# Tag Info

2

Usually each laser is specified by emitted maximum power and light wavelength. Using energy-power relation $$E=P\,t$$ and the fact that laser energy is the sum of energy of all photons emitted : $$E=n~h\nu$$, one can calculate total number of photons $n$, given laser power $P$, wavelength $\lambda$ and time window $t$ : $$n = \frac {P\lambda}{hc} ~t$$ ...

1

The formula for the geometric effects of the pinhole is due to the fact that each point will me mapped to a range of points on the screen $x_{geo}$ as admitted by the pinhole. The blur is due to diffraction which states that point from the source will be mapped to a "blob" (actually an Airy disc1). The width of this disc is roughly given as you ...

1

First, notice that the geometric blur increases linearly with the pinhole diameter, whereas the diffraction spot size increases inversely with the pinhole diameter. This makes it pretty clear that the effects have independent causes. Now, when it comes to the total resolution of the image, you're really adding the geometric blur of the light entering at ...

1

However, white light is a mixture of different frequencies of light in 'some' proportions, so how do we assign an energy for white light? We can't attribute an energy to a single 'white photon' by means of: $$E=h\nu$$ because 'white photons' don't exist. The only meaningful and useful way to talk about the energy of white light is to talk about the power ($\... 1 Assuming it's the intensity (rather than the field strength) of the pulse that's being treated as a delta function, it means the duration of the pulse is very short, and the total energy in the pulse is finite. 3 I want to add to the answer given about Helmholtz resonator. What you are expected to calculate (in principle) is the resonance of sound waves in a tube which has one side closed (your bottle). In this simple case the frequency of the different nodes depends on (among others) the length of the tube. As you fill water to your bottle, you decrease the length ... 1 This is a complicated phenomenon that depends on the size and density of the moisture droplets formed on the glass by condensation. Like a rainbow, this requires a mixture of refraction, diffraction, and total internal reflection to describe completely. The reason the pattern looks different near the edges of your image is probably because the droplets are ... 4 This is called a Helmholtz resonator, a kind of mass-spring oscillator where the mass is the mass of the air in the neck of the bottle, and the spring comes from compression of air in the main volume of the bottle. A formula for the resonant frequency is given here: https://en.m.wikipedia.org/wiki/Helmholtz_resonance 1 Take the path-length from the focal point to the edge of the zone plate,$\sqrt(f^2+r^2)$and subtract the path-length from the focal point to the center of the zone plate,$f$, then divide the result by the half-wave-length of the light,$\lambda/2$.$N = [\sqrt(f^2+r^2)-f]/(\lambda/2)$1 Sure, light could have a different speed in the same direction; that’s what happens at normal incidence (where there’s no refraction). The thing that’s actually conserved is the in-plane component of momentum,$k_x$. At normal incidence,$k_x=0$regardless of the refractive index, so there’s no need to refract. But at oblique incidence, without refraction in ... 27 This is just to add an illustration to noah's and Ralf Kleberhoff's answers which correctly point out that refraction is the main reason. Note that although most of the light rays do make it through the water drop, most of them do not continue on the path with the rest of the light bundle, but end up somewhere else. As a result, right behind the drop, the ... 24 Although there are already some good answers, I'd like to give it another attempt. Water is transparent in the sense that most of the light that enters some volume of water, also exits at the other end (unless we talk about multiple meters of thickness or lots of dirt in the water). But that doesn't mean that it passes water unchanged. Wherever light enters ... 0 It seems similar with Newton's experiment described in Second Book of Opticks, Part IV. He used concave mirrors of 1/4 and 5/62 inch thicknesses and light beam through a hole of the board which corresponds to the hole at the center of the rings on the screen in your drawing. I suppose that there is no special name for these rings but they would be made by ... 0 At the air-water boundary, most light is transmitted through, but a small amount is reflected. When you pour water from a container, it usually mixes with air and forms a bunch of bubbles. This creates a lot of air-water boundaries along the straight path of light, which results in significantly more reflection, with the reflection directions being chaotic. ... 3 The fact that you can see water at all means it's not completely transparent. Since not all the light goes straight through, there will be a shadow. 1 The meaning of "transparent" is that you can see through the object and light can go through it, however, the light can't smoothly pass through the object. It is called Refraction which means that light bends as it goes through water this bending of the light can create a illusion where there may be a shadow BELOW the glass of water but the actual ... 7 There are two factors at work here. The first is whether water is "clear", meaning, allows most light to pass through it. And it is, whether it's solid or in a bunch of tiny droplets. The second is whether you can see through it. In order for a substance to be "transparent", it has to have two properties. First, it has to allow a non-... 