Tag Info

New answers tagged

0

As an analogy, consider the photon that strike your face and reach my eyes, we say that that photon carries information about your face which then helps me to identify you, You are confusing the individual photons with the electromagnetic wave that is light, which is composed out of a zillion photons. but don't these photons collide midway with air ...


0

Assuming the outer ring is stationary (an arbitrary assumption just to establish a range of reference), the inner magnet will either move up or down until the bottom of the inner magnet (or top if it moved down, but we'll assumed it moved up) is aligned with the top of the outer magnet. In an ideal experiment, the inner magnet would need to experience ...


0

As with Fermat, light won't "choose" a longer path as that would violate Feynman's principle of least action. It just doesn't happen.


0

The mirror isn't drawn right. The angles should be equal. But it is a local minimum. Fermat's principle says minimum time. If the ray stays in one medium, it travels at constant speed. Minimum time is minimum distance. That is, there are many paths from A to mirror to B. The path a ray takes picks the point on the mirror that has the shortest path. You ...


2

A monochromator, by definition, must separate the light into different wavelength beams. Mirrors are only weakly wavelengthdependent, although some very sophisticated optical dielectric multilayers mirrors can have frequency dependence. Even so, the wavelength dependence of even a multilayer on direction is too weak for this application Gratings on the ...


0

You are making an incorrect assumption in your question: There is no physical evolution from a number state (aka Fock state). This evolution happened purely inside physicists' heads as it was realized that laser light is not properly described by number states. The problem is your assumption that the particle number ever is well-defined. Lasing action is an ...


1

The wire is a cylindrical reflector. The laser light that hits the top of the wire is reflected upwards; the light that hits the side is reflected sideways. This is simple reflection - no need for a diffraction explanation.


0

Light has properties of waves and particles and doesn't necessarily travel in "straight lines". There is an interesting experiment (I will look up the original scientists who did it and post that info here) where they set up basically a cardboard circle in front of a wall and then shine a flashlight at it. Interestingly enough, there is a shadow in the ...


0

You got the correct answer. Maybe you want it in terms of $\cos^{-1}$ instead? In that case the expression looks like: $$ \theta_{\tt max}=\cos^{-1}(\sqrt{\frac{n^2-1}{8}}) $$ This is for the case of two internal reflections (the picture on the left). Edit: I think I just realized that you have written down the correct equation and the correct maxima in ...


1

(Following @garyp's comment) We'll indeed have different focal points for different wavelengths: $$\frac{\partial f}{\partial \lambda}=-\frac{2 r_{N} \Delta r_{N}}{\lambda^2}$$ taking typical values of $r_N \sim 10^{-2}\mathrm{m}$, $\Delta r_N \sim 10^{-3}\mathrm{m}$, $\lambda\sim 5\cdot 10^{-7}\mathrm{m}$, we get: $$\frac{\partial f}{\partial \lambda} ...


2

I think the key point that you are missing is that the object is very far away compared to all of the other distances. Combine your final two equations to eliminate $y_0$ and then let $d_0\rightarrow\infty$, i.e. $\tfrac{1}{d_0}\rightarrow0$.


0

Answer B is Correct An aeroplane passenger, given a proper vantage point, will see circular concentric rainbows Rainbows viewed from airplanes are created like rainbows viewed from the ground, except that you're simply viewing it from a different (flying) perspective. That is, the water droplets there use internal reflection and dispersion to shoot light ...


1

The efficiency of the pumping source is $x$ means that $x$ amount of electrical power is converted to energy which is useful for pumping the laser medium. The absorption of the pump is $y$ means that $y$ amount of the energy from the pump source is actually pumped into the medium to generate the population inversion necessary for lasing. The total amount ...


1

A ray which is incident normal to an interface will have zero refraction (in first order optics). The angles in Snell's law are measure with respect to the normal of the interface and since $$n_1\sin\theta_1=n_2\sin\theta_2$$ if $\theta_1=0$, $\theta_2$ must also be zero. That means that for the ray to strike the flat interface of the semi circle at an ...


3

Searching that book for "lineshape function" will return this page, which explains what that means. Essentially, the atoms in the gain medium are usually able to respond to frequencies $\omega$ which are close to, but not necessarily exactly equal to, the central frequency $\omega_0$. The response is strongest at $\omega_0$ and then it tapers off over some ...


0

Apart from the overdrive (saturation), the imperfections you see are laser speckles. It is an interference effect produced by imperfections inside your laser, possibly involving some dust or back-reflections. If you use cheap diode lasers, you will almost always get some of these, even from just the bare laser chip. There is a trick to clean up your beam: ...


0

Something like what you're proposing is theoretically sound. For a one-photon state, you can define the following probability amplitudes that uniquely specify the one-photon state: $$\begin{array}{lcl}\vec{\phi}_E(\vec{r},\,t)&=&\left<\left.0\right.\right| \mathbf{\hat{E}}^+(\vec{r},t)\left|\left.\psi\right>\right.\\ ...


