# Tag Info

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The light is reflected from a mirror. If the mirror moves a distance $\frac \lambda 2$ then the incident ray of light has to travel an extra distance $\frac \lambda 2$ to reach the mirror and then the reflected ray of light has to move an extra $\frac \lambda 2$ a total extra distance of $\lambda$. If the path length changes by one wavelength then there is ...

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Maesumi has already provided references, I would just like to point out how you can solve the problem yourself: the law of reflection most certainly holds in the rest-frame of the mirror. So to find out what happens when it is moving, transform to that frame, apply the law of reflection, transform back. As correctly pointed out in the comments you will find ...

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This problem is considered in the following papers 1) A. Einstein, Zur Elektrodynamik bewegter Korper {Ann. Phys. (Leipzig)} {17} (1905) 891–-921. http://onlinelibrary.wiley.com/doi/10.1002/andp.19053221004/pdf https://www.fourmilab.ch/etexts/einstein/specrel/www/ Reprinted in Einsteins Miraculous Year: Five Papers That Changed the Face of Physics, ...

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It doesn't, and IMHO it shouldn't be a laser. Lasers produce light that is (1) coherent, and (2) of gaussian intensity distribution, both of which cause eye strain. The coherence leads to speckle; the gaussian spread means the focussed spot is also gaussian, and your eye keeps trying to improve the PSF. Another reason a laser is bad: it's a very small ...

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The light from a typical laser emerges in an extremely thin beam with very little divergence. Another way of saying this is that the beam is highly "collimated". An ordinary laboratory helium-neon laser can be swept around the room and the red spot on the back wall seems about the same size at that on a nearby wall. The high degree of collimation ...

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Semiconductor light emitters are made of such materials, which have quite large index of refraction. This makes it hard for light to exit the emitter — due to Fresnel equations and low index of refraction of air. In a laser the light mostly goes back and forth between two mirrors, and reflections only help the lasing. So the light either exits from a tiny ...

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Consider a hollow insulated body with a very small hole, a ray is passed in it. The body is made in such a way that the possibility of that ray to get out of that body is almost zero. So this body completely absorbs that ray. Reference

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Just to add to the above answers, and since to did not limit your question to the visible range - if you define black as absence of light (photons emitted or reflected), then there is no such substance, because according to black body radiation model, everything with a temperature above absolute zero (which is essentially truly everything in the universe:) ...

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If you have access to some physics review, I suggest you look at a series of articles by Anthony A. Tovar and Lee W. Casperson, "Generalized beam matrices: Gaussian beam propagation in misaligned complex optical systems," J. Opt. Soc. Am. A 12, 1522-1533 (1995) in which they describe the ABCDGH transfer matrix formalism. Here is the link to the first one on ...

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Apologies everyone, I misunderstood this question greatly. It's talking about the first dark fringe (or the zeroth order dark fringe) which occurs in between 0 and 1. I understand now.

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It's not m=0. The sin of d ( or the angle that you will find the dark spots) is equal to ( m+1/2) times the wavelength. m=0,1,2,3.........

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Very reasonable question. I will try to answer it in an intuitive way. If you have a scattering medium, photons are reflected in random directions; but when you have a refractive medium, something else happens. The photon is not absorbed and re-emitted: instead, the photon interacts with the electrons in the medium, and since these electrons are somewhat ...

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That's right. There are materials that possess these properties. They are birefringent, see Wiki Page Birefringence. Crystals can have different refractive indices along their crystal axes which leads to the phenomenon that you describe.

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The point-source response is also called the point-spread function (at least for telescopes). This defines how an idealized dot of light at infinity is imaged by the optics of the telescope (or eye). Instead of appearing as a perfect dot (presumably on a single pixel, assuming sufficiently small pixels, for a camera), the dot is imaged as some complicated ...

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It's maximizing the angular entropy in the sense that far more "available" angle vectors are inhabited. I suspect you need to be careful with the word "entropy" here. For example, the photons are not down-converted into a larger number of photons of longer wavelength (aka 'heat death of the universe').

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Your understanding of the degradation mechanism is correct. However, you can compensate for this to some degree by training your pupils to contract underwater, thereby reducing the refractive variation of light transmitted to your retina. This famous study by Anna Geslen describes this in detail, and also showed that the remarkable underwater visual acuity ...

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The observation you mention is a consequence of the fact that under the aforementioned conditions, the electric field of the s-polarisation (or, "TE") always remains parallel to the optical axis, no matter what the incidence angle is. If one substitutes the effective permittivity along the optical axis into the Snell law, it holds (for this polarisation ...

