Newest questions tagged wavelength - Physics Stack Exchange most recent 30 from physics.stackexchange.com 2019-07-16T02:38:55Z https://physics.stackexchange.com/feeds/tag?tagnames=wavelength&sort=newest http://www.creativecommons.org/licenses/by-sa/3.0/rdf https://physics.stackexchange.com/q/491666 1 Why does violet light bend the most? [duplicate] Bhaskar Das https://physics.stackexchange.com/users/236884 2019-07-15T02:45:34Z 2019-07-15T09:10:33Z <div class="question-status question-originals-of-duplicate"> <p>This question already has an answer here:</p> <ul> <li> <a href="/questions/65812/why-do-prisms-work-why-is-refraction-frequency-dependent" dir="ltr">Why do prisms work (why is refraction frequency dependent)?</a> <span class="question-originals-answer-count"> 4 answers </span> </li> <li> <a href="/questions/71126/why-does-the-refractive-index-depend-on-wavelength" dir="ltr">Why does the refractive index depend on wavelength? [duplicate]</a> <span class="question-originals-answer-count"> 1 answer </span> </li> <li> <a href="/questions/65156/refraction-of-light-and-frequency-dependence" dir="ltr">Refraction of light and frequency dependence</a> <span class="question-originals-answer-count"> 2 answers </span> </li> </ul> </div> <p>When white light passes through a prism, refraction occurs and it splits into its seven constituent colours. If the spectrum is obtained on a screen violet light appears much more bent than red light. How and why does this occurs?</p> https://physics.stackexchange.com/q/491385 -1 What is the derivation of Rayleigh's Equation? How do i get to Rayleigh's equation? [on hold] Jakob Hearn https://physics.stackexchange.com/users/236777 2019-07-13T11:50:12Z 2019-07-13T11:50:12Z <p>im looking for an answer of the derivation of rayleigh's equation, ie; how to get to the equation through derivaiton, thanks</p> https://physics.stackexchange.com/q/489659 -1 Why sky is blue? [closed] Cang Ye https://physics.stackexchange.com/users/227290 2019-07-04T07:16:45Z 2019-07-04T09:08:43Z <p>The major factor of different scattering is the ratio of wavelength to the size of particles which are working as microscopic scattering mirrors.</p> <p>In a sparse particle medium like air, the longer the wavelength, the easier it is to transmit and harder to be scattered. The shorter the wavelength, the easier it is to scatter and more difficult to transmit.</p> <p>Given the size of mirrors in air, the shorter the wavelength, the stronger the scattered light, the weeker the transmitted light.</p> <p>In sky, the mirrors are the air molecules. Red light gets more transmission, blue light gets more scattered.</p> <p>So the sky is blue. </p> <p>This is my understanding. Anything wrong with my above words, please give advise and corrections.</p> https://physics.stackexchange.com/q/487799 8 Why does the additive color model use red, green and blue instead of yellow, green and violet? Vun-Hugh Vaw https://physics.stackexchange.com/users/194169 2019-06-24T09:20:29Z 2019-06-29T13:32:36Z <p>Long cone cells in the human eye are most sensitive to 570-nm wavelengths which are more like spectral "yellow" than spectral "red" and short cone cells are more to 440-nm wavelengths which are more like spectral "violet" than spectral "blue" <strong><a href="https://en.wikipedia.org/wiki/RGB_color_model#Physical_principles_for_the_choice_of_red,_green,_and_blue" rel="noreferrer">1</a> <a href="https://en.wikipedia.org/wiki/Visible_spectrum#Spectral_colors" rel="noreferrer">2</a></strong>. Then why does the additive color model use red, green, blue instead of yellow, green and violet? Is it harder to make red by mixing spectral "yellow" and "violet" than it is to make yellow by mixing spectral "red" and "green"? Would a hypothetical "YGV" model allow for a wider or a narrower gamut than the RGB model? Even if it would allow for a wider one, would we be able to perceive such a gamut?</p> <p><strong>Note</strong>: By 'spectral "yellow"', I really mean 'spectral "yellow"' in white light, not "yellow" as in the subtractive model. 