# Tag Info

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The electric and magnetic fields are always in-phase if the wave can be treated as a plane wave (which simply means it cannot be too close to the source), and in vacuum or any medium with linear response, such as air. Boundary conditions of wave guides change this relationship, and must be solved for each specific case. If the wave guide is large enough, ...

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You're simply meant to solve your equation $n_1sin(\theta_1) = 1$ for $\theta_1$ at one wavelength of light. $n_1$ is (assumedly) known at that wavelength, and $\theta_1$ is the unknown critical angle. I don't think I've ever seen an undergraduate physics question - at least not about Snell's law, that asks a student to work with wavelength varying ...

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Using Snell's law when light travels from glass to air, $$1.5\sin(i) = sin(r)$$ If $\sin(i)=2/3$ then the refracted ray grazes the surface as $r=90^o$. If the value of $i$ is greater than this value, total internal reflection (TIR) takes place. For any $i$ greater than around $42^o$ in glass to air interface TIR would take place.

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It's a poorly written question. As you have discovered, a ray incident on the glass side of the interface at 60° to the normal would be totally internally reflected and there would be no transmitted ray. Well, not unless it's glass with an unusually low refractive index. It's possible there is a misprint and they mean the angle is 60° to the surface i.e. ...

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I) There are already several good answers. OP is asking about the momentum of the non-relativistic string in the transversal model, which in textbooks usually is given as $${\cal L}_T ~:=~\frac{\rho}{2} \dot{\eta}^2 - \frac{\tau}{2} \eta^{\prime 2}. \tag{1}$$ II) Let us fix notation: $\rho$ is the 1D mass density; $\tau$ is the string tension; $Y$ is the ...

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Of course. This is how antennas work. The Maxwell equations make this possible. The metal as a conductor has electrons in what is called the conduction band, and these are free from the ions in the metal lattice. The oscillating electric field $\vec E$ or $\vec D = \epsilon \vec E$ given by $$\nabla\times {\vec H} = \frac{4\pi}{c} {\vec J} - ... 0 Light is an emergent collective phenomenon from zillions of photons. The relation of the energy of the photon to the frequency of the wave is E=h*nu. The photon itself just has spin and energy when measured. Its wavefunction though has a complex dependence that does contain information which will build up the emergent beam with frequency nu. A detailed ... 1 The expression is independent of \sigma and therefore there is no higher harmonics growth etc., only decay of them. I think your exponents have \sigma in them through the x terms, e.g.,$$ e^{-n^{2} \alpha \ x} = e^{-n^{2} \alpha \ \bar{x} \ \sigma} = e^{-n^{2} \frac{\sigma}{\Gamma}} $$One could also find \sigma from \Gamma = \tfrac{\sigma \ ... 0 A progressive wave is a function of space and time where the dependence on space and time can be modeled by a function of one variable, composed with an “argument” function, which combines space and time and describes the geometry of the wave. Wavefronts are the loci where the “argument” function is constant. Let’s give a few examples. Plane wave If ... 2 You seem to have some fundamental misunderstandings about the nature of electromagnetic radiation. The electron transition describes a change in energy that leads to the emission of a single photon. The energy of a photon is inversely proportional to the wavelength of the light. So the light "gets" its wavelength from the energy it carries and not from the ... 7 TL;DR: There is of course a continuous transition between seeing an interference pattern and not seeing one as the width of the source is increased. So it does not really matter if it is < or \leq in the condition, since the relation only represents a qualitative statement about when the contrast in the interference pattern has significantly ... 4 There are already good answers here, but I'm afraid that to the best of my knowledge, Diracology's (and indeed Halliday-Resnik-Krane's) expression of the potential energy is not correct. I would like to point to this paper by Lior M. Burko which focusses on the subtleties of the derivation of the kinetic and potential energy of the string as a whole and ... 0 I presume that when you talk about travelling evanescent wave, you are referring to surface plasmons waves. There was an interesting paper(I can provide you the paper if you cannot find it) on whether growing waves could have a physical significance : "Surface polaritons like waves guides by thin lossy metal films " In a word, it is mentioned that a ... 1 I'm having a problem visualizing the transverse waves with 3 or 4 nodes you mention. All videos I've found show standing waves with only two nodes, the slackline moving up and down between the anchor points (or between an anchor point and the person walking the slackline). The rotation waves I saw (maybe torsion oscillation is a better name) also had only ... 6 You are absolutely right in everything you said. The momentum is non zero only if the wave has a longitudinal mode, which is in fact the realistic case. Moreover when this is the case, the wave equation is not that simple. Let me try show this. Longitudinal Mode Let us assume that when in equilibrium the string, of density \mu, is along with the x\equiv ... 9 A fake derivation We can rather easily compute a horizontal velocity for the string fi we assume that the total velocity vector is everywhere normal to the string (this assumption is not always valid, see below). The following picture then illustrates the computation: Take two infinitesimally separated points x and x+\mathrm{d}x and let the wave ... 4 You're wondering why pressure nodes form at an open end of a tube. The answer is, they don't! It's just a reasonably good approximation. Physically, consider the air molecules at the center of the tube. Since they're far away from the edges, there's no way for them to "know" exactly when the tube ends, so the sound wave must "leak out" slightly. The ... 6 The explanation is not a very full one. As you correctly note, you're taking a limit, so the assumption \sin\theta \to\theta as \delta z\to0 becomes exact. So Eq 16-23 contains no approximation. The assumption creeps in subtly when one assumes that the force calculated in Eq 16-23 is at right angles to the z axis. That is, that ... 3 Right off the bat in eq (16-23) it's assumed that the restoring force is linear in the displacement. That's only true for small displacements. 6 From the Wikipedia article on sound: In physics, sound is a vibration that propagates as a typically audible mechanical wave of pressure and displacement. To fully understand how is air vibrating in an open pipe, you have to consider not only the acoustic pressure wave,$$\frac{\partial^2 p}{\partial x^2}=\frac{1}{c^2}\frac{\partial^2 p}{\partial ...