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Note it is not exactly periodic, it only touches zeros for some equally spaced values of L, but its maxima become lower with decreasing coherence length. Note also you are using the non-depleted-pump approximation. The evolution of the SHG wave is determined by two factors when propagating through the nonlinear medium - it gains energy from the pump wave, ...

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The answer is purely a matter of engineering and cost, I think. I've bought prescription masks in the past, and they are as you say: basically a stock faceplate with a correcting lens attached to the inside face. This allows the manufacturer to stock a standard (flat) faceplate and pull units from stock to build specialized prescription items. Now, if ...

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Bandwidth is the most important thing here. In order to send information you need distinguishable pulses, which can be interpreted as 0 oder 1 for no pulse or pulse respectively. The optical bandwidth and pulse duration are related by the time bandwidth product $$0.44 < \Delta t \Delta f \approx \frac{c}{\lambda^2} \Delta t \Delta \lambda$$ Note the ...

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Huygens worked with scalar, longitudinal waves. This was proved to be incorrect in 1821 by Fresnel, who showed that polarization requires transverse waves. A (relevant but incomplete) history of light and how well different models explain certain experiments is given in my recent lecture ''Classical models for quantum light''; see ...

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The answer to this question depends on various aspects. For instance when you say white object, do you mean perfectly white? or white with respect to only visible light? Same happens for a mirror too. The most direct way to look into the problem would be following. Mirrors are nothing but extremely fine and optically flat white surfaces at the back of ...

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Yes, your myopia is relevant in the sense that you notice immediately a huge improvement in your long distance vision when wearing goggles underwater. Short-sighted people have difficulty in focusing distant objects (or nearly parallel light rays); the eye is "too long" for the lens and the focus falls in front of the retina. Corrective lenses for myopia ...

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There are several broadening mechanisms, and you have to know how they "add" together. Since a Voigt profile is the convolution of a Gaussian and Lorentzian profile, you rightly calculated both widths rather than just the overall width of the Voigt peak. For the Lorentzian portion, the width is the arithmetic sum of the individual widths: $\Delta ... 52 Is blurred effect due to turbulence? No, it is not. The turbulence has a little effect here. Even if there is no turbulence, one see everything blurred underwater. The reason is explained below. An eye is a natural lens. A clear shot of something you see depends on how well the image is focused on your eye. The most of the refraction in the eye occurs ... 6 This phenomena is called Heiligenschein effect. But there are many other names for it. This is generally seen on dew covered grass. The dew drops act like lenses, focusing sunlight on the grass leaves and illuminating them. Read these for more information and pictures: Dew heiligenschein (more images) Optical effects: Heiligenschein Watch this video ... 0 It depends on what error you want to quantify. You can take several images of the same beam at different times (frames of a movie), then for every pixel you find the time-average and standard deviation. This will give the time average and uncertainty, related to the stability of the laser intensity. 0 You do not have enough information. If you could relate the intensity in the picture with the number of photons detected by the CCD, you could use square root of that number. So say 49 in your graph correspond to 49 photons. Then the error bar on that point is 7. But if the same intensity correspond to 4900, the square root is 70, and your error bar is 0.7 2 The reason for these is mostly reflection inside the optics of the camera, both from the surfaces of the lens elements and from the structure of the lens body, diaphragm, camera body and so on. There can be other reasons for artifacts like this: diffraction effects when the lens is stopped right down and effects due to light bouncing around inside the film ... 1 Focal length of Lens 2 is$d_5$, because parallel rays converge at a distance of d5 from lens 2.Focal length of lens 1 is$d_1$beacuse diverging rays become parallel to optical axis at a distance of d1. Focal length of system is$d_2+d_3+d_4+d_5$FROM LENS 1. 0 This is a phenomenon called dispersion of light. White light is a mixture of all colours which get separated when they pass through a prism. The refractive index or simply for lay usage the angle at which light bends when it enters from one medium to another depends on the wavelength of light. So the colours are not newly created, but they are just ... 6 When light enters glass (or another transparent material), its frequency stays the same and its wavelength changes. In a comment, you say that you are using "color" to mean "wavelength". Well, I think you are using the word "color" incorrectly. According to a normal definition of "color", the color of light does not change when it enters glass. But the ... 0 Good question! There is an answer, but it's not as simple/satisfying as "yes they do!", and is pretty involved. There may be another way to do this, but this is how I approach similar problems. Whether or not they are described well by the Airy functions will depend on the exact geometry of each of the droplets/splats and also on the amplitude and phase of ... 0 The index of refraction for diamond is much larger than for ordinary glass which means the critical angle in air is much smaller. More rays inside the diamond will experience total internal reflection than would in glass. The dispersion value for diamond is much higher than for glass. In both glass and diamond red and blue light will refract at slightly ... 0 If the angle of deflection is at all large I think the corrector plate will be very thick at the centre, possibly a large fraction of the focal distance. If the beam was parallel rather than focussed before the mirror, the corrector would become a large lens which focusses the beam on a flat field. 1 The angles XCF and N'CB are marked as being the same angle, but that is clearly nonsense since that would makes line CB parallel to line FB. 0 Simply, if you have an object that appears Cyan in White light, then that must mean it absorbs Yellow and Magenta. If you then illuminate that object only with Yellow light, it will appear Black (dark) because the object absorbs Yellow light. 0 The angle of minimum deviation occurs when the incident angle and exit angle are equal, so no. 0 No. Light field cameras are indeed fascinating and they can be used perform impressive feats such as refocusing after the image data has been recorded. They are not, however, magical, so they are restricted to providing information which was already encoded in the photons incident upon their objective lens(es), as you noted at the end of your question ... 0 The only way a collimated beam will focus on the rear surface of a sphere is if the diameter of the sphere divided by the refractive index is equal to the focal length. Since the focal length is equal to the radius of the sphere divided by the change in refractive index (n-1), this can only occur when: 2R = nf = n (R/(n-1)) 2 = n/(n-1). Thus n=2. ... 2 You will want to refer to the blackbody curve, which will have a broad spectrum and a peak wavelength. Consider the sun for a moment. At a temperature of around 6000 Kelvin the peak wavelength of its blackbody curve will appear around 500nm, which is blueish-green light. But that is obviously not the only wavelengths that the sun is emitting. The ... 7 It sounds like you know some of the most important summary points about blackbody radiation, but here is a reference on the subject, since I will be talking almost entirely about blackbody radiation: https://en.wikipedia.org/wiki/Black-body_radiation Given any temperature, there is a certain emission spectrum (see ... 3 The first equation correctly states that $$v_g = \frac{c}{n}\left(1 + \frac{\lambda}{n}\frac{dn}{d\lambda}\right).$$ But if you look at the wikipedia page that you linked to, you'll see that the second equation should read $$v_g = \frac{c}{n}\left(1 - \frac{\lambda_0}{n}\frac{dn}{d\lambda_0}\right)^{-1},$$ where$\lambda_0$is the wavelength in vacuum, ... 2 It just geometry. If you want all incident rays parallel to the principal axis to reflect through a single point the mirror needs to parabolic in shape. For a focal length f the equation of the parabola (opening upwards) would be$4fy=x^{2}\$. A spherical mirror will have approximately the same shape as a parabolic mirror near the vertex. The focal point ...