39 A large amount of water (i.e. if the path of light through it is long) will simply start absorbing light, as it's not completely transparent. For smaller amounts, as when pouring it from one container to another, this is mostly negligible. However, there is also surface reflection. A small amount of the incident light will be reflected off by the surface. ... 5 I guess some light will get reflected at the interface air-water, which will lower the intensity of light through the stream. That and possibly lens-like phenomena which leads to a non-uniform distribution of light intensity on the other side. 0 Look at chapter 4 here, though some details of the derivation are not completely clear to me https://s3.cern.ch/inspire-prod-files-f/fd149f5c80b07118f03732f683bd7ee5 15 Regarding water droplets collecting on the surfaces of your glasses: Those water droplets backscatter the incoming light in random directions, including ones away from your eyes. This means that any glass lens surface populated with water droplets will appear less bright than it would without the droplets, and the random scattering will obliterate anything ... 39 Mist is a suspension of tiny water droplets in air. Light traveling through the mist gets randomly scattered, mainly by bouncing of the droplets. That makes mist far less transparent than bulk water. I don't think mist is literally gray in colour but the fact that mist is far less transparent than pure air (or bulk water) causes it to look the way it does. ... 0 Something to think about: the intensity of laser light that’s eye-safe can be as much as a factor of$10^7$lower than the incoming beam, depending on the laser details. When you buy real laser goggles, they will have the OD (orders of magnitude reduction) listed on them for their wavelength ranges of effectiveness. Since you may need greater than$10^2$(OD ... 0 Don't count on it. UV is absorbed by most materials; Even a clear window blocks but that is not enough to protect because eyes are so vulnerable. Don't use a false analog to judge UV transmission. It is cargo-cultish. For example, most opaque black plastics are transparent to near IR. The fact 450nm is so close to the spectrum you can detect, the fact that ... 2 A nice way to answer this question is to do the calculation in the rest frame of the mirror. Light propagating in the$y$direction in the lab frame will propagate in some other direction in the rest frame of the mirror. It will then reflect off in the ordinary way in that frame (angle of reflection equals angle of incidence). Finally, one transforms back to ... 0 This is variation in number of fringe widths with wavelength 1 1- Yes the stimulated transition is from virtual level |u>-$\Delta$to |f>. 2- Decay from |f> is just spontaneous emission. 3- Yes, within the cell, the THz field is reflected at both quartz walls and creates a standing wave pattern. 4- Fluorescence is emitted in all directions (spontaneous emission) 5- No, Raman means the use of virtual states, and ... 2 Alcohol has a smaller tension surface than water, so when applied it's less likely to form "little bubbles" that works as little lenses and tend to "blur" the light passing trough them (this is caused by the refraction of the light rays changing material). Your assumptions about "filling the holes" is correct. Infact glass and ... 0 Is there any material which has same refractive index as the air? No there isn't. There are more answers to this question which address that. If there is then what would happen if light passes through it? Let's assume that such an object, say a fabric, exists. According to the lens maker's equation,$$\frac1f=\left(\frac{\mu_2}{\mu_1}-1\right)\left(\frac1{... 0 The dark condition is that one that you shall find, that is under what circunstances the light do not enter in te 2nd enviroment with the restrictions a) and b) , one may be that in de$n_2$all the light is absorbed, so in the 2th enviroment will be no ligth and from the 1rs material you will see dark (but this is more complicated because you need to ... 0 Generally line width of a generator (optical or not) is the width of its spectrum, i.e., the spread of the emitted frequencies around the central frequency. The shape of spectrum might be however strikingly different for different systems: e.g., radio frequency generators or high quality gas lasers may actually have a very continuous spectrum around the ... 1 Yes, it is possible. Optical fibers confine light via total internal reflection, with very low losses. Still, the losses are enough that the light intensity decays substantially after a few tens of milliseconds. 1 I think this paper may answer your question, although I couldn't find the whole paper available anywhere, unless you have privileges to download from the American Journal of Physics: Monochromators as light stretchers. It showed that indeed the prism does hold on to light for a time period that is commensurate with the uncertainty relation that you wrote. ... 0 This is called a Moiré pattern and is forms when two repeating patterns (here, the pixels of the screen and of the camera) almost, but don't quite, line up. It's related to interference and to aliasing, if those concepts happen to be more familiar. I can't say why they take on a hyperbolic pattern here, but I'm sure it's a product of the mathematics of two ... 