New answers tagged

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@Shimanshu Gupta There is more to it than just the wavelength. First of all I assume you mean the path difference at the detection screen So you need to say how far that distance is, A typical slit experiment could be 1 meter (1,000,000,000nm) from the slit to the detection screen. Also you need to know where on the detection screen you are measuring to. It ...


0

I think you did the Fourier Transform (FT) correctly, the first plot is just shifted incorrectly. I'm not familiar with Mathematica's FT but in MATLAB this would be computed as fftshift(fft(data)). What you have shown is what you would see if you only plotted fft(data). That should address the first three questions. The answer to question 4 is yes, your ...


1

At the risk of giving an answer which is to complete, I'm going to offer interpretations for the indicated terms. Let me start by pointing out that the mathematical expression you are using describe the rotation of vectors in the complex number plane. This is convenient for dealing with phase differences, but we normally assume that only the real ...


1

A balanced beam splitter provides a 50% probability for the light from each path to end in each of the two output paths. An asymmetric beam splitter provides different probabilities. I do not know of a distinction between balanced and symmetric beam splitters, but I can envision some authors caring about the relative phases that the light picks up by ...


3

Moonlight is quite reddened compared to Sunlight. The lunar albedo is roughly 0.07 at 400 nm and double that or 0.14 at 700 nm. It doesn't look terribly red because our blue and red receptors aren't pinned at those extremes, but for any given plant you would have to look at the specific receptor molecules at play and what their sensitivities are. See for ...


6

In addition to the other answers, an important physical difference is the perceivable inconsistency between the apparent intensity of the light source (moon vs. sun), and the spectral (color) distribution. Of course, this requires the (obviously wrong) conjecture, that both are thermal light sources, which behave approximately according to the black body ...


15

By far the primary physical difference is intensity- moonlight is much less bright than sunlight. Such flowers bloom when the light intensity is low- it does not matter if they are in moonlight or a greenhouse.


0

The physical difference between sunlight and the light scattered off of the moon aka moonlight is that the scattered light is polarized in nature. However, I am not sure if night blooming cereus uses this.


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Actually, both the other answer and Veritasium himself are wrong. The shape shown is not due to the pinhole effect (the holes are way too large for that). In fact, the resulting image isn't even a circle! If the sun were a point source - an infinitesimally small source of light - the triangle hole would indeed result in a triangular image. (Projection of a ...


1

If the beam splitter is a non-polarising device and the setup is ideal, then statistically 50% of the photons go in one direction and 50% in the other. But this does not mean that at any point in time the ratio is 2:2. All other scenarios - with lower probability - are possible: 4:0, 3:1, 1:3 and 0:4.


1

Two things you might want to do: your data aren't perfect, in particular, there are two vertical jumps in them, one at approx -2.6 and another at approx 1.5 (not sure if this is what you refer to as "dead time"). These vertical jumps are in fact phase jumps of your oscillations. You won't be able to fit the entire data set with a $\cos^2$ function ...


1

You only see the color that aims at you. From the higher drops that is red. From the lower drops that is violet.


1

To do this efficiently, you want an optical device known as an X-cube. This is a cube that has three input faces: for red, green, blue. The output face combines all of those. Look up "X-cube optics" on the web, (so you don't get X-cube toys). Since you also want a fourth channel, that is a complication. People would typically add a dichroic plate ...


0

In a medium there in general is dispersion, so that the group velocity depends on the frequency. Note that the phase velocity can even exceed $c$, namely when $n<1$ such as occurs for x-rays. A pulse of light contains many frequencies, each with their own amplitude and phase. These components all travel at their own velocity and with their own extinction. ...


0

It sounds like you want a projector. Normally you use it to project a multicolored image on a screen. You want to project 4 different uniformly colored images on a small screen, one at a time. Or perhaps you just want to put 4 different LEDs some distance from the screen and turn on one at a time.


5

This is going to be really hard to do with LEDs since they're not collimated. If they were instead lasers then you would have some options at your disposal. If you have control over the polarization then you can combine the beams using polarizing beam splitters with lower power loss (but you would have differently polarized light at the output, may or may ...


4

Suppose an EM wave is emitted by some source in a step fashion. That is, ideally, prior to some time $t_0$ there is no emitted wave, and after that time there is an emitted wave with some constant non-zero amplitude. An initial wave will leave the emitter at the speed of light in a vacuum, even if the light is passing through matter. This initial "front&...


