# Tag Info

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I'm not convinced that variations in thickness are the cause. Variations in gloss (areas of specular reflection and areas of diffuse reflection) seem more likely: the "distribution requirements" are the same (in both cases the "defect" has to repeat at about equal distances), but the "thickness" hypothesis also requires that the curvature of the sections ...

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This is caused by chromatic aberration in the eye: short waves (blue) are refracted more (by the cornea and the crystalline lens) than long waves (red light). This isn't visible in normal conditions, but can be observed using your method. A good overview of the eye can be found in this ref., see page 49 for chromatic aberration. Because of it, a ...

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The equation holds true to all object, no matter their distance from the lens, HOWEVER when f is fixed, and the object are at different distances from the lens, then it follows that if one of the object is forming an image, then the other must not! If you had a lens with fixed focal length, and two object at different distances from it, but now you have a ...

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As @CuriousOne says, both the effects you refer to are explicable with ray optics. But both are subtle insofar that you won't find easy expressions for them in any undergrad physics course: you actually have to do accurate numerical ray simulations to see that spherical lenses impose geometrical distortion (barrelling and so forth). Very roughly, geometric ...

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The requirement is that $\chi^{(2)}$ be non-centrosymmetric. That's a bit different than having a particular parity. The states involved must be neither odd nor even; the parity must be mixed. That way the dipole matrix element exists between all three intermediate states involved in calculation of the susceptibility.

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Because in the FIR, $\omega \rightarrow 0$ and therefore the $4\pi i \sigma/\omega$ term dominates the $\epsilon_\infty$ term. It can therefore be safely neglected. You are right that the $\epsilon_\infty$ represents the contribution to $\epsilon(\omega)$ of the bound (or dipole-like) electrons.

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TL;DR: There is of course a continuous transition between seeing an interference pattern and not seeing one as the width of the source is increased. So it does not really matter if it is $<$ or $\leq$ in the condition, since the relation only represents a qualitative statement about when the contrast in the interference pattern has significantly ...

0

Some minor points: First, the units of $\lambda^2$ will be $(\mu m)^2$. Second it seems you are asking for $a, b$ so that you can determine $n$? In fact, what is more often done is to plot $n$ versus $1/\lambda^2$ to determine $a, b$. At the moment you can't do anything (unless you follow Frobenius and look up $a, b$, however this means you know the ...

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I believe the dyes behave similar to glow in the dark materials; They fluoresce a certain wavelength over time after being radiated. In the case of these shirts you'll have different chemicals for different colours, but which are all "activated" by UV radiation. There is a pretty nice explanation on Wikipedia, but basically this: You can excite an ...

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To discriminate between the other two answers. Lifetime spectroscopy would be capable of distinguishing between photochromism (in this case, reversible ultraviolet-switchable reflectance, example: hexaarylbiimidazole where the transition time is milliseconds to seconds) and ultraviolet fluorescence (where the "transition time" is zero, although the excited ...

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There is an explanation on the Del Sol web site but it omits the technical details. This is probably because Del Sol regard it as commercially valuable confidential intellectual property, and I have to concede that they are probably correct. Anyhow, some Googling has turned up a suggestion for how it works but there is no proof for this idea so treat it as ...

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A quick look at the Del Sol How it Works page shows their explanation, HOW DOES IT WORK? The Spectrachrome® crystal reveals color upon irradiation by ultraviolet waves; i.e., sunlight. When a flower blooms, the result is the exposure of the inherent color of the flower. A Spectrachrome® crystal is similar in that an energy-shift occurs causing the ...

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Here is a two slit calculator / simulator I designed in Excel. https://www.dropbox.com/s/kkopfv4xbratc9r/Alsept%20%202%20Slit%20Calculator.xlsx?dl=0 Download,save, open, enable editing The green cells are where you input distance, wavelength, slit width and separation.The yellow cells give you the fringe pattern spacing. The graph represents the right half ...

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Calculate the Fraunhofer diffraction integral of your slits and take the absolute square to get intensities.

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It is not clear to me what you meant by " I measured the size of the image on the surface of the lens". You calculated value of approximately 2 was done on the assumption that the object was in the focal plane of the convex lens of focal length $f = 12$ cm. If the height of the size is $h$ the best the naked eye can do is to view the object at the least ...

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If the question is about why red and blue light refract at different angles, this is because the refractive index, $n$ for many materials depend upon the frequency of light and they are referred to as dispersive. More often, the variation is given in terms of the wavelength ($\lambda$) and in the optics industry people quote the refractive index for the ...

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You have it backward, faster speed of light in a material corresponds to less refraction, not more. In the limit that the speed of light in a material was the same as the speed of light in vacuum, there would be no refraction at all. It can be shown that in a material the index of refraction is the speed of light in vacuum, $c$ divided by the speed of light ...

