# Tag Info

28

Why will a blue ray bend lesser than a red ray through a slit of the size a little bigger than the wavelength of the blue ray? Don't think of bending. Think of diffraction like this: if you have a plane wave incident on a slit, then you can think about the space in the slit as being a line of infinitely many point sources that radiate in phase. If you ...

15

This is not an advertisement. Under the rubric of "do try this at home", I wanted to share one more thing that I discovered after writing my previous answer - but it is so unrelated to that answer that I thought it better to write this as a separate post. I discovered two interesting things. First, when you spin a coin on a hard surface, it "rings" with ...

7

Picture a straight line of balls confined to a line with no gravity, friction or anything else. The balls are free to move around the line apart from the fact that every two neighboring balls are connected with an elastic spring which pushes them apart when they get closer and pulls them together when farther than a distance $a$. When we take one ball on ...

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My answer will be quite close to that of PhotonicBoom although a bit more graphical. When it comes to light phenomena, there are different ways to comprehend them: we can use a wave picture (Hyugens-Fresnel), we can use the most modern picture we have (QED) or we can use something a bit more intermediate which is the picture of light rays travelling from ...

4

I agree it looks like a transverse wave - like the ripples on a pond. But I believe you are fooled by a simple thing: the waves you are looking at look like "illuminated ripples" but are in reality just changes in temperature (changes in brightness of the sun's surface). If you have a shock wave traveling out across the surface of the sun, what happens? The ...

4

The answer lies in QED and is quite complicated mathematically. As a simple explanation this is what happens: Photons follow all possible paths from the source to the screen and each path has its own probability amplitude associated with it. Summing up all these paths cancels out most terms in the summation and you end up with your desired final "classical" ...

4

Huygen's principle alone will not answer your question, however the Huygen-Fresnel principle modifies this to include wavelength. It states that every point in an unobstructed beam acts as a secondary source of wavelets with the same wavelength as the primary wave. The amplitude of the optical field at any point is then the superposition of all the wavelets. ...

3

There are a lot of good answers already here, but I think I'll add a few pictures that show the ways I have to force myself to think to keep track of the differences between light waves and light rays when thinking about diffraction patterns. Above we have some light coming through the center of the slit. If we draw it like this it seems to make sense with ...

3

I don't mean to take anything away from the previous great answers, but the "simple and to the point" answer is, a very qualified, yes. By qualified, I mean one must know the coin's composition, thickness, diameter(or shape), density distribution, country of manufacture, etc. If we make assumptions and restrictions, then it becomes possible to calculate ...

3

Take the wave equation $$\nabla^2\vec{E} = \frac{1}{c^2} \frac{\partial^2 \vec{E}}{\partial t^2},$$ and let $\vec{E}(\vec{r},t)$ be a solution. Indeed taking the real part $\Re(\vec{E}(\vec{r},t))$ yields the physical significant values. The initial values at $t = 0$ are $\Re(\vec{E}(\vec{r},0))$ and the problem arises here: this does not give you enough ...

3

I guess you have this image in mind: This hods just for a single photon, or elementary packet of light. An unpolarized beam of light contains a bunch of photons with many different polarizations. The total electric field will randomly jump all around, still the interactions with matter typically involve a single photon at a time. This means that if you ...

3

Not a complete answer, but this is a classic "sloshing" problem. The interaction between the fluid and the container wall, under the influence of the external (periodic) force sets up a (self-reinforcing) and harmful resonance. This is of immense practical interest: jet-fuel sloshing inside airplane tanks, for instance.

2

The pattern of destructive and constructive interference depends on frequency. A single tone will show exact cancellations and amplitude doubling, a narrow band signal will show slight damping and gain, and a wide band signal will show a position dependent filtering. The positions of gain and dampening also depends on reflections from the surroundings but ...

2

The other big advantage of $\omega_0 t$ is that it is the time dependence of energy eigenstates. Other time dependencies can be constructed, but they won't have such nice properties as $E=h\nu$, etc. The usual definition of a "photon" has it being an energy eigenstate, so in a sense the frequency must be fixed by definition. Other electromagnetic wave ...

2

The phase-shift occurs due to the complex nature of the reflection coefficient. The so-called critical angle is given by: $sin(\theta_M) = \frac{n_2}{n_1}$ As long as $\theta<\theta_M$ we have only partial reflection and a real valued reflection coefficient $R$. As soon as the critical angle is exceeded $(\theta>\theta_M)$, we have $\mid R \mid=1$ ...

