines indicate maxima where constructive interference occurs while between them there is destructive interference.

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The inverse square law applies to point sources. For extended sources becomes accurate at distances that are large compared to the size of the source. At large distances the source looks like a point. What "large" means depend on the application. In the case of light fixtures, the Illuminating Engineering Society and other organizations have made ...

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The inverse square law applies to point sources. A real emergency light is not a point source, and therefore the law appears to not apply at close distances, because any real point is at a varying distance from different parts of the emergency light.

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TL;DR: the pictures given are at least inconsistent, if not wrong. It is not even clear what is plotted. Let d be the separation of your slits (i.e. from centre to centre), a be the width of a slit and N be the number of slits. Then in the using a scalar theory of diffraction in the Fraunhofer limit, we can write the amplitude $\phi$ for an plane wave of ...

2

The first thing you should do is cover your detector with an optical filter that only allows through light with the same wavelength as the laser you are looking for.

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A 40W incandescent light bulb has a luminous efficiency of 1.9%. That means only 1.9%, or 0.76W, of the energy consumed by the bulb ends up as visible light. LED bulbs have an efficiency of around 10% - the efficiency depends on the design and can be as high as 15% or as low as 8%. So a 6W LED bulb will produce between 0.9 and 0.48W of visible light. The ...

31

You're mixing power needs with luminous effects. According to the advert you posted, that LED bulb consumes 6W (power) to get a luminous flux of 500 lumens (lumen is a photometric unit, like candela or lux): https://en.wikipedia.org/wiki/Luminous_flux On the other hand, an incandescent bulb would need to consume 40W (power) to get the same luminous flux. ...

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The advertising suggests that the new 6W bulb generates as much light as a 40W incandescent light used to. Most of the energy in an incandescent light bulb is converted into infrared radiation that we can't see or it's at the red end of the spectrum where the human eye is not very sensitive. This is a typical incandescent spectrum and you can see how ...

-2

My friend once had the same question. We came up with this: Glass is transparent. It is transparent due to the fact that we can see light from the other side of the glass. In other words, there is little absorption by the glass particles of any wavelength of light (visible) . So all (most) of the light gets transmitted to our eyes. Hence transparent.

1

Contrary to what the other answer assumes, fog is not made of vater vapour, and light attenuation is not the reason why you can't see through fog. Fog is a suspension of microscopic droplets of liquid water in air. This is the same material we know as "cloud" when it doesn't reach all the way down to the ground. The fog is opaque not because light is ...

2

The light is reflected from a mirror. If the mirror moves a distance $\frac \lambda 2$ then the incident ray of light has to travel an extra distance $\frac \lambda 2$ to reach the mirror and then the reflected ray of light has to move an extra $\frac \lambda 2$ a total extra distance of $\lambda$. If the path length changes by one wavelength then there is ...

1

Things don't really have "colour". Colour is a perception that we have of light entering our eyes. We use our eyes to perceive the world, and in our eyes are a series of receptors (photoreceptors). These are the rod, and cone-shaped cells which exist in your retina. These cells send electrical signals to our brain, via the optic nerve, and it is a ...

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Every Element/atom has a different electron configuration. This gives the valence electrons unique energy levels and arrangements. When electrons absorb energy they are excited to certain and again unique energy levels. When The energy is released it gives a photon a certain frequency which we perceive as a certain color.

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A lot of things can affect the color of an object. As you've mentioned, absorption plays an important role in determining what part of the visible spectrum gets subtracted from the color your eyes perceive. Optical bandgap arising from the microstructure of materials determines what portion of the spectrum is absorbed. It is closely related to the electronic ...

1

It's an issue of contrast; in the classical wave experiment there is plenty of data, and the contrast between the peaks and valleys is very clear; but when you are counting one-by-one the pixelation remains obvious. Pixelation can be reduced by (a) more gray levels in each pixel, and (b) more pixels per unit of area. You can simulate this by taking off ...

5

This is one of the first examples of energy levels for electrons within the atom! If we take the Bohr model, which imagines that electrons circle the nucleus on set orbits Each of these orbits has a corresponding energy. The electrons are more stable at lower energy levels, and thus, prefer to be there. When you provide energy to the electrons (in the ...

2

The spatial coherence is due to the fact that even for a single emitted photon it's the same wave that reach the 2 slits. I'm noot too sure what you mean by that. Spatial coherence has nothing to do with photons, it comes from the apparent size of the source as seen by the observer. Every source you might want to use in an interference experiment (a ...

2

As for your last question, a similar experiment has been done though it doesn't involve a double-slit. It's called the Michelson experiment, and using mirrors it tests the interference pattern created by light when the light-waves are combined with time-delayed versions of themselves. By changing the distance of one of the mirrors, the time-delay can be ...

1

Just to add to the above answers, and since to did not limit your question to the visible range - if you define black as absence of light (photons emitted or reflected), then there is no such substance, because according to black body radiation model, everything with a temperature above absolute zero (which is essentially truly everything in the universe:) ...

0

You can get it as "clean" as you want - use sensors that only respond to a narrow wavelength (eg something so unlikely that it almost must be from your source...not IR for example!), set up your experiment so that the source will, statistically, emit a single photon every day or two to avoid the possibility that you're getting interference from the source ...

2

The key is in your words "to ... appear". I believe that it's a perceptual issue with how your brain processes the two kinds of images: a smooth rendering or a pixelated rendering. There is another possibility. In order to be sensitive to single photons, the detector is also going to be sensitive to very low-level noise. An image taken with a bright ...

5

Newton's 2nd Law of Motion gives the impressed force as $F=dp/dt$, so a physical theory for a massless particle exerting a force requires that the particle have momentum, $p$. First we will discuss mass, momentum, the force law, and Special Relativity. In Newtonian physics mass is identified in two ways: by it's inertia, or as the quantity of matter. The ...

1

Maybe there IS an effect similar to what you describe. In photoelectric absorption, the permissible departing electron states in a material determine the probability of absorption (the rate). This is easily seen in X-ray absorption near a photoelectric excitation threshold (like a K-edge), and is the basis of XAFS (X-ray absorption fine structure) ...

-4

According to the relation $$\mathrm{Force = Mass \times Acceleration}$$ force is defined as the phenomenon, which creates acceleration in a body with mass. The mass is for the body on which it acts and not the body which acts. Theoretically it is possible for light to exert force since it has energy and momentum. Hope your doubt is clear

32

Yes, photons can. See https://en.wikipedia.org/wiki/Radiation_pressure (and photons are certainly massless). PS In fact, any massless particle has momentum (which has a fixed value since they can only travel at the speed of light) and if it is scattered on a body, it changes its own and the body's momentum, which is what a force does.

1

You are very close. First consider how the different lights colors interact with each other, please see additive color for an explanation. Then if you consider a double slit experiment as shown on wikipedia with coherent monochromatic light. Now we know that sunlight is composed by continium of wavelength of slightly different colors, but you can think of ...

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