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Whoa...whats that?sound waves travel faster in liquid than air; 4.3 times faster dependent on the temperature, The higher the elastic constant of the medium the faster sound travels and The tighter the molecular bond the higher the elastic constant. Light travels faster in air than water, simply because air is less denser than water.The denser the medium the ...

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For each wavelength you have been given (exactly) two values $\theta$, so \begin{align}n\lambda &= d\sin\theta_1\\ (n+1)\lambda &= d\sin\theta_2\end{align} subtracting these two equations, we get $$\lambda = d\left(\sin\theta_2-\sin\theta_1\right)$$ You should be able to figure it out from there...

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If you are talking about a spherically symmetric em wave created by a point source then the intensity of the wave (namely the power transferred through unit area) drops according to the inverse square law (this follows because the total power radiated over a sphere of fixed radius remains the same, so as you increase the size of the sphere the power per unit ...

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Standing wave or Stationary wave is a result of two waves (incident wave and reflected wave) propagating in opposite direction with same amplitude and same frequency. This phenomena is a result of interference of wave where energies of two waves add or cancel out depending on the phase. (energies add up in case of complete in-phase and cancel out in case of ...

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One may describe the waves in terms of $\Delta x$, the deviation of the position of strings' atoms from the equilibrium locations. Because they're attached, $\Delta x(\sigma)=0$ for $\sigma$, the coordinate along the string, equal to any of the end point values, $\sigma=0$ and $\sigma=L$. But $\Delta x$ obeys a wave equation, so the eigenstates of the ...

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The second derivation is correct, as explained by Diracology. However, the first derivation is 'sort of' correct, in the sense that the location of potential energy can be ambiguous. For example, consider the three following systems. A mass on a stretched spring. A mass sitting on a table. A charged mass next to another charge. These three systems have ...

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The energy of an element of a traveling wave is not constant. Halliday-Resnick-Krane is right. For a string of density $\mu$ and tension $T$ the kinetic energy of an element $dx$ is $$dK=\frac 12\mu dx\left(\frac{\partial \xi}{\partial t}\right)^2.$$ For the potential energy we have $$dU=Tdl,$$ where $dl$ is the stretched amount of the string. A small ...

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The equations you have here relate the amplitudes to the left on the interface with the amplitudes to the right of the interface. The amplitudes (of the rope or electric fiel) are given as complex numbers, $\bar A=A\exp(i\delta)$, with it's absolute value $A$ (a real number) and the complex phase $\exp~i\delta$. (case $v_2>v_1$) Using Eulers formula you ...

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This is actually much more subtle, the question is very deep and whole books are written about theory of coherence in optics, quantum mechanics etc. What is important is that you might have an ensemble of sources (say in a lightbulb) which all emit light of the same frequencies but in random moments, making the relative phases between different light pulses ...

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Yes. Omega is the time derivative of phi. Phi1 dot = phi2 dot means omegas are the same. See my other answer a couple days ago on the subtleties of coherency. There is phase noise on any transmitter and freq as a result has drift and random noise. It depends on the time proof, it could be coherent to 1 part in 10^6 for milliseconds and 1 in 10^5 for a sec. ...

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No, the frequency will not change. If the wind is blowing at constant speed and the distance between source and observer remains constant, then the time it takes for a sound wave to get from source to observer will be constant. So the time interval between wave peaks (period T) when they are detected by the observer remains equal to the interval between ...

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As a diagram screen $A$ is closer than screen $B$ and so the fringes are closer on screen $A$: Fringe separation $\Delta x = \dfrac {\lambda \; D}{d}$ where $d$ is the slit separation, $\lambda$ is the wavelength of light and $D$ the distance from the slits to the screen.

