# Tag Info

2

One way to study this case is through the numerical analysis of diffraction, as described in my other answer to you. You can also do this pretty much as you describe through Huygens's principle or as Feynman describes in his popular QED book. If you set up an equation to describe what you've said, you'll see that the amplitude at a point with transverse ...

2

To understand this explanation, you need to understand Fourier decomposition of the electromagnetic field. In any homogeneous medium, any electromagnetic field can be thought of as a linear superposition of plane waves, all in different directions. Because they run in different directions, the phase delays they undergo in propagating from, say, your ...

-1

So this was the answer given in the key:

1

Your coloured object is absorbing light, i.e. light is changing into mechanical energy, while the atoms in the Bunsen burner are emitting light, i.e. mechanical energy is changing into light. If you have, for example, sodium atoms in a flame those atoms are continuously colliding with air molecules. The velocities of the air molecules are a function of ...

2

In Bunsen burner atoms get heated up which means they absorb energy, and go to some excited states (or even get ionized). And then they de-excite in any way they can which means radiative decay is dominant, as there is almost no other way to dissipate energy. There is no crystal lattice, and compared to luminescence rate collisions are infrequent. So yes - ...

2

The phosphors lining the glass tube of a fluorescent light do a pretty good job of smearing the atomic mercury line emission spectrum into something closer to black body radiation. There could be a phosphor mix that accomplishes what you want...

2

When you scatter light off of a material there is a photon-phonon interaction which will shift the photon frequency depending on the phonon energy (Raman scattering, Brillouin scattering). The effect is quite small, however. How much broadening do you need? Rayleigh scattering through a warm, high density gas will probably go quite far in messing up the ...

3

This might not be feasible for your setup, but you could try rapidly rotating your light source, which would Doppler broaden your spectral lines. You're correct in that the speed would need to be a substantial with respect to the speed of light. As an example, if your frequency is 500nm let's say. If you'd like it to spead out on the order of a single ...

1

The material may become opaque for radiation that has frequency close to resonance frequencies of the material. The electrons in matter are sensitive to radiation in certain ranges of frequencies (absorption peaks or bands) and can get excited. This is accompanied by stronger absorption of the radiation. The electrons are sensitive to certain resonance ...

1

I don't know where you measured 39°. The blue angle is 36.87° (the sides of the red triangle are 3-4-5), as it should be.

1

It depends on "how monochromatic" a source you need for your current use. Further, you can have multiple modes of a single wavelength. Using a Fabry-Perot etalon can clean up things a bit. But if your question is not how to achieve, but rather how to evaluate, your source, then you will be limited by the resolution of your spectrometer, or the peak-spacing ...

0

You should keep in mind that true monochromatic light is not possible due to uncertainty principle. The emission will be always a band with a certain width which depends on temperature and other technological factors. The best thing to do is to use a high resolution spectrometer and take a spectrum of your laser, taking the necessary precautions not to ...

1

In the optics regime, every time a wave impinges on a surface it is modifying the angular momentum of an electron. Since the electron PE is usually comparable to the visible regime. As for measurements with coherent waves. I don't think this is an easy task, though I think it is possible. I mention waves, because depending on the energy level of your ...

3

Instead of scattering, think of it as diffuse reflection. The bidirectional reflectance distribution function (BRDF) describes optical surface properties. It's application is as well in computer graphics, as in-depth ray tracing simulations. It depends on angle of incident light $\vec \omega_i$ (2 dimensions) and angle of observation $\vec \omega_r$, also 2 ...

3

The reason is not quite as intuitively put as for ropes, but it is essentially to make the fields consistent with the electromagnetic boundary conditions, which in turn can be traced to (1) Kirchoff's voltage law and (2) no conduction currents can flow in a dielectric. Consider a tiny, thin rectangular loop running parallel to the interface with one half ...

0

The simple answer is that your LiNbO3 crystal will almost certainly be cut such that the coefficient of importance is $r_{33}^T$. This means that the electric field is applied along the $z$ axis, the polarization of your light field is along the $z$ axis, and the direction of propagation through the crystal is along the $y$ axis (or maybe it is the ...

4

I like Brandon's very physically intuitive answer: mine is a little drier. It is simply that three waves $E_j(t);\,j=1,2,3$ mix through $n^{th}$ order nonlinearity by way of $n^{th}$ power term $\left(\sum_{j=1}^3 E_j(t) e^{-i\,\omega_j\,t} + E_j(t)^* e^{i\,\omega_j\,t}\right)^n$ in the Taylor series for the input to output transfer function. So in the ...