8 It looks like there are at least two ways to go about this: Bistable MEMS MEMS (Microelectromechanical systems) are very small structures, with features from$1-100\mu m$in size, generally made with semiconductor-like processes. They can be made so they are bistable, if you apply a pulse of power, their state will change and remain stable until another ... -1 The white light is the combinations of seven 7 colors each having a different frequency What is the added frequency and the added amount of light in all of the colors it is not only adding but it is also taking it's What is the added frequency and the added amount of light in all of the colors it is not only adding but it is also taking it's frequency so ... 2 The angle of refraction is not 90${}^\circ$. The angle between the refracted ray and what would be the reflected ray is 90${}^\circ$. The reason is that the source of reflected light is the polarization of the medium. The direction of the reflected ray is fixed, determined by the direction of the incident ray. Brewster' angle is that incident angle that ... 1 At Brewster's angle there is still reflection of TE polarised light. Also, by Snell's law the angle of reflection equals the angle of incidence (+$\pi$). 2 There are many questions here. I will take on the color printer one: In (inkjet) color printing, you define a superpixel on the page which for example could be divided up into 9 subpixels in a 3x3 array. To get almost any color you want for the superpixel, you populate the superpixel with anywhere from one to 9 spots of (for example) magenta, 1 to 9 spots ... 1 "Does the magnification solely depend on its shape/material (in which case they would all be the same)?" In the realm of simple geometrical optics, the magnification of a lens is given by q/p, where q is the image distance and p is the object distance. So it's not really a property of the lens but rather of the image conjugates in any particular ... 0 I suspect the purpose of this problem is to see if you know how to use the lensmaker's formula, rather than to design a practical lens system. So IMHO the right approach would be to construct a formula for total magnification, with the given parameters as constraints, set magnification at$10^6$, then solve for the lens spacing as a function of lens focal ... 0 Look at a piano keyboard. The space that each octave takes up (a C key to the next C key) is the same. This is actually logarithmic spacing. If you spaced the keys linearly according to their frequency, the keys would be spaced much further apart in the higher octaves. Each C note actually has twice the cycles per second as the previous C note. So some ... 0 There are different phase matching conditions. Here it seems that (a) represents type I phase matching and (b) and (c) represent type II phase matching. Then there are also different combinations of wavelength that would satisfy the energy conservation:$\omega_{pump}=\omega_{signal}+\omega_{idler}$. The different wavelength produce different cone angles. ... 2 When you look at a spectrum, you're using some kind of optical effect to spread out the different frequencies of light. Most commonly this is done by a triangular prism, where the difference between the speed of light in air and the speed of light in the prism material ('refractive index') causes different frequencies of light to bend by different amounts. ... 2 ...the "width" of any given color is different This is to be expected, since the unit of the horizontal axis might be different. Two separate bands that are equally wide on a frequency (energy) scale will always have different width on a wavelength based chart. 0 You can just list the possible outcomes of the setup, in this case$\{|0\rangle, |H\rangle,|V\rangle\}\$, and write the corresponding projectors as you did. I would note that for this to be a POVM, i.e. in particular for the projectors to sum to the identity, you need this list to be an orthonormal basis for the space (whether this is such a basis depends on ...

7

Others already pointed out the effects of sensors and pigments (or emitters) that cannot perfectly mimic the response of the (standard) human eye. So one would need to look at a real spectrum. The excellent answer by Jonathan Jeffrey is to point to a prism. But there one has the problem that the dispersion increases towards the ultraviolet, and that relative ...

48

Most computer monitors aren't capable of displaying any spectral color. Some of the RGB monitors could display at most three of them: some red wavelength, some green and some blue. This is because the gamut of the human vision is not triangular, instead it's curved and resembles a horseshoe: In the image above, the black curve represents the spectral colors,...

2

However, in each of them the "width" of any given color is different. That is because the images were recorded with different methods and materials. What does the "true" visible light spectrum look like, then? It can't be that each and every image search result is correct. "True" needs the measurement of the frequency on the ...

17

If you're really curious, buy a cheap prism, and take it outside in sunlight. You'll be dispersing the frequencies present in sunlight, and in addition, your eyes are more or less sensitive depending on the frequency, but that's a good start for being able to see what a "real spectrum" of visible light is. A monitor does not produce all frequencies ...

Top 50 recent answers are included