1

It depends on what you mean by whether "geometrical optics can be applicable". Geometric optics will work fine with a system with a grating like the one you mention, although you need to know the rules for calculating the directions and strengths of the transmitted / reflected rays. There are three main points to heed here: In general, several separate ...


2

It’s not the same as having two mirrors, the camera-screen setup doesn’t involve the same physics. In the camera-screen setup, the only optical phenomena taking place is the absorption of photons by the camera’s sensor. Once the image is formed in the camera, all other effects are electronic, digital or electromagnetic in nature. The representation of the ...


1

This was commonly used by students to make artsy videos to music that were played on U.S. PBS TV stations. If the camera and the screen are not perfectly aligned, each frame in the regression is rotated relative to the previous frame. This gives a spiral effect and simply twisting the camera to the rhythm of the music caused the spiral rate to change. ...


1

It is quite similar - but there is a delay. The camera "sees" a frame, then processes it, and displays it. It is like the dual mirrors, but while the dual mirror setup "instantaneously" shows an infinite array of mirrors-in-mirrors, there is a delay (typically 1/30 of a second) for the camera from frame to frame, so the frame-in-frame happens more slowly. ...


0

Do you have access to a dynamic signal analyzer or similar? My most recent setup involves locking in to a signal from a split photodiode that is either singly or doubly modulated. We chose our base modulation frequency by looking at the noise spectrum -- our photodiode when illuminated with a DC signal looks something like this: There's a ton of ...


3

The only natural lasers I know of are astrophysical. For example a natural infrared laser in the vicinity of the star MWC 349 was discovered in 1995. If MASERs count then a number of them are known. There's a review in this paper.


0

I think your notation makes sense. You define a vector $ C \in \mathbb{R}^n$ as $ C= (c_1,c_2,...,c_n) $. The absorbance is a function of $n+1$ variables: $ \nu $ and $ c_{mol}$ for $mol=1,2,...,n $, than can be expressed in a more conpact way as $A(\nu, C)$. Your last equation it is perfectly valid since you have defined $c_{mol}$ as the $mol$-th component ...


0

See in Wikipedia the topic "Phase velocity" http://en.wikipedia.org/wiki/Phase_velocity and also Snellius law. Let $M_1$ be a less refractive medium, and $M_2$ a more refractive one. Let $n_1$ respectively $n_2$ be their refraction indexes. It is known that in any medium the light wave has a phase velocity specific to that medium, and this velocity and the ...


0

As mentioned here also remember that light has wave behaviour. When a wave of water travels over shallow water, it slows down. This corresponds to light reaching a material of more "resistance" against its' wave motion (we simply measure that by measuring the speed of light in that material - the refractive index is the proporty $n=c/v$.) This link shows a ...


1

No. The only effect that the red color would have would be to attenuate all the colors that are not red. With white light in, you'll still get a spectrum out, but the blues and greens will be dimmer than they would be with a clear prism. (The positions of the colors might be shifted by a very small amount because absorption comes with a small change of ...


1

It all depends on what you want to resolve and how far away it is. For the Andromeda galaxy at a distance of about 2.5 million light years, HST will resolve visible light objects about 0.6 light years apart. If we take our separation from Proxima Centauri (4.3 ly) as typical of separations in the spiral (certainly it's less in the center of the galaxy), ...


0

There are some other options I can add. By using two cylindrical lens you can build a beam expander that only expands in one direction, thus correcting for the elongation of the beam. Once you have corrected for this you can then focus the beam on to a pinhole spatial filter. This would only partially work since the laser needs to have a Gaussian profile ...


0

Before the photon is absorbed, i.e. whilst the photon quantum field is in a pure, one photon state, you can define the following probability amplitudes that uniquely specify the one-photon state: $$\begin{array}{lcl}\vec{\phi}_E(\vec{r},\,t)&=&\left<\left.0\right.\right| \mathbf{\hat{E}}^+(\vec{r},t)\left|\left.\psi\right>\right.\\ ...


1

You only see something if light travels from it to your eye. When you project an image on a wall, light travels in a straight line from the projector to the wall. It does not enter your eye on the way, so you don't see the beam. when light hits the wall, it bounces in all directions. Some of the light bounces toward you. Again light travels in a straight ...


0

Imagine an electron being a sphere with an infinite number of lines extending radially from it's surface. These lines extend an infinite length. As you move the sphere towards an observer electron, the number of lines interfering with the observer will not change. They miss the observer electron. There will be one possible line of interference but your ...


0

Maybe the explanations you got until now were sufficient, but I would just want to add something simple. Let's represent a light-wave traveling in the direction $x$ as $Ae^{i\phi}$ where the phase of the wave is $(\text i) \ \phi = kx - \omega t = 2\pi\left(\frac {x}{\lambda} - \nu t\right).$ Consider a wave-front (a surface on which the phase is ...