3

Why not imagine the third-order process as a two-stage second-order process like so:

2

The best way to understand this phase shift is to solve and study solutions of the Helmholtz equation near the boundary between two dielectric mediums. You don't quite have to solve the full Maxwell equations: the assumption that the light field can be modelled by one scalar field (approximately equal to one transverse component of the electric field) rather ...

1

In one sense you are right: the only free space "perfectly collimated" optical field is the plane wave in the sense that these are the only eigenfields of Maxwell's equations, being fields which conserve their form under propagation and only undergo scaling by an eigenvalue in such propagation. Since Maxwell's equations conserve energy in free space, ...

0

If we look at an actual, finite, laser incident on an interface, it is not a plane wave. It has finite extent, so it is a superposition of different plane waves which all have different angles of incidences. In optics, we can model the total field by adding all the incident plane waves and all the reflected plane waves together. Because of the different ...

1

When you derive a Huygen-Fresnel Propagator (which is how actual wavefronts propagate according to Maxwell's equations) a Fresnel zone is really the difference (in phase) between surfaces of equal phase on the propagating wavefront and a plane slicing or tangent to that surface of equal phase. These Fresnel zones are defined when propagating a plane wave ...

0

LED's are not like lasers. You should treat them as point sources(like a pinhole with a bright source on the other side). I would recommend you curve your mounting for your LEDs. That way, you could make one lens see them as one bad point source, rather than trying to refocus a bunch of good point sources. what you would really want is a lenslet array, they ...

0

I am not sure I understood your query completely (handheld and underwater communication??), but I will try my best to help you out: 1) For a single LED, in general, the generalized Lambertian pattern is widely used as the radiation pattern. In this pattern, the illuminance at a location is a function of the distance of the location. You can find more on ...

1

I might add a few commas to that Wikipedia sentence, as "A perfectly collimated beam*,* with no divergence*,* cannot..." to show informative rather than additional parameters. To answer your question about "collimated" vs. "plane wave" , consider two point sources at th plane of focus of a lens. Each point source gives off spherical waves; the lens ...

2

In Fraunhofer diffraction, the farfield pattern is proportional to the spatial Fourier transform of the input field, so the tilt on the input field simply translates the diffraction pattern transversely. A tilt corresponds to multiplying the input field by $\exp(i\,\vec{k}_0\cdot\vec{r})$ where $\vec{k}_0$ is the wavevector showing the nominal propagation ...

1

Since the data is captured in Red,Green,Blue and you know the correction filter's transmission in each of these bands you can simply scale the RGB output to give you the colour shift you want. All you need to know is the Red,Green,Blue bandpass of the Bayer filter on your camera's chip. You probably need to do this with your camera's raw mode. Other modes ...

2

I heard a few times that using them as sunglasses is hurting the eye since UV light is not filtered, but the pupil is wider than it would be w/o wearing them because the visible light is dimmed. IFAIK, there is no evidence for this claim. See this paper (unfortunately, it's behind a paywall): The supposition that, because of pupil dilation, there ...

1

I am assuming for simplicity that either the waveguides are one-moded or, if not, the input to the system is in one mode alone and the system design is such that coupling between modes is negligible. Usually with resonant ring systems like this one used as interferometers, we are using them to sense changes in the ring's optical phase delay. Actually, if ...

0

I'd like to add a bit of detail to mpv's answer. My take on the different optical microscopy techniques are: Intensity imaging methods which suffer from a divergent noise contribution from out-of-focus information, in exactly the same way as the night sky should be uniformly bright given an infinite universe as described by Olber's Paradox. These methods ...

3

There are many types of microscopy and it would be difficult to summarize them all in a single answer. But the basic types are: Optical microscopy: the specimen is observed by visible light, the optics is made from optical lens. The image sometimes can be observed by naked eye in an eyepiece, or there can be a CCD sensor. Electron microscopy: the speciment ...

0

When you move the mirror, you change the way the light has to go along, thus changing the time it needs to get to the interference surface. This means you change the phase shift of both light beams, too. It has nothing to do with different frequencies (you use only one light source, as I assume?).

1

The main idea behind polaroid sunglasses is that reflexion from water, snow and other glary reflectors is mainly polarized in one direction. To understand this, witness the behaviour foretold by the Fresnel Equations (the graph below taken from the Wikipedia "Fresnel Equations" page): so that you can see for a wide range of scattering angles from these ...

1

Coherent states, although strictly quantum, are "isomorphic" to classical states. They are also isomorphic in the same way to one-photon states. There are bijective maps between any pair of the following three sets: (i) the set of all quantum coherent states (ii) the set of all one-photon states and (iii) and the set of all solutions of Maxwell's equations. ...

