# Tag Info

1

there are other parameters like the number of free electrons in the atoms of the material, atomic size etc. Close. While density of particles does matter, it also depends on the material property. More precisely, it is closely related to how the electrons react when situated under electromagnetic oscillation. Each bound electrons has its natural ...

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I want to know whether the amount of refraction of a given monochromatic light depends solely upon the density of the of the medium ( increase the density to increase the angle of refraction), or there are other parameters like the number of free electrons in the atoms of the material, atomic size etc There are a number of factors at play, the ...

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Hamilton's principal of least action would apply which is more general than Fermat's principle. If Energy remained constant then action would be proportional to time so probably something like Fermat' principle would apply, but this is just a guess.

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You need to learn about the Eikonal equation and the equivalent ray path equation, which I talk about in my answer to the question Physics SE question "Ray tracing in a inhomogeneous media", and, if you need to know how it comes as the *slowly varying envelope approximation" from Maxwell's Equations, I talk about this in my answer to the question , "Optics: ...

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A polariser is quite good at blocking EM radiation, but not perfect. No polariser is perfect, and does not block 100% of the radiation. This is typically specifed in terms of the extinction ratio. The best laboratory grade polarisers have extincition ratios of the order 100000:1 [1]. Such good polarisers are only possible over certain wavelengths, and for ...

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Using your diagram as a reference, the rays from the source are parallel to the (vertical) median of the prism (C). The surface(s) they are incident on are inclined by angles $a$ and $a'$ to this median. So the incident ray makes an angle of $a$ and $a'$ with the respective surfaces (note: not with the normal). From the law of reflection (equality of angles ...

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This answer assumes you use can use more than one lens. The lens at the far aperture needs to be very short focal length ($f_1$) convex or Plano-convex lens. This will be able to accept light from a wide field of view and focus it to a point $A$ inside the tube. Next you place another lens of focal length $f_2$ at point $B=A-f_2$. This will collimate the ...

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Imagine you start right next to a star. As you move away from the star, the intensity of the light $I$ (in $W/m^2$) goes down depending on distance $r$ following an inverse-square law: $$I\ \alpha\ \frac{1}{r^2}$$ but the solid angle $\Omega$ (in $sr$) also decreases in the same proportion: $$\Omega\ \alpha\ \frac{1}{r^2}$$ Therefore since radiance $L$ ...

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If you combine the two statements In the paraxial approximation for a monochromatic light field, the complex light field in the back focal plane of a lens is the Fourier transform of the complex light field in its front focal plane. The lens in the human eye is a lens that can accomodate its focal length so that the retina lies in the focal plane of the ...

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Yes light can change propagation direction when scattering off of a grating into pre-defined directions that depend on the angle of incidence and the grating period (as well as the refractive medium the grating resides in). This is just Bragg's Law. The change in direction (or momentum) is determined by the above parameters. You can think of the change in ...

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We first consider the relation: $$n\delta{\lambda} = d\delta{\theta}\cos{\theta}$$ It's content is that the $n^{th}$ order maximum of a wavelength $\lambda + \delta{\lambda}$ is displaced from the corresponding maximum for a wavelength $\lambda$ by the angle $\delta{\theta}$, related to $\delta{\lambda}$ by the above equation. Now, we can ask the question, ...

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I could not address all issues with wltrup's answer in one comment. Firstly, the reflection number 1 is not responsible for the the top image of the photo. This is actually reflection of the environment, and would be the same independently of the picture below. Here, the light follows the path: ambient->surface->eye To actually capture an image of the ...

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Polarization maintaining fibers are (as far as I know always) single-mode fibers. It is inherently difficult to efficiently couple light into single-mode fibers because, by definition, they will only support one specific optical mode which you have to carefully match with your input beam. So this mode matching will most likely be your main challenge. Getting ...

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This system is actually a little more complicated that I first thought because the path length to both eyes must be the same to "clone" the light source.

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It might be clearer if you read it as "detect THE single particle passing through the slits" . If you use a strong light then there are lots of photons passing through the slits and the interference could be completely classical with different photons going through each slit and interfering at the screen - just like water waves. This was the picture for ...

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The equation you are quoting gives the power of a lens in terms of its geometry and refractive index. Simply rearranging the terms (dividing by $n$) gives you an expression for $\frac{1}{f}$ which is known as the power of the lens and is expressed in diopters. For the usual situation of a lens in air, we can put $n=1$ which leaves you with an even simpler ...