'Subtractive "yellow"' is not the same as 'spectral "yellow"': the former is the result of 'spectral "red"' and 'spectral "green"' in white light filtered out by the ink pigment and perceived by the human eye as "yellow", while the latter is actual 'spectral "yellow"' in white light. Common parlance has always been the worst to describe the concept of "color", because people can mean very different things by "red", "green", "blue", "yellow", "brown", etc. For example, "brown" is a "color" different from "orange", but in fact it is simply a dark shade of the "orange" hue (which is 'spectral "orange"'), and hue may also be referred to as "color".</p> https://physics.stackexchange.com/q/487542 1 Are there accepted spectral lines, or wavelength of light emitted, for the various neural and ionized atoms? If yes, where can I find them? [duplicate] Pedro de Oliveira https://physics.stackexchange.com/users/235092 2019-06-22T23:25:58Z 2019-06-23T22:37:52Z <div class="question-status question-originals-of-duplicate"> <p>This question already has an answer here:</p> <ul> <li> <a href="/questions/455637/what-are-good-reliable-databases-of-atomic-spectra" dir="ltr">What are good, reliable databases of atomic spectra?</a> <span class="question-originals-answer-count"> 1 answer </span> </li> </ul> </div> <p>I am working with the redshift phenomena and analyzing the spectral lines of various emissions by galaxies. However, when I came to analyze the change in wavelength I was confused on what to compare to in terms of rest emitted wavelength. I found various values for the same ionized atom that varied tremendously. I am wondering if there were accepted values for the wavelength emitted by the atoms?</p> https://physics.stackexchange.com/q/487500 0 Why frequency does not change when light passes through the denser medium? [duplicate] A.H.Kaidan https://physics.stackexchange.com/users/220186 2019-06-22T17:41:02Z 2019-06-23T11:44:02Z <div class="question-status question-originals-of-duplicate"> <p>This question already has an answer here:</p> <ul> <li> <a href="/questions/21336/what-determines-color-wavelength-or-frequency" dir="ltr">What determines color &mdash; wavelength or frequency?</a> <span class="question-originals-answer-count"> 11 answers </span> </li> </ul> </div> <p>as far as I noticed always people in physics have a predefined assumption that frequency is constant. whereas we know that the c is the outcom of product of wavelength and frequency. we have different wavelength and frequency iv whit light. the energy depends on both of them wavelength and frequency. if you consider the speed of light as a global constant parameter then both attribute of that must have that properties. but we know when light passes through a glass for example; it loses energy as we can measure the changes in temperature of glass that is increased. then I think the change of speed of light is because of interaction between light as an electromagnetic wave and fields inside the material would occure and frequency and wavelength both would change by 1 divided on square root of <span class="math-container">$n$</span> or refracting index. by the way we lose the speed but when the light enters the air again it behaves as before then we have the same speed. you van not measure the speed at the boundary condition as speed needs distance and time to be measured. then why we must suppose frequency is constant?</p> https://physics.stackexchange.com/q/486916 2 Components of wave vector Professor Kirby https://physics.stackexchange.com/users/232485 2019-06-19T11:54:02Z 2019-06-24T11:15:53Z <p>Is 3-dimensional wave vector defined as <span class="math-container">$$\tag{1} \mathbf{k}=\frac{2\pi}{\lambda_{x}}\mathbf{\hat{x}}+\frac{2\pi}{\lambda_{y}}\mathbf{\hat{y}}+\frac{2\pi}{\lambda_{z}}\mathbf{\hat{z}} ?$$</span> If it is, then it's magnitude would be <span class="math-container">$$\tag{2} |\mathbf{k}|=2\pi\sqrt{\frac{1}{\lambda_{x}^{2}}+\frac{1}{\lambda_{y}^{2}}+\frac{1}{\lambda_{z}^{2}}} .$$</span> But on the other hand, we know that the magnitude of a wave vector is given by <span class="math-container">$$\tag{3} |\mathbf{k}|=\frac{2\pi}{\lambda}.$$</span> So in order for these two to be equivalent it can't be true that <span class="math-container">$$\tag{4} \lambda=\sqrt{\lambda_{x}^{2}+\lambda_{y}^{2}+\lambda_{z}^{2}}.