2

They will look all over the place. If you take a particular photon, you will see it in a particular polarisation state, but it will have nothing to do with the photon next to it, or the one before. Actually, getting a perfectly unpolarised source of light is quite difficult. One of the best cheap options are the sodium discharge lamps, used extensively in ...

2

The gaps just need to be smaller than the wavelength of the UV radiation. Waves can only pass through gaps larger than their wavelength. In symbols $$L < \lambda$$ Where $L$ is your gap length, and $\lambda$ is the wavelegnth of the UV radiation. Typically this from is $400$ to $10$ nm, with UVA in the $400$ to $320$ nm range, UVB in the $320$ to ...

2

I start my answer with the the second question (similar situations). This is eerily similar to the free-particle dispersion relation (or the relativistic energy-momentum relation) in natural units ($c = 1$ and $\hbar = 1$), $$\omega^2 = k^2 + m_0^2$$ This system of units is commonly adopted in particle physics and quantum field theory. It can be clearly ...

1

The off-axis intensity is not sero, it is simply quite low. This article on the Wolfram web site shows the 2D diffraction pattern from the sort of aperture you describe along with the equation for calculating the intensity: $$I = 16C^2a^2b^2\left(\frac{\sin(\theta_xka)}{\theta_xka}\right)^2\left(\frac{\sin(\theta_yka)}{\theta_yka}\right)^2$$ where the ...

1

Another way to look at it. If we have two media with two different transmission coefficients, $Z_1$ and $Z_2$, we know that we have to account for all three parts of our incoming wave, so $$1 + R = T$$ where on the left of the boundary we have our incident wave (1) and the reflected bit $R$ and on the right of the boundary we have our transmitted wave ...

1

The interference between all the rays emitted from the aperture to a fixed point on the screen can be constructive or destructive, depending on the various path lengths involved (measured in wavelengths). If you change wavelengths, the path lengths (measured in wavelengths) change. What is constructive interference between paths at one wavelength can be ...

1

A wave is a perturbation in a system that propagates. The wavelength is the typical length along which a wave is coherent, which means that what happens at some position affects the wave behaviour in the vicinity if this point at distances of a few wavelengths. The reason for that is that the medium in which the wave propagates has some rigidity and the ...

1

The velocity of light is 299792458 meters per second. That is the velocity of displacement of a peak to peak as the wave passes. the wave in slow motion It is not clear what you want to do with your experiment but it is not possible to carry it out the way you envisage. There is no technology that could spin anything that fast. Now in your title you ...

1

This is a hard question to answer, in the end. However, be assured that long, long, ago we started looking for nuclear expositions by looking for X and gamma radiation using the Vela satellites http://en.wikipedia.org/wiki/Vela_(satellite) . These did not find much in the way of violations of the nuclear test ban treaties but did discover astronomical ...

1

This is a general property of (pseudo)Riemannian geometry. I do not think there is anything specifically physical about it beyond the geometry. It holds even if $\varphi$ in not Lorentz invariant. In (pseudo)Riemannian geometry the covariant derivative $\nabla_i$ replaces partials $\partial_i$. The Laplace-Beltrami operator $$\Delta \triangleq ... 1 (I am refering only to the updated question) This is a question which has been asked numerous times by physicists after the formulation and popularization of Maxwell's equations. They have previously known the wave equations on strings where the matter of the string was vibrating, or on water, where the water surface was the one carrying the wave. How ... 1 This is a question that was drastically changed and the other answer and the comments to the question and the answer are discordant with the edited question. I already advised in the comment that you read a simplified article in wikipedia on electromagnetic radiation. Classical electromagnetic radiation cannot be simplified easily by analogies. I will try ... 1 First things first: waves that have already reached close vicinity to the beach DO displace water towards the shoreline - just notice how the water moves back and forth at the point it's ankle-deep. This is related to the phenomenon by which they lose their wave form and get a crest. There are many forces acting on a surfer, but two of them are the ... 1 In an ideal experiment (say, a large room with sound-absorbing walls) you should be able to readily identify the null points. In practice, however (e.g., in your house) the sound from your speakers will impinge on and reflect from the objects in the room. The reflected wave may hinder the ability for you to readily detect null points. This will depend ... 1 This is the generic form of a dispersion relation with a low frequency cutoff. It is a good model for the dispersion relation of electromagnetic waves in a plasma, i.e. the ionosphere, where:$$ \omega^2 = \omega_0^2 + c^2 k^2  Which is the same as your form, though I've put back in $c$ for the nondispersive phase velocity and changed $k_0 \to \omega_0$ ...

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