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Dunno what book you're quoting, but you should realize that the index of refraction of air is $n = 1+ \epsilon$ (where I'm using the mathematics standard of $\epsilon$ being a tiny number). Thus the power in air is $1/F$

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Have a look at my answer to Slit screen and wave-particle duality because this covers a lot of topics relevant to your question. You're correct that if we imagine the photon as a little ball then if the arms of the interferometer are different lengths the two "halves" of the little ball cannot arrive at the detector at the same time. But this is not how the ...

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When one says "photon" one is in the quantum mechanical frame. Quantum mechanics does not follow the rules of classical mechanics if one tries to consider the photon one classical entity, like a bag of energy flowing. The photon is a point like elementary particle in the standard model, it has no extent and when it hits the detector it registers at a ...

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Does a spherical wave-front, as you describe it, require at minimum 2 spherical bursts from a single atom? If spherical wave-fronts form as energy bursts that are independent in nature, with time greater than 0 (t>0) between each burst, and each burst being equal in energy, then in order to adhere to E=hv, might determining the frequency and energy of a ...

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If a single red photon hits your telescope from the direction of a planet in the Andromeda Galaxy, then all you know that the planet emitted a red photon. Was it caused by a fire? A scattering of starlight through its atmosphere? An Andromedan with a laser pointer? A single photon of light from an unknown source has about as much information as a random ...

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Suppose you are using a CCD or a photographic plate to record your image. The interaction with the light occurs when the detector absorbs a photon, and this happens at a point. So the image is built up from a collection of points - one for each photon that is detected. In everyday life, e.g. taking pictures with the CCD in your phone, the intensity of the ...

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Plane wavefronts ensure that the field at the apertures are in phase and coherent and thus the interference pattern is produced (see Fig.1). Non planar input to an aperture equals a case where the point oscillators at the apertures are not in phase and depending on the coherence length of the incoming field, might not be coherent (see Fig.1). Thus, the ...

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The classic example of when the correct path should maximize the time is inside of a mirrored ellipse. There are four possible paths for a light ray which begins and ends at the center (shown below). Two of those paths are maxima and two are minima. The fact that the original statement of Fermat's principle does not account for this is probably what Hecht ...

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You can actually do this a bit more simply (or at least without integrations). Luminance is invariant in geometrical optics. That is, the brightness of an image cannot be brighter than the source. The radius of the sun is 0.6958 x 10^6 meters. The radius of the earth's orbit is a mean of 149.6 x 10^6. Then the brightness at the surface of the sun is the ...

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All of these criteria are more or less the same. I'm not aware that there is one "standard" criterion. More importantly, I think, is that these numbers obtain for ideal conditions, but real systems will suffer from various imperfections and aberrations meaning that the actual resolution and spot size will be degraded. Furthermore, specific applications ...

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The sun is an extended source. This means that it occupies a definite solid angle in the sky $\omega = 6.8\times 10^{-5} Sr$. To visualise this (not to scale), let say that the black area in the following diagram is the angular extend of the sun as seen from the surface of the Earth (ignore the other labels), What happens when we concentrate sunlight ...

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Wikipedia has a good article on hybrid photovoltaic-thermal systems. As you proposed they consist of a solar cell with a thermal collector at the rear. Solar energy conversion is a fascinating topic from a thermodynamic perspective and has been summarised beautifully by the work of De Vos, The Thermodynamics of Solar Energy Conversion, ISBN: ...

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$\omega_{3}$ and $k_{3}$ are the frequency and wavevector of the generated wave. As you mentioned above, we are sort of looking at: $\omega_{3} ~=~ 0~,~ \pm (\omega_{2}-\omega_{1})~,~ \pm 2 \omega_{1}~,~ \pm 2 \omega_{2}~,~ \pm (\omega_{1} + \omega_{2})$. The names for these, respectively, are: optical rectification, difference frequency generation, ...

2

You will see transmission peaks when integer multiples of the half-wavelength fit into the cavity length $L$. So, this means the condition (n is an integer): $L = n \lambda /2$, or: $\lambda = 2 L / n$. This is often expressed using "wavenumber" $1/\lambda$ (proportional to frequency), so that the spacing is even: $\frac{1}{\lambda} = \frac{n}{2 L}$. ...

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A year later, here is a probabilistic (pseudo QM) explanation. I am confused by the diagram that appears to show unpolarized laser light - I thought that most lasers by their nature produce polarized light; after the first polarizer that question is moot, so let's start there. A polarized photon can be thought of as being in a mixture of states - when it ...

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If there are two weakly-coupled single-mode waveguides, then the whole system is almost guaranteed to have two modes (you can call them even & odd, bonding & antibonding, symmetric & antisymmetric, whatever notation you like). The best (and usually only) way to calculate mode indices is electromagnetic simulation software.

