# Tag Info

## Hot answers tagged optics

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 ...

5

I'd like to add to Peter Shor's comments. An ideal partly silvered mirror comes very close to conserving energy. This actually implies quite a lot about the phase shift imparted by both mirrors, particularly if the mirrors are symmetric in their action. The total power of the two outputs must sum to a constant and so, as the relative phase of the ...

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 ...

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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

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 ...

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 ...

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 $$\cos(\angle S_2S_1B)=\frac{S_1B}{S_2B}=\sin(\theta).$$

3

The primary rainbow is at 40.7-42.4 degrees, where ~42 degrees is the critical angle for red light to reflect back to the observer, and 40 degrees for blue light. what this means is that for a droplet higher than 40 degrees (assume horizontal sun ray, 0 degrees), it can no longer reflect blue light to the observer. by the same logic, no red light can be ...

3

The imaging is not being done by focussing transmitted light as would be done in an optical microscope. Instead it's detecting light emitted by the nanoparticles as they fluoresce. This means there is no lower limit to the size of the particl;e detected, except that when the particles get very small they emit too little light, i.e. they are too faint, for ...

3

You are asking about retroreflectors. Some background: In general (implying 3D) and with the assumption that no functional refractive elements are allowed, a hollow corner cube reflector would be the tool of choice. Nevertheless equally useful glass corner cubes exist. Both types are colloquially known as triple-prisms. So that's where your first answer "3 ...

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 ...

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 ...

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 ...

2

Under normal circumstances, what you are seeing is the steady state condition where the rate of absorption is equal to the rate at which energy is conducted away or radiated away again, so the material doesn't heat up. As I understand, unless a material fluoresces, the de-excitation happens in the infrared. Also, you should realize that just because a ...

2

If we assume the wave-fronts that initially enter the system to produce this image are symmetric, than the result is intuitively symmetric as all the components used to produce it are too. And anyone who looks at the images you have shown, will clearly see a symmetry about the Y-axis. Assuming they do not look for minuscule errors due to wave-front ...

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 ...

2

It is indeed a topic that is discussed in many books but only a few give a rigorous mathematical description of the phenomena. For stringency in non-linear optics topics I always trust HARTMANN ROMER: Theoretical Optics, An Introduction. 2005 WILEY-VCH Verlag GmbH & Co. KGaA, Weinheim. A book which is also mathematically rigorous is BOYD, ROBERT W: ...

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 ...

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 ...

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 ...

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 ...

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 ...

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 ...

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 ...

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. ...

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 ...

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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 ...

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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 ...

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