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9

In a classical electrodynamics picture there is an exponentially dampened wave penetrating the medium with the smaller index of refraction. See e.g. http://en.wikipedia.org/wiki/Total_internal_reflection. This wave can indeed be detected. The easiest way to see this is by bringing a third optical surface close to the interface. There will be non-zero ...


5

Yes, the concept of a standing wave is frame dependent, and a standing wave selects a specific frame just like the concept of a rest frame. My initial answer was similar to BenCrowell's, and I made the same oversimplification mistake that his initial answer does. The are no spatial coordinates for the standing wave (even in its rest frame) in which the ...


5

Good question. You're right, you can't really throw out one of the solutions. The author of this PDF is using a poorly explained shortcut, which goes like this: if you expand out the full complex solution, you get $$\begin{align} \Psi(\theta) &= c_1 e^{i\omega t} + c_2 e^{-i\omega t} \\ &= (a_1 + b_1 i)(\cos\omega t + i\sin\omega t) + (a_2 + b_2 ...


4

For a start, and in context of question (a), you can simplify the situation by thinking it as two glass slabs parallel to each other with an air interface of neglictible thickness between them. I'll consider only what happens in the first slab as by symmetry, what happens in the second one is the same. Now, first: All the light you want to consider is the ...


4

Yes it can be done, and indeed it's a well established technology called active noise control. The idea is based on destructive interference. If at some point two sound waves have the same intensity and frequency and they're 180ยบ out of phase then they will sum to zero and the sound intensity at that point will be zero. Your phrase negative sound just means ...


3

This isn't a direct answer, but adds to Ben Crowell's answer and Curious Kev's answer and is simply something to help you have confidence in the correctness of your physical reasoning, which is perfectly sound. Think of your standing waves in a box with perfect mirrors at either end, initially at rest relative to an observer $A$, and then you give the box ...


3

I hope you'll find this paper useful: J Weiner, Rep. Prog. Phys. 72 (2009) 064401. Although it mainly discusses light waves, the idea (and even the results) may be generic. For very small apertures (sub-wavelength apertures), strong reflection is expected to happen. According to Bethe's theory of light transmission through small holes, the transmission ...


3

I think maybe it's just a bad photo. The double slit diffraction pattern is a convolution of the pattern from a pair of delta functions with that of a single slit. i.e. You have the regularly spaced double slit pattern, but this is modulated by the pattern produced by a single slit of width equal to that of the slits used for your double slit experiment. In ...


2

Imagine you have a monochromator that you can tune to give you whatever wavelength you like. Now send light at a wavelength of 568 nm towards the interface. What has been calculated is that destructive interference takes place, so that less of the incident radiation is reflected. Let's take an extreme case where the reflection from the first and second ...


2

Well, I am not sure if your statements are entirely accurate because the fast mode can approach the lower hybrid resonance. In fact, in this regime, it becomes effectively indistinguishable from an electrostatic whistler mode. At low frequency and oblique angles, the fast (or magnetosonic) modes are right-hand polarized (with respect to $\mathbf{B}_{o}$) ...


2

What you have to imagine is the fact, that waves when are (ideally) reflected from a wall lead to a standing wave. If in the wall is a tiny slit the components of the reflected waves left and right the slit dissipate in the direction of the slit (spherical waves in each point according to Huygens description). These components undermine the incoming wave at ...


2

The wave mechanics dispersion relation you cite is for EM waves propagating in free space. In other media, the dispersion relation is not necessarily linear (it can be quadratic or have some more complex dependence). So in this context, there's nothing special about quantum mechanics. More generally, the dispersion relation tells us about the phase speed ...


2

Here is/was my attempt at a solution to this problem - thanks to the contributions of others here, I am now reasonably sure this is correct. The setup is two plane, polarised waves travelling in opposite directions (let's say $x$). e.g. $${\bf E} = E_0 \sin (kx + \omega t)\hat{\bf j} + E_0 \sin(kx - \omega t) \hat{\bf j}$$ The associated magnetic field ...


2

The shock wave from a supersonic object is a cone composed of overlapping spherical wavefronts. As individual wavefronts form, they propagates radially outward at speed $c$ (speed of sound) and have a radius $ct$. At the same time the object traveling at speed $v$ moves forward $vt$. The angle of the vertex of the of the shock wave is known as the Mach angle ...


2

Here are some topics to read about: Frequency doubling, also called second-harmonic generation as Johannes mentions. Here, you put one wave into a medium, and some fraction of it is converted to a wave with a different frequency. By carefully engineering the medium you can get quite a high conversion percentage. Other nonlinear optical processes, not just ...