0

There are a lot of intertwined ideas here. Let me try to tackle just part of it. When a photon interacts with a medium, it causes local polarization - that is, electrons are displaced by the E/M field of the photon. This interaction leads to a slowing down of the wave - and, as you pointed out, a shortening of the wavelength. However, at this point the wave ...


0

When you say "free carrier absorption" I think intra-band transitions. These transitions do not conserve crystal momentum, so they must occur somewhere where translational symmetry is broken. An example of such a place is the region near near the surface of a semiconductor where the bands are bent, or when an electric field is present. The ...


5

First, for clarification: As Jon Custer already mentioned, THz radiation is not difficult to generate. It's simply part of the black body radiation. What is in fact difficult is to generate coherent or at least narrow-band THz radiation. Regarding emission from semiconductor material: There is not so many materials, which would offer a bandgap in the THz ...


1

The most common colour for the sky on Earth is a white sky. Clouds cover around 70% of the Earth, and the nature by which they are scattering shortwave radiation makes them appear white or grey. This is because photons coming from the Sun are likely to be scattered multiple times, and because the dependence on wavelength is not strongly increasing or ...


2

This is an extremely hard question to answer definitely; its answer is heavily dependent on three things (1) the makeup of the atmosphere in question (2) the density of the atmosphere in question and (3) the surface temperature, and therefore the spectrum of output light, of the star in question. If the atmopsphere's gasses are themselves are coloured (e.g. ...


0

Hmm... Mars has a red sky. But that's cause of the iron content in the soil absorbs other parts of the visible spectrum and reflects back the red part, so the sky looks reddish. I do not think a blackish sky was too far-fetched cause carbon tends to be black, smoke is blackish grey. Perceived color depends on different elements absorbing different portions ...


1

You're correct in that the angular average $\langle \cos^2 \theta \rangle = 1/2$; the simplest way to see this is to take the average of the trigonometric identity $\cos^2 \theta + \sin^2 \theta = 1$, and use the fact that $\langle \cos^2 \theta \rangle = \langle \sin^2 \theta \rangle$. Thus, if your textbook defined the irradiance as $I\equiv \langle ...


0

For what it's worth, you might want to look up the original paper on the theory of helical diffraction: "The structure of synthetic polypeptides. I. The transform of atoms on a helix" W. Cochran, F. H. Crick and V. Vand, (1952) Acta Crystallographica 5, 581-86. This is the seminal work that allowed Crick and Watson to deduce the DNA structure. It is not a ...


1

Heat is random motion of atoms. In doppler cooling, lasers are slightly below a transition frequency when viewed in the lab frame of reference. Atoms moving faster than average toward the beam see it blue shifted just enough to absorb a photon. These atoms all receive a kick that reduces their kinetic energy. Now they are excited. They decay by emitting ...


2

Break the problem into two parts. First consider the convex mirror. A point source at the center of curvature would be reflected onto itself. We can equally well describe the rays coming out of the mirror as parallel beams refracted by a lens with focal length $f_1= r$. Now if we put a second lens in front, the combined lens will have a power given by ...


0

In principle, yes, but for clear air the scattering is very very weak, and the scattered light would probably be drowned out by other background sources, especially the blue sky. It would be easier at night. If the air is not clear, but instead is carrying dust or water droplets or smoke, the beam would be easily visible and recordable, again much more ...


0

Yes The image below shows the Keck telescope's laser guide star. It is designed to excite sodium atoms in the mesosphere. These excited sodium atoms flouresce and act as an artificial star which allows the telescope to correct for optical aberrations caused by Earth's atmosphere.


1

Yes, an object standing responsible for polarization can have two refractive indices. For example, calcite is a bireferingent material which splits up a plane polarised light into an e ray (one which doesnt obey Snell's law) and an o ray (one which obeys Snell's law) once incident on its surface. Since the velocity of both the rays will be different, there ...


5

Terrific photo - good that you were able to get it and don't apologize. This diagram might explain it: The sun is "below the horizon" as demonstrated by the green dashed line. A mirage can be formed by rays following the blue line (exaggerated scale, showing a layer where light can be reflected because of a sufficiently large change in density). If the ...


0

High quality, monochromatic laser beams are governed by diffraction instead of geometrical optics. Talking about rays doesn't really tell the full story. The parameter of a laser beam which expresses how well collimated it is is called the Rayleigh range, $z_R$. The units of $z_R$ are units of distance, and you can think of it roughly as 'the beam will ...


1

I'll leave the practicalities to the two other great answers (Floris's and boyfarrell's because I want to focus on one of your statements: "...focusing and collimating ...which I guess are contradictory objectives, or are they?" Absolutely not. In a weird kind of way they are the same thing: or kind of dual concepts. They are if you like two extreme ...


4

For a laser beam to be narrow and stay narrow, you need parallel beams. The usual approach is to focus the beam onto a very small pinhole (say 10-20 µm or so), then focus a second lens on the pinhole (expander). The fact that all rays have to pass through the very small point means that any diverging components of the beam will be intercepted by the pinhole; ...



Top 50 recent answers are included