Tag Info

New answers tagged

3

You're not missing anything. You are right, $k=\omega/c$. The argument $\sqrt{\frac{\omega ^2}{c^2}-k_z^2}$ in the Bessel function is the projection of the wavevector onto the radial direction. The use of Bessel functions beclouds what's going on a bit. Recall that a plane wave with wavevector $\vec{k}$ has the functional variation $\psi(\vec{r}) = ...


0

So the length of a wave is the distance between two compressed regions as shown in this representation of a longitudinal wave: is in general not true but rather it is a pictorial representation for simple cases in one dimension. A wave is any solution of a wave equation of the form $\Box \psi(\textbf{x},t) = 0$ that can be expressed in the form $$ ...


4

You don't need a particular point on the wave. You only have to make sure it's the same point on each successive wave. If you have a microphone, hooked to an oscilloscope, you can measure the time between peaks (or troughs, or zero-crossings), multiply by the speed of sound, and that's your wavelength.


0

There are a couple of ways of knowing the wavelength of laser pointer. 1) Using Snell'law (law of refraction): A light passing the border between two media whose refractive indexes vary. The incident light PO of wavelegth, $\lambda_{1}$ travelling in a media of refractive index, $n_{1}$ is refracted in to another media of refractive index, $n_{2}$, with a ...


1

In quantum mechanics observables are represented by (some classes) of self-adjoint operators on some Hilbert space. Saying that you can precisely measure a quantity given by the operator $A$ means that your state can be one of the eigenstates of that operator. Likewise, if you want to precisely measure two quantities $A, B$ together your state needs to be an ...


1

Some thoughts on the subject: The key difference between a microphone and an antenna is that the microphone is sealed from the back - it senses a pressure difference between the front and the back of the membrane regardless of the extent of that pressure region. If you have a small membrane that is not sealed from behind, then at low frequencies it will ...


2

A microphone is a transducer that converts variations in air pressure from sound waves into electrical signals. Air pressure varies as the wavefront passes into the diaphragm (or the ribbon, or the condenser) of the microphone. The diaphragm needn't be as long as the wavelength, as it senses the wave from a "head-on" perspective rather than "looking at it" ...


1

Microphones transform the pressure wave of sound to an electric signal. The wavelength of the sound wave tells us the distance over which the wave's shape repeats itself in space. The frequency measures the changes in the medium in time. As the sound wave passes, the molecules of the microphone vibrate in place ,according to the frequency, like a harmonic ...


99

Your iPhone is a pretty good grating. I just did a simple experiment with an iPhone, a green laser pointer and a sheet of graph paper. This was the result: The display of the iPhone 6 has a resolution of 326 ppi - meaning we have a "grating spacing" of 25.4/326=0.0779 mm. Different models have different resolutions - make sure you find out what your ...


8

If you've got a grating with known distance between the slits, you can use diffraction: let the light fall in perpendicular to the grating and place a screen a few meters further away. You'll find the maximum intensities (the light dots if you've got a grating with 100 or more slits per mm) under an angle of: $$\sin\theta_m = n\frac{\lambda}{d}$$ with $n$ ...


0

As far as we know, light is mediated by a particle without rest mass. Special relativity says that such a massless particle's speed must always be observed to be $c$, irrespective of the observer's motion. In general relativity, a massless particle's speed as measured locally is also always $c$. No experiment so far has detected a measurable difference ...


0

Along w/ the info in the other answers, keep in mind that "white" is not a fixed item. In very dim conditions, your cones don't work, and the rods in your retina only report intensity, not color, so everything looks white/grey. In very bright conditions, your retina overloads, and all you sense is 'white.' Now, one can define "white light" analogous ...


1

"It is known" that each atom has a characteristic atomic emission spectrum, as long as the atoms are isolated from one another. Emission spectra are usually observed in gases at low pressure. But when the atoms are compressed into solids or liquids, the close proximity of the atoms distorts the environment in which the emission takes place, and shifts the ...


2

Yes and no. Fusion inside the sun produces light - but the atom are moving so fast that their electrons are not attached - it is a plasma. As such, you would be hard pushed to find emission lines in the sunlight. You will see some absorption lines - the colder hydrogen and helium further out will absorb little bits of the radiation. What you are left with ...


0

The relationship between wavelength and distance is similar to the relationship between frequency and duration, and no: neither pair is the same. You can see by using dimensional analysis. Wavelength is distance divided by cycles. Frequency is cycles divided by time. Multiply the two, the cycles cancel out, and you get distance divided by time, or velocity. ...


0

Velocity is a more widely used term, usable for all moving objects and waves, while wavelength is of course only usable for waves. Wavelength is the minimal distance between two points of a wave with the same phase. Take for example a sinusoidal wave: the wavelength will be the difference between two maxima or two minima. The velocity of a wave is used ...


1

A wavelength is a particular distance, corresponding to the length travelled during a period, which is a special time. Since $v=d/t$ holds good for the distance $d$ travelled by a constant velocity object over any given time interval $t$, a fortiori this relationship holds for the special, particular time known as the period. So, yes, $v=d/t$ is how you ...


0

The lambda is the distance between 2 points having the same phase like two successive crests the velocity is the wave can be conceived as how many crests for example passes through a reference in a given time you can use both equations but c=f*lambda is used if you have lambda , its proof is V = distance / time , if a crest traveled a distance = wavelength ...



Top 50 recent answers are included