$$</span> But doesn't this follow from Pythagoras's theorem so it should be true? I would guess that definition (<span class="math-container">$1$</span>) is not true but <a href="https://en.wikipedia.org/wiki/Particle_in_a_box" rel="nofollow noreferrer">this</a> Wikipedia article about particle in a box and its section 'Higher-dimensional boxes' seems like it uses definition (<span class="math-container">$1$</span>). Or maybe it doesn't, but then is it not true that <span class="math-container">$$\tag{5} \lambda_{x}=\frac{2L_{x}}{n_{x}} ?$$</span> If not, could you explain why?</p> https://physics.stackexchange.com/q/486831 0 Deriving Planck's Constant from Wien's Displacement Law Eric https://physics.stackexchange.com/users/165493 2019-06-18T22:18:44Z 2019-06-19T03:42:45Z <p>So I'm reading an introductory book on Quantum Theory (David Park, 3rd ed.) and I am having trouble with the following question:</p> <p>"According to Wien's displacement law, the wavelength <span class="math-container">$\lambda_m$</span> at which blackbody radiation at temperature T has its maximum intensity is given roughly by <span class="math-container">$\lambda_mT \simeq$</span> 3 mm K. Assuming that the quantum energy at this temperature is of the order of kT where k is Boltzmann's constant, estimate the value of Planck's constant."</p> <p>My attempt at a solution is as follows:</p> <p><span class="math-container">$\lambda_mT \simeq$</span> 3 mm K <span class="math-container">$,\quad$</span> <span class="math-container">$\lambda = \frac c \nu$</span> <span class="math-container">$\quad \rightarrow\quad$</span> <span class="math-container">$\frac {cT} \nu = 3\cdot10^{-3} mK$</span> <span class="math-container">$\cdot (\frac k c )\quad \rightarrow \quad$</span> <span class="math-container">$\frac {kT} \nu = 1.38\cdot10^{-34}Js$</span></p> <p><span class="math-container">$(\frac 1 2 mv^2)_{max} = kT_{max} = h\nu - e\phi$</span></p> <p><span class="math-container">$\frac {kT_{max}} \nu + \frac {e\phi} \nu =h$</span></p> <p><span class="math-container">$1.38\cdot10^{-34} + \frac {e\phi} \nu = h$</span></p> <p>At this point I get stuck, the orders of magnitude and units seem to be right but I'm not sure what to invoke/what is given in the question that can solve/eliminate this last term to get h, or if I'm even on the right track.</p> https://physics.stackexchange.com/q/486773 0 Why an open tube at both ends suffers resonance? [duplicate] Guilherme Correa Teixeira https://physics.stackexchange.com/users/203110 2019-06-18T15:47:36Z 2019-06-18T18:44:53Z <div class="question-status question-originals-of-duplicate"> <p>This question already has an answer here:</p> <ul> <li> <a href="/questions/150929/question-on-open-organ-pipe" dir="ltr">Question on open organ pipe</a> <span class="question-originals-answer-count"> 3 answers </span> </li> </ul> </div> <p>Why an open tube at both ends suffers resonance when subjected to a sound that propagates through the air with length of where <span class="math-container">$L / 2$</span>?</p> <p>I already know the methodology to calculate the harmonics in an open tube. So, I'm not expecting an answer that is based on the calculations. I want to know why this phenomenon occurs from the molecular point of view of the air, taking into account that air could be considered as an ideal gas whose molecules have no interaction with each other.</p> https://physics.stackexchange.com/q/485425 0 Can application of force contract wavelength of particles Tanmay Siddharth https://physics.stackexchange.com/users/234171 2019-06-11T03:45:03Z 2019-06-12T12:07:16Z <p>When I put the de broglie relation for momentum in Newton's law F=dp/dt</p> <p>I saw that in some way F is inversely proportional to wavelength. So if we apply greater force, the shorter the wavelength gets.</p> <p>Does it mean that the applied force is contracting the indivisual constituent particle?...I am not talking for whole object.