0

The point 1 , where incoming parallel rays all meet, is the focal point. At this point all the incoming rays from a far object are just a single point , not an image (this is the distance you would hold a magnifying glass away from something you wanted to burn with rays from the sun) . The image of a far away object will focus on a screen at some point 2 ...

4

Coherent states are quantum states, but they have properties that mirror classical states in a sense that can be made precise. To be concrete, let's consider coherent states in the context of the simple harmonic quantum oscillator which have precisely the expression you wrote in the question. One can demonstrate the following two facts (which I highly ...

1

You have separate Fresnel equations for s- and p-polarized light. The two polarizations reflect/refract separately. You can reconstitute them on the other side to recover the new polarization vector if you want.

2

It is all about what meaning you put into the words "quantum" and "classical". Fock space and elements of this space are notions that belong to quantum theory of radiation and have no direct relation to states of radiation in classical electromagnetic theory, so the coherent state may be called "quantum" with good reason. However, coherent states have ...

3

If coherent state are indeed the most classical states (which means that the mean value of the EM fields obeys the classical Maxwell equations), the state used in the paper you mentioned are not coherent state (at least in the arXiv paper), but cat states ! The state $|\alpha\rangle+|-\alpha\rangle$ is not a coherent state ! It is the superposition of two ...

2

Don't worry, I did research in surface plasmons and even then I was more than a year into it before I truly understood, on an intuitive level, how the light gets a 'kick' from the grating. You are correct that it is diffraction at a 90 degree angle to the normal, but there is an easier way to think about it. You say you've never taken a formal course in ...

0

$\sin \theta = (m \lambda)/d = (5 \times 1228^{-6})/(1/600) = 3.684$ $d = (1/600)\times 3 = 0.005 \lambda = (d \sin \theta) /m = (0.005 \times 3.684)/4 = 0.004605\: \mathrm{mm} = 4605\: \mathrm{nm}$

1

The condition comes from "phase-matching" - or in other words that the wavevector of the SPP ($\beta$ in your example) is matched to the wavevector of the in-plane component of the incident light. Now before the light hits the surface, this in-plane wavevector is given by $k \sin \theta$, but when it hits the grating, it receives a momentum "kick" of $\pm ... 6 There is some evidence of polarization perception. Many people are able to perceive polarization of light. It may be seen as a yellowish horizontal bar or bow-tie shape (with "fuzzy" ends, hence the name "brush") visible in the center of the visual field against the blue sky viewed while facing away from the sun, or on any bright ... 1 Our eyes cannot see any difference between ordinary (i.e., unpolarized) and polarized light. You can check it yourself, if you look through a polarizer (for example, some sunglasses have one). All you can notice is that the world gets slightly darker (because you block roughly half the incoming light). In addition, some reflections might be reduced ... 1 Ok, I've found this: http://www.cv.nrao.edu/course/astr534/Brightness.html I proves that it's not possible to build such optical system. The conservation of brightness also applies to any lossless optical system, a system of lenses and mirrors for example, that can change the direction of a ray. No passive optical system can increase the ... 0 Though it is not common to use Fresnel lenses for electricity generation but 100 MW power plant is nearing completion in Rajasthan state of India using linear Fresnel lens technology. So to say that this technology is not feasible for large scale use is not correct and time may come if that above mentioned power generation goes smoothly, the scene may change ... 3 Applying the law of cosines to the triangle$\triangle S_1S_2P$will yield $$r_2^2=r_1^2+a^2-2ar_1\cos(\angle S_2S_1B).$$ The angle that appears is complementary to$\theta_m$, so you can either use$\angle S_2S_1B=\frac\pi2-\theta_m$and trigonometric identities, or simply see that $$\cos(\angle S_2S_1B)=\frac{S_1B}{S_2B}=\sin(\theta).$$ 0 Here is some possibly useful information from Goodman's "Statistical Optics." (Sorry about the lack of symbol quality -- so much for cut/paste from a PDF) Light from a thermal source is regarded as unpolarized if two conditions are met. First, we require that the intensity of the light passed by a polarization analyzer, situated in a plane perpendicular ... 1 A compound microscope never forms an image at infinity. The image formation is always a variation on the drawing below. For big magnifications, the image position can be quite a long way from the viewer. What you may be getting confused by is that many (indeed almost all) modern microscopes use infinity conjugate optics. This means that they are made up ... 1 Your approach is correct, but you really need to draw even a crude sketch. The lamp is$x$m from the mirror; the image is$5$m from the lamp, which puts it$(5+x)\$ from the mirror. These are the d values for your magnification formula, which is correct...

Top 50 recent answers are included