2

If your system is equivalent to a single lens with focal lenght $f$, then you can proceed like if you had one only lens with that $f$, so $M_L=\frac{s'}{s}$, with $s$ the distance to the object


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How can an experiment built on the premise that light actually slows down in matter work, when light actually travels through the apparatus at the speed of light in a vacuum and only the phase-velocity of the superposed wave is lowered? Indeed. The correct functioning of the experiment is certainly good evidence supporting the premise upon which the ...


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I don't have Feynman's lectures available, but I am not sure if you have interpreted him correctly. The fact that the speed of light is different in matter (and varies with wavelength) is the foundation for such profound effects as refraction and dispersion. Just take any prism and see the effects, and you'll know that the speed of light in matter is real.


0

This Numerical simulation by GONDRAN Alexandre is for $R=10\lambda$ The figure on p.5 here goes out to 30R. At $4000R$ you are out of the Fresnel regime. (See the condition on the Fresnel number in Wikipedia). I can't find an image of Fraunhofer diffraction from a disk, but by Babinet's principle it looks (roughly) like the negative of the Airy disk from ...


0

To locate the light circle, look at reflection angles from various parts of the mirror.


1

A metal is reflective because it is conductive. Conductivity will limit reflectivity as it gets thin and narrow. There are considerations beyond this of course if you want to make a solar sail. A metal wire mesh as thin as possible would be weak and fragile. Strength matters. Manufacturability of giant a sail matters. Ability to survive Earth's atmosphere ...


2

An ideal lens takes a point source that is located on its focal plane, say $\mathcal {F}$ emitting monochromatic homocentric rays/spherical waves and transforms them into parallel rays/plane waves whose direction (propagation phase) depends on the location of the point source in the focal plane relative to the symmetry axis. By reciprocity it also goes the ...


2

Assuming everything is correct, and this red light is not a reflection of some other light source or something: one possible way of achieving this is that the medium is absorbing green light and emitting the red light through photoluminescence. So in this case it's technically not light changing frequency in the medium; it's "new" light produced by ...


1

Notation in Figure-01 : \begin{align} n_1,n_2 & \boldsymbol{=} \texttt{indices of refraction} \nonumber\\ \theta_1,\theta_2 & \boldsymbol{=} \texttt{angles of incidence and transmission} \nonumber\\ \mathbf i & \boldsymbol{=} \texttt{unit vector on incident ray} \nonumber\\ \mathbf t & \boldsymbol{=} \texttt{unit vector ...


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The inverse Fourier transform differs from the Fourier transform only by a flip operator, so for many purposes they can be treated as the same thing. Hence you can apply an inverse transform simply by using a second identical lens.


1

Yes. If you put your thumb in the direction of propagation of light, then as time passes, the electric field vector can move clockwise or anticlockwise only with the help of either your left hand or your right hand. Whichever hand it is gives the sense of polarisation. This tells you the direction in which field vector moves as time passes.


1

There is actually a very intuitive physical difference between left circularly polarized light and right circularly polarized light: left circularly polarized light gets transmitted through a left circular polarisation filter, while right circularly polarized light does not right circularly polarized light gets transmitted through a right circular ...


1

This is a deep question that can be treated at different levels of sophistication. I won't attempt to provide a complete answer, but I can chime in with some information that I hope is insightful. The most basic level of understanding comes from the classical picture of light as a wave in the electromagnetic field. (1) electrons (and all charged particles) ...


1

First off, you must be careful to avoid too sticking too much to a classical picture where the atom is mostly "empty" and the photon is a point particle that you imagine should "pass through it easily". Texts that state that the photon needs exactly the right energy to excite the electron may be trying to keep things simple, but I ...


1

Some substances do absorb photons of just the right frequency. When you shine a white light on them, you see sharp absorption lines in the spectrum of the reflected or transmitted light. The excited atoms may decay and emit the light again. If they do, they emit in all directions. Sometimes they decay in more complex ways. For example, a fluorescent atom may ...


1

First, you confuse permittivity with refractive index. The refractive index of silver for the visible wavelengths is almost 0. Secondly, all laws still apply (ie Maxwell eqs). However, I think its easier to think of it this way: metals have conducting electrons, and they oscillate mostly freely. They are not bound in a potential which affects their mobility ...