</p> <p>And now suppose if there were two partcles as two added quantum waves, appearing as one. </p> <p>Now if we apply force then when they will contract onto themselves we will see as if one particle disintegrated/brokedown into two. </p> <p>Am i getting it correct? Please answer and comment</p> https://physics.stackexchange.com/q/485027 1 Heisenberg's Uncertainty Principle - finding uncertainty in wavelength mhold https://physics.stackexchange.com/users/153111 2019-06-08T22:26:29Z 2019-06-09T13:23:48Z <p>I am confused about this problem: <a href="https://i.stack.imgur.com/DoHJM.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/DoHJM.png" alt="enter image description here"></a></p> <p>I needed to find the uncertainty of a wavelength using Heisenberg's uncertainty principle. In the solution, they differentiated <span class="math-container">$λ=c/f$</span> with respect to frequency to get <span class="math-container">$\frac{-c}{f^{2}}$</span>. λ is then substituted back in to get <span class="math-container">$\frac{- λ^{2}}{c}$</span>. <span class="math-container">$Δf$</span> had already been found, so the final equation to find <span class="math-container">$Δλ$</span> was:</p> <p><span class="math-container">$$Δλ = \frac{- λ^{2}}{c}*Δf$$</span></p> <p>Which honestly makes no sense to me. Why is it necessary to differentiate λ? I was thinking that the final equation would be something like:</p> <p><span class="math-container">$$Δλ = \frac{c}{Δf}$$</span></p> <p>since <span class="math-container">$λ = \frac{c}{f}$</span>. Why wouldn't my method work?</p> https://physics.stackexchange.com/q/484253 1 Are kitchen microwaves in the audible range? SeanJ https://physics.stackexchange.com/users/213565 2019-06-04T13:33:26Z 2019-06-04T14:01:26Z <p>The waves of a typical kitchen microwave oven have a wavelength of <a href="https://en.wikipedia.org/wiki/Microwave_oven#Principles" rel="nofollow noreferrer">12cm</a> while the audible spectrum is between <a href="https://en.wikipedia.org/wiki/Sound#Sound_wave_properties_and_characteristics" rel="nofollow noreferrer">1.7cm and 17m</a>, so one might think that they overlap and that kitchen microwaves should be audible. </p> <p>Are they, are they drowned out, or blocked by the appliance's cage?</p> https://physics.stackexchange.com/q/484251 1 Formation of Red and blue bands in the sky Kingshuk Mondal https://physics.stackexchange.com/users/232007 2019-06-04T13:27:41Z 2019-06-04T14:42:39Z <p>Today at the time of sunset I saw this <img src="https://i.stack.imgur.com/Gs7Lj.jpg" alt="The upper part of the photo points towards East and the lower part west"> The sky is divided into two parts one In red and one in blue which are very distinctive Can anyone tell how? The upper part of the image points towards the East and the lower part west </p> https://physics.stackexchange.com/q/483968 2 Sizes of Elementary Particles ShroomZed https://physics.stackexchange.com/users/197015 2019-06-03T00:56:10Z 2019-06-03T04:34:02Z <p>Present observation shows that elementary particles have no internal structure, and have no real size as they are described by wavefunction. </p> <p>Something that therefore confuses me is that on a lot of online "size comparison" videos and animations, elementary particles are given sizes, some much larger than others. <a href="http://htwins.net/scale2/" rel="nofollow noreferrer">An example.</a></p> <p>Where exactly are the sizes for these animations coming from and how are they so precise? Are they just straight made up or are they from some other source that deals with theoretical sizes? Apologies if this question seems silly, but I've been puzzled about it for some time. </p> https://physics.stackexchange.com/q/483419 0 Wavelength of cosine-squared LetzerWille https://physics.stackexchange.com/users/169895 2019-05-30T22:43:01Z 2019-05-31T03:17:11Z <p>I am confused. Usually, the wavelength is the x-distance between the tops of two consecutive waves. Here is the graph. </p> <p>There is only 0.1 m between 2 crests. But the answer counts the wavelength as 0.2 m</p> <p><a href="https://i.stack.imgur.com/ewox0.