2

I think the problem lies in the fact that you integrate to $T = \frac{1}{f}$ instead of integrating over unit time. If you had integrated over unit time, the $f$ would have disappeared from the denominator. Energy per unit time makes more sense than energy per cycle of the oscillation. And obviously if energy per unit time is some value, then energy per ...


1

I cannot speak for how to match with a mechanical resonator but you can match arbitrarily well any EM radiator that is small relative to its wavelength at one (1) frequency. This means that the radiator (loop or dipole, monopole, etc.) will absorb all radiation coming from directions in which it can radiate as a source without any reflection. The rest is ...


1

If you insist on thinking of photons as waves (which is fine as long as you ignore absorption, though you should really think of it as a disturbance in an electromagnetic field), you can more or less think of all of their amplitudes as being equal, and this is why your premise doesn't make sense. More precisely, the amplitude of a single photon isn't ...


1

Some of the other answers have suggested that the amplitude of the photon's wavefunction is well defined, and that it has the same value for any two photons of the same energy. Whatever else we may say, this can't possibly be right. The amplitude of an electromagnetic wave is defined either by its electric field or by its magnetic field. (In a sensible ...


1

It means that the second wave is horizontally shifted by a half cycle (a 360 shift would not have any effect because it will be a full cycle). So in a 180 shift the peaks are in the positions of the troughs and viceversa


1

This wave displacement (the y-axis of your graph) is tracing out a simple harmonic motion, i.e. oscillating between -A and +A in a sinusoidal fashion. A more illuminating way (which is more natural in fact) of imagining this would be to consider a wave in general any sinusoidal wave and look at the motion of any specific single point on such a waveform as a ...


1

Possibly you are talking about the difference between the "far field" and "near field" solutions for the simple oscillating electric dipole. Often when dealing with such a system, if we are looking at the field more than a few wavelengths away from the dipole (or more formally, $kr \gg 1$ or $r \gg \lambda/2\pi$) then the solution looks like a spherically ...


1

That the phase speed can have a dependence on the wavelength/frequency of the wave. For instance, a whistler mode wave can have a cubic dispersion relation at low frequencies. In this limit, the higher(smaller) frequencies(wavelengths) propagate faster than the converse. It results in a sort of "spreading out" of the wave modes. This if often seen ...


1

The dynamic range quoted is the peak to peak range of the whole sensor, not an individual lenslet. The formulas are relating to the individual lenslets. For an individual lenslet WFS10-5C, the max angle is about $\frac{142 \mu m /2}{3.7mm} =0.019 rad\approx 1^\circ$. The maximum $\Delta Z= d \tan 1=2.7 \mu m$ or about $5.4 \lambda$ for $\lambda=0.5 \mu m$. ...


1

Before we begin, we should define some terms and parameters/functions that will be used later: Wave Number: $\equiv$ effectively the number of wave crests (i.e., anti-node of local maximum) per unit length $\leftrightharpoons$ ``density'' of waves $\rightarrow$ $\boldsymbol{\kappa}$ $=$ $\boldsymbol{\kappa}\left(\omega,\textbf{x},t\right)$ in general ...


1

When you say displacement, do you then mean displacement of each particle? Or the displacement of the wave / wavetops? I guess the latter. If this is a rope being swung to create the wave motion, then remember to keep a sharp distinction between particle motion and wave motion. As the particles in the rope move up and down (transverse) the wavetops are ...


1

Although normally considered as photon interactions, any inelastic scattering process will result in the alteration of the frequency of the electromagnetic radiation. An obvious example is Compton scattering, where high energy (X-rays+) light scatters from free electrons. The scattered light has lower energy and longer wavelengths than the light incident ...


1

Floris, great answer. The confusion in the question is a really common one that isn't emphasized enough in teaching. Intuitive View Here's an easy way to remember this: When the speaker pushes, the air touching the speaker moves one way. When the speaker pulls, the air touching the speaker moves the other way. Put more formally, for each movement of the ...


1

Yes. There are more complete developments based on electromagnetic theory. Exact solutions are known for a few cases. Almost all applications rely on approximations, many of which are very good approximations under the right conditions. But the more complete explanations end up looking like Huygens Principle, and serve to justify it. Huygens' original ...


1

If you have a small diapragm moving slowly then the air will just flow around it and you won't get any appreciable pressure rise in front of the diaphragm. That means there won't be any longitudinal pressure waves (i.e. sound waves) generated normal to the diaphragm surface. If you now make the diaphragm larger the air has farther to move to get to the ...



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