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I don't think there is a really strict definition of a ray. I would say that a reasonable definition is: electromagnetic energy density is confined to a finite area (e.g. gauss beam and not plane wave) in the transversal plane The Poynting vectors are reasonably parallel in this plane. (e.g. gauss beam and not spherical wave) The direction the ray moves ...

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I just found this source (1) which explains that rays are curves such that the direction a ray points is the same as that of the Poynting vector at a given point. References (1) p110 of 'Microwave Antenna Theory and Design' by S.Silver (link to google books)

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This is a problem in Fresnel diffraction. There is no closed-form solution, as there is for the PSF (point spread function) in the focal plane. There is an exact solution in terms of an integral. The details and the integral, along with series approximations good for small values of defocus can be found in Chapter 9 of Born and Wolf. Born and Wolf also ...

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Ok, I found an answer on my own: For the first definition the answer from L. Levrel is sufficient. For the second definition one has to follow the following set of equations: $$\begin{split} \frac{1}{v_g}&=\frac{dk}{d\omega}\\ &=c^{-1}\left(n+\omega\frac{dn}{d\omega}\right)\\ &=c^{-1}\left(n-\lambda\cdot\frac{dn}{d\lambda}\right)\\ \Rightarrow ... 2 Group velocity, for any kind of wave, is defined as$$\boxed{v_g=\frac{\mathrm dω}{\mathrm dk}}.$$Phase velocity is defined as v_p=\dfrac ωk, and refraction index as n=\dfrac c{v_p}. So ω=c\,\dfrac kn. Hence,$$\mathrm dω=c\left(\frac{\mathrm dk}n-\frac{k\,\mathrm dn}{n^2}\right)=\frac{c\,\mathrm dk}{n}\left(1-\frac kn\frac{\mathrm dn}{\mathrm ...

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