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/ewox0.png" alt="enter image description here"></a></p> https://physics.stackexchange.com/q/483106 0 Database of Color Emissions Lokasa Mawati https://physics.stackexchange.com/users/233234 2019-05-29T11:05:43Z 2019-05-29T11:05:43Z <p>I am reading about the color <a href="https://en.wikipedia.org/wiki/Green" rel="nofollow noreferrer">Green</a> and am wondering if there is a database anywhere listing the atoms/molecules/powders/minerals and their color wavelengths in various forms. Basically I would like a list of color wavelengths along with some things associated with those wavelengths (like atoms associated with the color green).</p> <p>For example, this lists several wavelengths in which lasers display green:</p> <blockquote> <p>Lasers emitting in the green part of the spectrum are widely available to the general public in a wide range of output powers. Green laser pointers outputting at 532 nm (563.5 THz) are relatively inexpensive compared to other wavelengths of the same power, and are very popular due to their good beam quality and very high apparent brightness. The most common green lasers use diode pumped solid state (DPSS) technology to create the green light. An infrared laser diode at 808 nm is used to pump a crystal of neodymium-doped yttrium vanadium oxide (Nd:YVO4) or neodymium-doped yttrium aluminium garnet (Nd:YAG) and induces it to emit 281.76 THz (1064 nm). This deeper infrared light is then passed through another crystal containing potassium, titanium and phosphorus (KTP), whose non-linear properties generate light at a frequency that is twice that of the incident beam (563.5 THz); in this case corresponding to the wavelength of 532 nm ("green"). Other green wavelengths are also available using DPSS technology ranging from 501 nm to 543 nm. Green wavelengths are also available from gas lasers, including the helium–neon laser (543 nm), the Argon-ion laser (514 nm) and the Krypton-ion laser (521 nm and 531 nm), as well as liquid dye lasers. Green lasers have a wide variety of applications, including pointing, illumination, surgery, laser light shows, spectroscopy, interferometry, fluorescence, holography, machine vision, non-lethal weapons and bird control.</p> <p>As of mid-2011, direct green laser diodes at 510 nm and 500 nm have become generally available, although the price remains relatively prohibitive for widespread public use. The efficiency of these lasers (peak 3%)[citation needed] compared to that of DPSS green lasers (peak 35%)[citation needed] may also be limiting adoption of the diodes to niche uses.</p> </blockquote> <p>Perhaps there is a database of laser types and their wavelengths, or astronomic objects and their wavelengths (stars and such), or emission spectra of the atoms laid out in text format.</p> <p>Ideally it would be in text format, or an HTML table, but PDFs work as well.</p> <p>Here is an example of the emission spectra I was imagining might be in text format like <code>hydrogen,300nm,325nm,etc.</code>.</p> <p><a href="https://i.stack.imgur.com/seDLU.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/seDLU.jpg" alt="Hydrogen emission spectra"></a></p> https://physics.stackexchange.com/q/482363 0 Wavelength and Lattice constant - need to be the similar magnitudes to have interference? mikanim https://physics.stackexchange.com/users/158294 2019-05-25T17:07:57Z 2019-05-25T21:51:33Z <p>I was taught that they need to have similar magnitudes but I did an exercise last week and the magnitudes were different by 3 decimal places. Before I also noticed them being either the same or maybe one decimal place different. </p> <p>So, is it actually true that the need to have similar magnitudes? And if so, to what extent? Lastly, <em>why</em> do they need to be of similar magnitude? </p> https://physics.stackexchange.com/q/481973 1 How can I get the wave number and wave vector? [closed] David Lee https://physics.stackexchange.com/users/232791 2019-05-23T21:58:18Z 2019-05-24T09:08:37Z <p><span class="math-container">$$\overrightarrow{E} = (-10 \hat x + 4 \hat z) e^{-j(2x+5z)}$$</span></p> <p>I recently started to study electromagnetics, but I'm having a hard time following up.</p> <p>May I ask how to calculate the wave vector <span class="math-container">$\overrightarrow{k}$</span> and the wave number, please?</p> <p>Also, how can I calculate the magnetic field vector in the frequency domain?</p> <p>I really appreciate your help guys.