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(the following answer is included essentially in "The Feynman LECTURES ON PHYSICS-Mechanics, Radiation & Heat ,Vol. 1, 26-3 Fermat's principle of least time.) Suppose you are at point A in the land and a screaming girl is at point B in the sea. You can run with a speed $\:v_{1}\:$ on the land greater than the speed $\:v_{2}\:$ you can swim in the ...

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Comparison of fabrication techniques for hollow retroreflectors describes, in great detail, the two obvious methods: The use of a precision solid prism corner cube as a mandrel to hold the glass plates prior to gluing of the edges. An adjustable set of precision mirror mounts, designed to hold the three plates In both cases interferometric techniques ...

1

If you just have collimated light going into the system and nothing at the image plane then, yes, there will be nothing but collimated light coming out of the last lens and you could put the image screen where ever you want (and not see an image anywhere). But, if there is something at the "input" plane it scatters the incoming collimated light and its ...

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The absorption of IR lights are still more than x-rays which decrease the resolution of the imaging process. It is usually used in surface imaging in medicine. On the other hand, if we increase the IR in order to increase the resolution the temperature will be the problem.

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It may be possible using all of: 1) The thin lens graphing formula 2) The small angle approximation: $sin(x)$ ~ $x$ for small $x$. 3) The circle of least confusion. Hovever a better approach might be to graph depth of field [ using whatever accuracy or lens definition criteria you are interested in ] For specific distances and f-stops using your ...

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Your understanding of a real caustic (I presume you call it real as opposed to the imaginary caustic that you mention later) is correct. First the easy part: an imaginary caustic is a caustic located on the extension of the light rays beyond the optical system from which they arrive. For instance, in the presence of a convex lens, imaginary caustics may ...

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In Wikipedia the caustic is defined as follows. In optics, a caustic is the envelope of light rays reflected or refracted by a curved surface or object, or the projection of that envelope of rays on another surface. You can think of the envelope of a family of curves as a curve that is a tangent to each of them. Here is a diagram on page 60 of "A Treatise ...

1

Coherent sources- Two sources of light are said to be coherent if the waves emitted from them have the same frequency and are 'phase-linked'; that is, they have a zero or constant phase difference. The calculation yielded 90.pi ; We know that 2.pi denotes a phase change of zero as the waves will come back to initial phase relation of the coherent ...

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Coherency has to be over a period of time. Instantaneously any two waves (or signals) at the same freq will have at at one time or point a phase difference. If their freqs stay exactly what they had at one time, they will stay coherent with each other, i.e. The phase differences won't change over the same period of time for the two. Over a period of time the ...

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We can prove this if you dont understand the formula. Suppose $M$ is a point object at an actual depth $MA$ below the free surface of water $XY$ in a tank. A ray of light incident on $XY$ normally along $MA$ passes straight along $MAA'$.Another ray of light from $M$ incident at $\angle i$ on $XY$,along $MB$ gets deviated away from normal and is refracted at ...

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I did a little simulation using this software I wrote. The first box is air-oil-water at normal incidence, the second is looking at 45 degrees: (Your youtube video is not truly air/oil/water, but rather air/oil/wet pavement, I think, which is a bit different. So don't take these pictures too literally. But they're illustrative.) The two images are only ...

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Of course, it all depends on what you are after, and what you can tolerate. But you can get an estimate by asking a similar question: over what range of defocus does the spot size not change appreciably? Or equivalently, what is length of the near field of the focused spot? That distance can be estimated (order of magnitude!) in the following way. One ...

1

There are some materials, called Birefringent in which the index of refraction depends on the direction in which light is moving (often also dependence on the particular polarization of the light). This is due to the crystallin structure of the material which allows faster propagation in one direction or another. Another example would be a magnetized ...

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Interferometers rely on division of amplitude, when most other interference devices (Young double slit, Fresnel mirrors, Fresnel double prism, Billet double lens, Lloyd mirror...) rely on division of wavefront. Apart from this difference, the way they work is the same: from a single primary source, make two secondary coherent sources. As a consequence, the ...

0

You could use an electro-optic modulator. These don't need kV supplies, can have very fast rise/fall times, and can be fully programmable by using a digital delay generator (these can also be triggered optically for extremely good accuracy/preventing timing drift between the delay generator and whatever source you're using). You usually need the following: ...

2

Yes, depending on the quality of the filter, and so long as you make sure it is reflective rather than absorptive (i'm sure you've already checked this...) One thing I would recommend is to move the hard spectral edge of the filter a few nm away from the signal band as clipping will give you ringing/kramers-kronig dispersive effects. This is particularly ...

