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53

You are seeing particles. However there's more to this than meets the eye so I need to explain exactly what I mean by this. Light is neither a particle nor a wave. Instead it is a quantum field. As a general rule while light is travelling it appears as a wave, but when the light quantum field is exchanging energy with anything it does so in quanta that ...


8

It depends on your definition of "a sound". If a sound is not a sound unless it is perceived as a sound (that is, processed in the auditory system of a sentient being), then the answer is "no". If a sound is a coherent disturbance in the pressure distribution of the air, and this disturbance propagates through the medium "at the speed of sound", then the ...


6

It would be physically impossible to be able to "see" light as anything other than a particle (photon). The only time photons, or any other subatomic particle for that matter, can be described as a wave is when we are NOT looking at them.


6

The "modes" in this case refer to the standing waves that can exist in a cavity. A very nice diagram / explanation is given at http://hyperphysics.phy-astr.gsu.edu/hbase/quantum/rayj.html To summarize: if you consider a cavity of dimension $L$, the modes that fit inside the cavity have wave numbers $n_1$, $n_2$, $n_3$ such that $$n_1^2 + n_2^2 + n_3^3 = ...


5

The expression you have for $\sin$ is real, as can be understood by the following: $$\begin{align}e^{i\theta}&=\cos\theta+i\sin\theta\\ \text{and so}\\ \frac{e^{i\theta}-e^{-i\theta}}{2i}&=\frac{\cos\theta+i\sin\theta - \cos\theta+i\sin\theta}{2i}\\ &=\sin\theta\end{align}$$ There is no complex number when you're done expanding...


5

The quantum wavefunction is not a wave in the classical sense. In particular, it is not a wave obeying $E = \hbar \omega$, as electromagnetic waves do. The term wavefunction is, in this sense, just a bad naming decision. It is no wave, there is nothing oscillating, and it has no connection whatsoever except the mathematical form to physical waves.


4

A picture is worth a thousand words. Here's how it looks as a function of space, evolving in time: Here blue is real part, and purple is imaginary part of the complex exponent $\exp(i(kx-\omega t))$. If you instead just look at $\exp(-i\omega t)$, you'll get this:


4

Acoustic waves travel through a medium (air, water, metal, etc), there is no known medium through which light travels Both the speed of sound and the speed of light have fixed values regardless of the speed of their source Acoustic waves can be longitudinal (in gases) or transversal (in solids) whereas light is only transversal. You can measure acoustic ...


4

Consider a cube of edge length L in which radiation is being reflected and re-reflected off its walls. Standing waves occur for radiation of a wavelength λ only if an integral number of half-wave cycles fit into an interval in the cube. For radiation parallel to an edge of the cube this requires $L/(λ/2) = m$, an integer or, equivalently $λ = 2L/m$ ...


3

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 $\lambda=0.5 \mu m$. ...


3

To answer the question "same wavelength, different frequencies, which arrives first?": naturally, the one with biggest speed, which is proportional to frequency AND wavelength according to the formula: $$v = \lambda f$$ So, for the same wavelength $\lambda$, the one with bigger frequency $f$ will have bigger speed $v$, thus arriving earlier. For the ...


3

The book is imprecise because there are other types of polarized light. Now, consider a point in the path of the wave, like the green point in animation you mentioned. In that image, the light is linearly polarized so the electric field draws a line at the location of the point. However, there is also circularly polarized light where the electric field draws ...


3

The only requirement for light is that the electric field must be perpendicular to the magnetic field at any given point in time or space. This assumption arises naturally from Maxwell's equations. The most intuitive way of thinking about light is with the picture you included of the light wave. However, you have to imagine an infinite number of light waves ...


3

Note that the numerator is the difference of a complex number and its conjugate. This difference is imaginary (or zero) for any complex number: $$z - z^* = 2i\mathfrak{Im}\{z\}$$ To see this, write $z = \sigma + i\omega$. Then $$z - z^* = (\sigma + i\omega) - (\sigma - i\omega) = 2i\omega$$ Thus, if we want the imaginary part of a complex number ...


3

An on-axis pointlike object, infinitely far away will produce rays that focus at the focal point of your converging lens. The sun and the trees are far enough away that "infinity" is not such a bad approximation, but they sun and the trees are neither pointlike, nor on-axis. An off-axis pointlike object, infinitely far away will produce rays that (nearly) ...