</p> <p><strong>ADDED: Thanks for the comments, guys.</strong> I noticed these answers. <a href="https://physics.stackexchange.com/a/189504/232791">https://physics.stackexchange.com/a/189504/232791</a> <a href="https://physics.stackexchange.com/a/287991/232791">https://physics.stackexchange.com/a/287991/232791</a></p> <p>So it seems like the wave vector <span class="math-container">$\overrightarrow{k}$</span> is <span class="math-container">$$\overrightarrow{E} (r) = \overrightarrow{E_0} e^{j(\overrightarrow{k} \cdot \overrightarrow{r})}, \overrightarrow{k} = k_x \hat{x} + k_y \hat{y} + k_z \hat{z}, \overrightarrow{r} = x \hat{x} + y \hat{y} + z \hat{z}, \overrightarrow{k} = -2 \hat{x} + -5 \hat{z}$$</span> And the wave number, the vector size is <span class="math-container">$$k = \frac{w}{v_p}, \overrightarrow{k} = k_x \hat{x} + k_y \hat{y} + k_z \hat{z}, |\overrightarrow{k}| = \sqrt{k_x^2 + k_y^2 + k_z^2} = \sqrt{(2\pi/\lambda_x)^2 + (2\pi/\lambda_y)^2 + (2\pi/\lambda_z)^2}$$</span> equal to <span class="math-container">$\sqrt{k_x^2 + k_y^2 + k_z^2} = \sqrt{29}$</span>, right?</p> <p>Since <span class="math-container">$\overrightarrow{E_0} = E_{0_x} \hat{x} + E_{0_y} \hat{y} + E_{0_z} \hat{z} = (-10 \hat x + 4 \hat z)$</span> has no <span class="math-container">$y$</span> dependence, meaning that this must line in the <span class="math-container">$zx$</span> plane,</p> <p>I thought <span class="math-container">$\overrightarrow{k}$</span> should be perpendicular to the <span class="math-container">$zx$</span> plane.</p> <ul> <li>Have I correctly got both the wave vector and the wave number?</li> <li>"Differentiating and integrating wrt time", do you mean I have to apply Faraday's Law <span class="math-container">$\overrightarrow{\nabla} \times \overrightarrow{E} + \frac{\partial \overrightarrow{B}}{\partial t} = 0$</span>?</li> </ul> <p>Or <span class="math-container">$\overrightarrow{\nabla} \cdot \overrightarrow{E} = 0, \overrightarrow{\nabla} \cdot \overrightarrow{B} = 0$</span> in free space?</p> <p>Just like your first comment, I don't see time, nor frequency as well. May I please ask for help?</p> https://physics.stackexchange.com/q/481376 0 Why doesn't the wave's frequency change as it gets refracted? [duplicate] Joonyoung Lee https://physics.stackexchange.com/users/232545 2019-05-21T02:03:16Z 2019-05-21T02:58:40Z <div class="question-status question-originals-of-duplicate"> <p>This question already has an answer here:</p> <ul> <li> <a href="/questions/263288/why-does-the-frequency-of-a-wave-remain-constant" dir="ltr">Why does the frequency of a wave remain constant?</a> <span class="question-originals-answer-count"> 3 answers </span> </li> </ul> </div> <p>I know that frequency means a complete wave produced per second. But when the wave gets refracted, it's wavelength decreases. If the wave's wavelength has decreased doesn't it means that the wave has been produced more which causes the frequency to increase? Please explain in simple words :)</p> https://physics.stackexchange.com/q/481342 0 Why aren't extremely-low-frequency (ELF) radio waves used for underwater radar? Kurt Hikes https://physics.stackexchange.com/users/182409 2019-05-20T21:35:33Z 2019-05-21T02:16:10Z <p>Since extremely-low-frequency radio waves are used by submarines for some simple, low-transmission-rate communications, why can't those same wavelengths be used for submarine radar? It may not be ideal, or highly precise, but couldn't it still provide some useful information?</p> https://physics.stackexchange.com/q/481274 2 What is wavelength at classical turning points using WKB Approximation? [closed] Powerful blaster https://physics.stackexchange.com/users/185446 2019-05-20T14:57:34Z 2019-05-28T11:01:53Z <p>According to what I know is that a classical turning point in Newtonian Mechanics is a point where a particle has a zero kinetic energy (Total energy is equal to potential energy) and must be instantaneously at rest. This means it stop its motion and reverse direction similar to harmonic motion oscillating back and forth between points <span class="math-container">$x=-A$</span> and <span class="math-container">$x=+A$</span>. In the equation given below doesn't the wavelength <span class="math-container">$$\lambda(x) = \frac{h}{\sqrt{2m(E-U(x))}}.