0

Thanks for all good comments and answers! After thinking again about the problem I'm posting an additional answer because of one major reason that wasn't clear to me and which is explaining why you see a faint "visual-like" image in thermal images: Diffuse illumination, pretty much like in normal visual images. In normal images you see things ...


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The suns rays here on earth are only approximately parallel due to the angular extent of the sun. Light coming from opposite limbs of the sun will be about half a degree from parallel. For some purposes, that's close enough and we consider them parallel. For other purposes it isn't. The atmospheric distortion isn't going to add much more than the half ...


2

Such fibers are waveguides which work as photonic crystals, that is, periodic structures with alternating layers of dielectrics. Photonic crystals have stopbands, i.e. spectral regions where the reflectance is high (and transmittance is low). These stopbands allow the waveguide to keep the light inside and actually conduct it (not lose it due to scattering). ...


1

Dispersion, $D$ is defined as: \begin{equation} D=\frac{\partial^2k}{\partial \omega^2} \end{equation} for the whole system, which is length independent (answering second question). For the first question, then you can see that for highly multimode fibres, waveguinding will not affect propagation much, and the dispersion relation is that of the core's medium....


0

There are quite a few questions here so I will only answer briefly. No, the emerging photons are not entangled. In fact, they do not emergy in pairs. Since the input states can both be described by canonical coherent states, so too can the outputs. 2,3. Still not entangled. SPDC does indeed generate entanglement while [beam splitters with coherent-state ...


0

Wire grid polarizers almost perfectly reflects the orthogonal polarization that does not pass thru the polarizer. There is a small amount of Joule heating in the wires.


1

You are correct, the angle of the transmitted beam depends on the order of diffraction and the frequency shifted. The effect can be significant, for example, we use AOMs in our lab to shift the frequency of a beam to image $^{87}$Rb atoms, we see a sudden drop in intensity after some change in frequency due to the shifted beam not coupling into the fiber ...


-1

The "duality" only comes in when you make a measurement. So the confusion about this duality is really just confusion about the measurement problem in Quantum Mechanics. The fact that light is a wave is reflected in the description of the electromagnetic field as an element of a complex linear vector space. The particle aspect is simply reflected ...


-2

While the classical electrodynamics studies light as a wave, the quantum theory takes account for wave-particle duality, i.e., it pretty much already takes a single prospective. The previous answers used term quantum electrodynamics, which makes sense in order to differentiate it from quantum mechanics, which does not quantize the light. This designation ...


-4

I interpret the question as follows: is there a field that studies electromagnetism either exclusively from the wave concept or exclusively from the particle perspective. The answer to this question is no. There are fields of applied physics that use only the wave perspective, but this is a practical choice. Ray optics and wave optics are tools for optical ...


0

$v=f\lambda$ refers to the relationship between the speed of a wave that is travelling through a medium to its frequency and wavelength. Now, forget the waves and come to oscillations for example like the simple pendulum. The position of the Bob is $x=x_0 sin(\omega t+ \phi)$ . If you differentiate this with respect to time, we get $\dot x = \omega x_0 cos( \...


0

A telescope is a system of two lenses Diagram of a simple telescope. Parallel light rays enter from the left, pass through the objective lens, come to a focus at the focal plane, and exit through the eyepiece lens. The focal length of the objective is F, and the focal length of the eyepiece is f. The magnification happens from the first lens on the left ...


0

If I understand the setup correctly, then the "fringes" you'll see would not be due to interference but because of different reflected orders passing through the lens at different angle producing different spots on the screen. The first beam enters the lens at an angle $\theta_1$ given according to Snell's law by $$ \sin(\theta_1)=n\sin(\theta) . $$...


0

As José Andrade pointed out, Amplitude is just how strong the light is. However, there is a problem with José's assumption that: the color does not change. They look the same. And @Not_Einstein is right when he writes: as the intensity is one component of color, the color will change. Maybe not the hue, but the color will. Bright red is not the same ...


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Yes, that theory is called quantum electrodynamics, or QED. It predicts all of the “particle” behaviors and all of the “wave” behaviors of light from a single unified mathematical model. QED was the first successful quantum field theory. Since its development other similar field theories have been developed for the strong and weak nuclear forces. So ...


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