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Hint : A key to the solution is what is meant by the complex wave 3-vector $\:\mathbf{k}\:$. This vector is not any complex 3-vector in $\: \mathbb{C}^{3}\:$ $$\mathbf{k} \ne \left(k_{1}, k_{2}, k_{3} \right) \in \mathbb{C}^{3}, \:\:\text{that is with} \:\: k_{\rho} \in \mathbb{C} \tag{a-01}$$ but $$\mathbf{k}=\left(k_{1}, k_{2}, k_{3} \right) ... 0 Yes, it is possible but it involves a lot more work. The problem is that the different components you want to separate are not similar enough for a simple subtraction to work. The frequencies are far from exactly the same, and even if they were approximately similar the relative phases would change quite fast. The phase information cannot therefore be used ... 0 As knzhou mentioned, this is true even for progressive waves in a string. Your mistake might be to think about simple harmonic motion instead of harmonic waves. I will show it for a progressive transverse wave in a string. It is easier to visualize. At the end I will give you a sketch for longitudinal waves. For a string of density \mu and tension T ... 0 You must first add the amplitudes and then square them to find the intensity. It is easier at first to assume that the amplitudes of the superposing waves are the same A. Then as the intensity is proportional to the amplitude squared the total intensity from the two sources individually is proportional to 2A^2. The interference pattern has a cosine ... 0 The physics in the 1st approximation are worked out here: https://en.wikipedia.org/wiki/String_vibration with the result that frequency depends on the length, tension, and mass density as:  f = \frac{1}{2L}\sqrt{\frac{T}{\mu}}  The 1st factor is why bass strings are long and treble are not (on a piano). The lower 3 strings on your guitar have a ... 0 I think the question is in regards to energy conservation as a function of position. The power (time average of the amplitude squared) of each individual wave: <A(x, t)^2> + <B(x, t)^2>  is not the same as the time average power of their superposition: <(A + B)^2> as a function of position. For instance, with counter propagating ... 0 I don't think that energy considerations are particularly useful if you want to know the amplitude of the resultant wave. For example for a standing wave, depending on the location of the particle it can get different potential energies. The correct way is as you suggest, to sum up the waves. For example, if two counter-propagating waves interfere$$A(x,t) = ...

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The speed of sound depends on the medium in which it travels. In air it can be affected by ambient temp, relative humidity , as well as atmospheric pressure. The speed usual quoted at a standard temp and pressure. http://byjus.com/physics/speed-of-sound-propagation/

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The wave can be any shape $f(x)$. But when you focus in on a sufficiently small element of the curve, you can do a Taylor expansion about the point $x_0$: $$f(x) = f(x_0) + (x-x_0)f' + \frac{(x-x_0)^2}{2!}f'' + ...$$ As the distance $(x-x_0)$ gets smaller, higher order terms can be neglected. If you consider a point with horizontal slope, then $f'=0$ and ...

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A single line isn't very useful for approximating a curve. You could use small segments, but then you'll need several, and the calculation would be more complicated. As noted in the comments, nothing stops you from using any other second order curve, i. e., a plane curve whose rectangular Cartesian coordinates satisfy an algebraic equation of the second ...

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In general, the term cannot be neglected. It's entirely possible that the term has to stick around when advection is significant. But, since you are asking how it goes away, I'm guessing you are in a situation where it does, in fact, go away. Like any other order of magnitude analysis, you have to non-dimensionalize your equation. This means you have to ...

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Your understanding of a real caustic (I presume you call it real as opposed to the imaginary caustic that you mention later) is correct. First the easy part: an imaginary caustic is a caustic located on the extension of the light rays beyond the optical system from which they arrive. For instance, in the presence of a convex lens, imaginary caustics may ...

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In Wikipedia the caustic is defined as follows. In optics, a caustic is the envelope of light rays reflected or refracted by a curved surface or object, or the projection of that envelope of rays on another surface. You can think of the envelope of a family of curves as a curve that is a tangent to each of them. Here is a diagram on page 60 of "A Treatise ...

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Coherent sources- Two sources of light are said to be coherent if the waves emitted from them have the same frequency and are 'phase-linked'; that is, they have a zero or constant phase difference. The calculation yielded 90.pi ; We know that 2.pi denotes a phase change of zero as the waves will come back to initial phase relation of the coherent ...

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Coherency has to be over a period of time. Instantaneously any two waves (or signals) at the same freq will have at at one time or point a phase difference. If their freqs stay exactly what they had at one time, they will stay coherent with each other, i.e. The phase differences won't change over the same period of time for the two. Over a period of time the ...

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It depends: 1) On whether you mean mass density or particle density, 2) on the temperature and type of gas. The main observation is that speed of sound is approximately independent of particle density, and mostly a function of temperature. 1) The speed of sound (squared) is given by the compressibility at constant entropy per particle $$c_s^2 = \left. ... 0 I think I've figured out why. It's simply because the critical density n_{crit} is a function of \omega. The critical density of the plasma is that which is required for the EM wave frequency to equal the plasma frequency \omega_p, so is dependent on the frequency of the radiation. So$$\omega=ck\left(1-\frac{n_e}{n_{crit}(\omega)}\right)^{-1/2} ...

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When the phase velocity is a constant (with respect to wavelength), the group velocity will indeed be equal to it, as you yourself have shown. What you've got wrong here is the assumption for this case that the product of the phase velocity and group velocity equal the square of the speed of light, which can be true in other cases but not for plasma ...

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