$$</span> tend to infinity when <span class="math-container">$E=U(x)$</span>?</p> https://physics.stackexchange.com/q/480917 1 What experiment would confirm De Broglie equation on photons? Manu de Hanoi https://physics.stackexchange.com/users/26275 2019-05-18T18:29:56Z 2019-05-19T05:39:30Z <p>If we want to check experimentally that, <strong>for a photon</strong>:</p> <p>λ=h/p (De Broglie equation)</p> <ol> <li><p>Has such experiment been carried out?</p></li> <li><p>What is/would be the experimental setup?</p></li> </ol> <p><a href="https://en.wikipedia.org/wiki/Matter_wave#Experimental_confirmation" rel="nofollow noreferrer">Wikipedia</a> doesnt show such protocol for a photon</p> <p>EDIT: my question is badly worded, De Broglie's equation for a photon would be called Planck–Einstein relation </p> https://physics.stackexchange.com/q/480043 2 Claim that DeBroglie relation doesn't work in crystal Leo L. https://physics.stackexchange.com/users/219318 2019-05-14T15:54:49Z 2019-05-15T08:17:43Z <p>In this Wikipedia article on Position and Momentum Space, <a href="https://en.wikipedia.org/wiki/Position_and_momentum_space" rel="nofollow noreferrer">https://en.wikipedia.org/wiki/Position_and_momentum_space</a></p> <p>there is a claim that "the de Broglie relation is not true in a crystal" in the sentence before the content box.</p> <p>Is this claim valid? If so, why? What is the implications for quasi-particles (e.g. plasmons and polaritons) in materials?</p> https://physics.stackexchange.com/q/479975 0 Focal Length, Wavelength Relationship deBoogle https://physics.stackexchange.com/users/198164 2019-05-14T09:10:29Z 2019-05-14T09:18:14Z <p>I have a cylindrical lens with a design wavelength of 587.6nm. The radius (S2) of the lens is 2mm, and the radius of the opposite face (S1), i assume is 0mm.</p> <p>The focal length is given as 3.91mm and the back focal length is given as 1.4mm.</p> <p>If I am using a wavelength of 905nm. </p> <p>The question is how do I calculate the change in focal length of the lens due to the increase in wavelength. I have read many articles that seem to relate the lens equation and something called cauchy equation but I couldn't see anything that made sense to my non-optic way of thinking.</p> <p>thank you in advance ddB</p> https://physics.stackexchange.com/q/479910 1 How are classical and quantum momentum related in an intuitive manner? Wanf https://physics.stackexchange.com/users/231910 2019-05-14T02:35:19Z 2019-05-14T17:15:19Z <p>I know that quantum momentum is inversely proportional to the wavelength of the probability or matter wave of a given particle, but I don't get how this relation of this abstract mathematical construct (the probability wave) relates to the actual observable property (momentum). I don't get how <span class="math-container">$mv = h/λ$</span> when mass times velocity is something very "real" and classical while wavelength times Planck's constant is not.<br> Basically, can someone please explain how the momentum of a probability wave (given by <span class="math-container">$p = h/λ$</span>) is the same as the momentum of the particle that the probability wave describes (given by <span class="math-container">$p = mv$</span>)? Please do not use too much math in your answer because I don't know too much of it.</p> https://physics.stackexchange.com/q/478686 1 Can you change the wavelength of light keeping frequency constant and can you do the opposite as well? [duplicate] Zinc https://physics.stackexchange.com/users/231442 2019-05-08T07:51:02Z 2019-05-11T21:07:05Z <div class="question-status question-originals-of-duplicate"> <p>This question already has an answer here:</p> <ul> <li> <a href="/questions/22385/why-does-wavelength-change-as-light-enters-a-different-medium" dir="ltr">Why does wavelength change as light enters a different medium?</a> <span class="question-originals-answer-count"> 3 answers </span> </li> <li> <a href="/questions/263288/why-does-the-frequency-of-a-wave-remain-constant" dir="ltr">Why does the frequency of a wave remain constant?