2

Before we go too far ... the wavelength of visible light is about half a micron. If you do succeed in setting up standing waves, and figure out how to visualize it, you won't be able to see the wave pattern themselves. The wavelength is too small. You might be able to visualize the increased intensity in the cavity, though. I've seen that in plastic ...


2

Wave equations have a long history in physics, they are usually equations involving second derivativs, the solutions are sinusoidal ( sines and cosines) and have been used to model classical waves, starting from water, sound, pressure waves, and finally light classically, with Maxwell's equations. When Schrodinger's equation was able to reproduce the Bohr ...


2

At the point where the wave is perfectly flat, what is in that exact moment pulling the rope downwards? Nothing. The force of the wall on the rope vanishes at that instant. In that moment the wave is not pushing the wall upwards anymore, so the wall will also not give any force. Right. Where to has the energy in the travelling wave been ...


2

Putting comment into answer, as requested: At the point where the wave is perfectly flat, nothing pulls it downwards. But to continue moving it doesn't need any force - it just obeys Newton's first law and moves with nonzero velocity.


2

Yes, it applies to particles. Your expression for the "particle version" is correct.


2

As dmckee's comment says, your question doesn't really have an answer. However it's exactly the sort of question that fascinated me in my time as a teenage physics enthusiast, and I think it touches on some really interesting aspects of physics. To get a grip on this you need to understand how matter is described by quantum field theory. This is not ...


2

This question can't be answered from the picture alone without additional observations/data but @David Rose has given a good list of hypotheses. The size of the waves appear to be within the regime of 'capillary waves' Capillary waves are surface boundary waves with wavelengths on the order of millimeters up to a centimeter or two, where the energy restoring ...


2

The only thing you have to ensure when using phasors is that the real (or imaginary part, depending on the convention used) part of the phasor reduces to the original entity, i.e. $$Re[\phi]=x$$ And since the real operator is linear, you can easily check that the real part of the sum of individual phasors reproduces your x, as you correctly speculated: ...


2

As stated in the comment, the "fish" is the mathematical symbol for "proportional to". In the case of the amplitude and energy, $$ E\propto A^2 $$ means that $$ E=CA^2 $$ for some constant $C$.


2

I suppose you could say this is cheating, but you could surround the object emitting the sound with a perfect vacuum. Sound waves are vibrations in a medium; because a perfect vacuum has nothing in it, it cannot "conduct" (for lack of a better word) sound waves. You could attempt to levitate the object with magnets; because of Earnshaw's theorem, the setup ...


2

Think of it like this: when is the kinetic energy a maximum in this oscillation? At the point when $y=0$, or at the point at which the rope is flat. If there were a force pulling it, it would violate conservation of energy as from that point thereafter energy starts to transfer to potential energy of oscillation. In short, the energy is transferring ...


1

You need to look at the idea of Separation of Variables for Partial Differential Equations. You consider a toy universe comprising oscillators in a box: let's think of a cuboid microwave cavity with electromagnetic fields losslessly confined within perfectly reflecting walls. The Cartesian components of the electric field all fulfill Helmholtz's equation: ...


1

Suppose we have a drum. When you bang on the drum it will vibrate. When you look at any point on the surface of the drum head, you will see it go up and down, similar to if to take a mass hanging from a spring attached to the ceiling and watch it bob up and down. However, there is also an important difference between these two situations. In the situation ...


1

A wave e.g $$\sin (kx + \omega t + \phi)$$ when reflected runs in the opposite direction. In other words gets a rotation by $\pi$ or what amounts to the same thing gets a phase shift by $\pi$. $$\sin (kx + \omega t + \phi + \pi)$$ Tentative proof: Let's say a wave $\psi \sim e^{i(kx-\omega t)}$ on reflection the wave should propagate opposite (or ...


1

It seems that "fish" refers to $\propto$. It is means "proportional to". Since @Kyle Kanos has provided useful links about this, I won't repeat them. But I want to add a useful tool for checking unfamilar symbols: http://detexify.kirelabs.org/classify.html You can draw the symbol and get its latex code, and then it's easy to find its meaning (usually the ...



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