0

The graph of angle of deviation vs angle of incidence is a $U$ shape. The fact that the angle of deviation is the same for these two rays means that a ray which is incident at $40$ degrees to the normal will emerge at $60$ degrees to the normal. This allows us to find the refractive index. $$\sin i_1 = \sin 40 = n\sin r_1$$ $$n\sin i_2 = \sin r_2 = \sin ... 1 You are probably talking about an object under water, watched from outside (air). Given the object's and observer's position, you have to then find an angle for the light originating from the object, refracted on the surface, to hit the observer. As the observer is above the surface of the water, the angle of the light in the water can only be below the ... 4 I suspect that it is due to the fact that the oil slick you observe has almost dried up on the road. The standard calculations assume a thin film of oil on a flat surface of water. Perhaps here the oil and moisture layers on the road take on the same deformations as that of the road. As a result, from whichever angle you look at it, the effective path length ... 9 The visibility of interference fringes from a double slit depends on the how correlated the fields from the source at the two slits are. The typical slit-distance for which the fringes are manifestly visible is called the transverse coherence length d_{\rm coh} of the source. For a spatially extended source, this quantity primarily depends on two ... 0 I'll try to build an answer based on the above comments. There are two things to consider : Temporal coherence, which has nothing to do with polarization. One simple way to modelize it is the following : imagine you have an incoherent pointlike source that emits EM radiation at some frequency \omega. This source can be a thermal one, or spectral for ... 10 I'll suppose you're familiar with the theory exposed in the Wikipedia page you're citing. Keeping its notation, path difference is proportional to n_2\cos θ_2. But θ_2 is a result of refraction of incident light in air of index n_1≈1. So \cos θ_2 is bounded below:$$n_2\sin θ_2≤1\sin θ_2≤1/n_2\cos θ_2 ≥ \sqrt{n_2{}^2-1}/n_2 Taking $n_2$ ...

1

I like to look at it the other way around: the minimum resolving angle depends on your wavelength. That's why people use X-rays to look at crystal structures for instance. In imaging, where one typically uses spherical apertures, the best focus that can be obtained is not a single spot but a so-called Airy pattern. There's a bright spot in the middle and ...

1

The refraction angle will be different for different materials provided these materials have a different index of refraction. Reversely, by measuring the angle you can determine the index of refraction of the material. You can look up Snell's law.

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If your first equation is correct at the point of splitting, then your second equation is incorrect. Both $x$ and $t$ will have changed. The two amplitudes just prior to recombination are given by the third (assuming that the two arms are identical e.g., no phase element in one of the legs, and identical path lengths as in the "classic" free space ...

2

The two waves are interfering after having followed different paths, so $x$ must be different between the two. But you are observing them at the same time $t$ which must be the same for the two waves. So answer 1 is the good one.

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Borrowing the concept of a "black box" from engineering, we have a photon "going in" into the box and a photon coming out of the box. We take a "picture" of the photon going in (its amplitude and wavelength), and we take a "picture" of the photon coming out (its amplitude and wavelength), and we compare the "pictures." If the "pictures" are equal, then we ...

1

Hint: Use the final equation from my other answer here: $$E(0,t) = E_0\exp(-i\bar{\omega}t)\frac{\sin[(N/2)\Delta\omega t]}{\sin[(1/2)\Delta\omega t]}.$$ This is already a function of time. Now, $I(0,t)\propto |E(0,t)|^{2}$, so just square that function for the intensity (normalize to remove constants if you like). That will ...

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Some other effects that might be at play: 1. Reflections from the end-faces of the fiber causing interference 2. Brillion Scattering 3. Check to see if in fact the fiber you're using has a cut-off wavelength shorter than the wavelength you're actually using.

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Check the pointing stability of the laser, which, together with mechanical vibrations, would make the coupling efficiency fluctuate. After making the setup as mechanically stable as possible, try to put small diameter tubes everywhere around the beam before the fiber. And/or enclose everything in a box. Air movement has an effect, and it helps to block it.

1

Your problem is more math related than physics. To find proper full-width at half maximum $\Delta t$ and $\Delta \omega$ you have to equate the functions to $x$, with some proportionality coefficient so that when $x=1$ you're at the maximum of the function. This way after solving for $t$ or $\omega$ you'll find the half-width half-maximum by plugging in ...

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I think that the following might help: The laser modes of a cavity of length $L$ have the following (angular) frequency spacing: $$\Delta \omega = 2\pi c/(2L) = 2\pi/(T_c)$$ Here, $T_c$ is the cavity roundtrip time. The frequencies of the cavity take the following form: \omega_n = \omega_{\text{offset}} + ...

1

Any static setup will produce a static output. As mentioned in the comments, a diffusing plate will mix things up reasonably well, but with annoying side effects such as beam angular dispersal. What we actually used in some setups was a rotating phase plate. First we built or bought (I forget which) a phase plate with more or less random phase patches, as ...

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