</a> <span class="question-originals-answer-count"> 3 answers </span> </li> <li> <a href="/questions/21336/what-determines-color-wavelength-or-frequency" dir="ltr">What determines color &mdash; wavelength or frequency?</a> <span class="question-originals-answer-count"> 11 answers </span> </li> </ul> </div> <p>Can you change the wavelength of light keeping frequency constant and can you do the opposite as well? I understood the basics but please don't hesitate to go deeper into the concept. Also, If you happened to have an elegant explanation please drop it here if you can. </p> https://physics.stackexchange.com/q/477878 1 Refractive index and optical fibre question Rudra Mutalik https://physics.stackexchange.com/users/170287 2019-05-04T15:56:33Z 2019-05-04T17:33:36Z <p>This is an A level AQA question: </p> <p>A signal is to be transmitted along an optical fibre of length 1200m. The signal consists of square pulses of white light and this is to be transmitted along the centre of a fibre. Explain how the difference in refractive index results in a change in the pulse of white light by the time it leaves the fibre. </p> <p>Table:<br> COLOUR | REFRACTIVE INDEX OF FIBRE | WAVELENGTH (nm)<br> Blue ------------------- 1.467 -------------------------------- 425<br> Red -------------------- 1.459 ------------------------------- 660 </p> <p>What I am confused about is why would blue travel slower?</p> https://physics.stackexchange.com/q/477084 1 Logarithms in uncertainties Jim421616 https://physics.stackexchange.com/users/90243 2019-05-01T01:28:56Z 2019-05-01T06:00:01Z <p>I'm looking at the following plot: <a href="https://i.stack.imgur.com/qHZE7.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/qHZE7.png" alt="enter image description here"></a></p> <p>The vertical lines show the upper and lower frequency bounds for each of the bands W4, W3, W2...and I'm trying to convert them to wavelengths using <span class="math-container">$\lambda=\frac{c}{\nu}$</span> so that I can show a similar plot, only in wavelength space, rather than frequency.</p> <p>As an example, I know that <span class="math-container">$\lambda_{W4}=22\mu$</span>m, so <span class="math-container">$f_{W4}=1.36\times10^{13}$</span> Hz. What I can't figure out, is the uncertainties in the graph. It's stated in the caption that the frequency bands are <span class="math-container">$\Delta log_{10}=\pm 0.05$</span>, but I'm not sure how to find the bounds from that. Is it:</p> <p><span class="math-container">$1.36\times 10^{13} \pm log_{10}0.05 = 1.35999\times 10^{13} \rightarrow 1.3600003\times 10^{13}$</span></p> <p>or</p> <p><span class="math-container">$log_{10}(1.36\times 10^{13} \pm 0.05) = 13.08 \rightarrow 13.63$</span> <span class="math-container">$=1.2023\times 10^{13} \rightarrow 1.3599\times 10^{13}$</span></p> <p>or am I completely off track?</p> https://physics.stackexchange.com/q/476165 1 Variation of Refractive index user226375 https://physics.stackexchange.com/users/226375 2019-04-26T13:56:58Z 2019-04-26T14:45:53Z <p>We know that refractive index, for any medium,</p> <p><span class="math-container">$$n=1/\sqrt{\epsilon\mu}.$$</span></p> <p>Also, according to Cauchy's relation</p> <p><span class="math-container">$$n=A+B/\lambda^2,$$</span></p> <p>where <span class="math-container">$A$</span> and <span class="math-container">$B$</span> are constants related to the medium.</p> <p>According to the first relation, refractive index isn't in any way related to the wavelength of the light, it is only related to the permitivitty and permeability of the medium. According to the second relation, however, it depends on the wavelength too. Which formula is right? Why does the discrepancy arise?</p> https://physics.stackexchange.com/q/474301 1 Phase difference in a standing wave? Olly Scargill https://physics.stackexchange.com/users/228718 2019-04-22T09:15:40Z 2019-04-22T09:29:49Z <p>What would the phase difference between P and Q be? I assumed that because they are 1/4 of a wavelength apart, it would be Pi/2, but supposedly the difference is 0.</p> <p><a href="https://i.stack.imgur.com/YX400.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/YX400.png" alt="enter image description here"></a></p>