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A waveguide cannot directly transmit a whole image--it gets scrambled beyond recognition even if all the light passes through. What you can do is form an image at the entrance to a bundle of waveguides, then each waveguide carries one pixel, so you still have an image at the other end. Analogous to an endoscope.

2

Qualitatively, the thing that happens under water (when you wear a diving mask) looks like this: The green lines represent the path the light would have taken without the water, and therefore the "apparent size" of the bubble. But as you can see, the refraction of the light away from the normal (transitioning to a medium of lower refractive index) causes ...

1

Wearing the mask underwater doesn't do anything to enhance your vision, but it does make objects in the water appear larger/nearer. This is due to the refraction of light at the air/mask interface (more info in my answer to this question). The objects are magnified by a factor approximately equal to the ratio of the indices of refraction ...

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Use your power meter and graph paper to map out the reflectivity as a function of angle. The Fresnel equations (plots shown below) are rather sensitive to the relative indices of refraction of the glass-foil interface. If the index of refraction of the foil is higher than the glass then you will hit a plateau above the critical angle at which all of the ...

4

I don't know the "official" answer but here is what I might try. I am hoping that others will contribute to make this a "good" answer. First - we were not told whether the wavelength of the laser is transmitted at all by the blue foil; but since blue foil typically absorbs red light, and most laser pointers are red (I have a blue one but they are ...

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There are closed form solutions for very simple ideal cases such as two planar waveguides, having no bends, with step-index profiles, made of a uniform material, having no losses, ... But real couplers have none of that. The waveguides have to bend in order to bring the two waveguides near each other, leading to bending losses at the coupler. The ...

3

I don't think it's very likely, but one other thing I can think of: when the sign before the $\omega$ is a minus then the wave represents a wave travelling to the "right" - positive x direction - and maybe your TA wants only waves travelling to the right. ^^ In case you don't know why the minus sign represents a wave travelling in the positive x direction: ...

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You may have lost marks for leaving out the imaginary unit $i$. If not, then understand that either $e^{\pm i\,\omega\,t}$ can be used to represent the real signal with positive frequency $\omega$ - it's wholly a question of convention. But once you have made the choice you must stick with it and the choice has implications throughout all the equations of ...

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I think that you are missing some $i$ or $j$ (complex $\sqrt{-1}$ in your expression, like in $exp[j(\omega t-\phi)]$. consult this page of the book: The Light Fantastic: A Modern Introduction to Classical and Quantum Optics and read the equation 5.10 and the side note: The choice of the complex form ... is made for convenience. The frequency $\omega$ ...

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Due to the Structure of Glass, No Interference. To determine the thickness required to cause thin-film interference, both in light and in oil or soap bubbles, you rely on the following equations: $$2n_{film}d_{film} \cos{\theta} = m\lambda$$ $$2n_{film}d_{film} \cos{\theta} = (m-\frac{1}{2})\lambda$$ These being the equations for constructive and ...

0

It's true that laser's traditional applications are related to their monochromatic (one wavelength), collimated (one direction) and coherence ("one phase" or phase matching) characteristics. Laser beams have low dispersion, can be amplified and focused to reach very high photon densities and pulse shape can be sculpted at will. Many applications are only ...

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Pieced together from other discussions: The only way to get large amounts of monochromatic and/or collimated light is through stimulated emission, which happens to also produce coherent light. For example, if we want a more collimated LED then we we end up with a diode laser. And if we want to more collimate the output of an arc lamp we have to introduce ...

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That's right, the peak intensity is at $z=0$. By the way, you should have written how $\omega$ and $z$ are related.

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Perhaps part of the problem is that the statement $2 ( l - l') = n \lambda$ is not correct in general. It applies to the specific situation where both the incident wavefront and the refracted wavefront accumulates a path difference of $l - l'$ between each layer of your crystal. That is true if the diffraction causes a reflection back in the same direction ...

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Rather than a definite answer, I can, at this stage, only present a partial collection of information regarding or related to the possibility of using laser power from a launching site on the surface of the Earth and achieving orbit in a single stage launch profile: From: http://www.astrobio.net/news-exclusive/beaming-rockets-into-space/ Beamed thermal ...

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See, first of all our eyes are not a good device to determine the main color of a group of photons. the main color that we see is actually the intensity of a specific range of frequency in the light wave. it means there are all kind of photons coming out of the sun, but the amount of "Yellow" photons are much much more. that's because we see the sun ...

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