# Tag Info

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You can use two parabolic mirrors (either one concave + one convex or two concave) in an afocal Gregorian-telescope arrangement but with multiple reflections. Every second reflection would give you an extra magnification. For example, if you take two parabolic mirrors with 100mm and 20mm focal length (5:1) then you will get after the second reflection 5x, ...

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Your picture of the eye is incomplete. You need to add the retina and to consider how the lens refracts the rays coming from the virtual object $B'$ so that they actually arrive at the same point of the retina.

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If the example reflector were a flat mirror, one image of your nose is behind the mirror surface (and if the mirror were on a wall, that'd put that image inside the wall). Another image of your nose is... your nose. You can see both, but the mirror image is virtual (because the light from that image does NOT pass through the position inside the wall ...

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The two rays which are drawn entering the eye are refracted by the cornea/lens system and form a real image on the retina of the eye. It is the real image which is formed on the retina which you detect and then surmise that it came from an “object” located at $B’$ which we call a virtual image. Put another way the eye focuses the two diverging rays on ...

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Flat metal surfaces reflect radio waves the same way that mirrors reflect light beams. for the case of a flat metal surface on a plane being hit with a radar beam from an antenna on the ground a distance away, almost none of the incident beam bounces off backwards in the direction of the receiving antenna and so a plane built of such flat surfaces yields a ...

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This sketch may give you an idea for the case of the ideal thin lens: The ray AC will continue in a straight line (going through the center of the lens), while BD, which passes through focal point F, will refract to become parallel after the lens. Simple construction shows that the point where these lines intersect is the focal plane. A spherical mirror M ...

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First let’s define „unconditionally“. The answer is yes in the sense that, as long as the mediums concerned and (a) linear and (b) isotropic the standard derivation of the Fresnel equations works, that is, (1) Assume a plane wave solutions to Maxwell’s equations for (a) an incident wave, (b) a reflected wave on the incoming wave side of the interface and (c) ...

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After gaining access to the literature above and some reading I found this passage: At the boundary we assume the radiation to be refracted by Snell's law, and the beams to be separated into reflected and transmitted parts as given by the Fresnel formulas. [...] this formulation is rather restrictive in that it does not consider volume or surface ...

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Yes, this calculation is absolutely possible. On your conventions we cannot answer, because you do not fully define your terms, and in any case the calculation is for you to make. The scheme is fairly simple, though: Use Snell's law to calculate the angle at which the ray is transmitted into the prism. Use the inner triangle formed by the ray and the ...

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It's a fact that the action for a free relativistic massive point particle happens to be (proportional to) proper time $\tau$, which superficially sounds like the Fermat's principle of stationary time. However, there are important differences and shortcomings: On one hand, the massive point particle action stops being proper time if we try to include ...

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Almost certainly the different color you see along the beam path inside the plastic is due to fluorescence: the green light in the incident beam causes the plastic to fluoresce. This is a very common effect. You often can see it in glass, too. Fluorescent light is emitted in all directions, so it can be seen even if there is no scattering.

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Derived from the Radiation transfer eqn. one has the so-called time-stationary Photon Diffusion eqn. (which is the screened Poisson eqn.) $$D \Delta I(\vec{x})- \mu_a I(\vec{x}) + S(\vec{x})=0,$$ where $S$ is the light source, $D = \frac{1}{3(\mu_a + \mu_s)}$ is the diffusion constant and $\mu_s$ the reduced scattering coefficient. This is the usual ...

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Unlike attenuation, scattering is difficult because energy which is scattered out of the beam can be scattered back in. If scattering coefficient is very weak over the path length of interest, or the scattering strongly favours small angular deviations, (green path in diagram) you can add the scattering and attenuation coefficients together to estimate the ...

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In general, an optically active material rotates the plane by an angle that is proportional to the distance traveled through the material. As noted in the Wikipedia article you linked to, the angle that the polarization vector rotates through is given by $$\Delta \theta = \frac{\pi \Delta n L}{\lambda}$$ where $\Delta n$ is the difference in the indices ...

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This is beacause, it is based on the assumption that only one optical and mechanical mode interact. Each optical cavity supports in principle an infinite number of modes and mechanical oscillators have more than a single oscillation/vibration mode. The validity of this approach relies on the possibility to tune the laser in such a way, that it populates a ...

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There is a bit of a subtlety here. Usually one would consider a medium with higher refractive index is optical thick, which means that a thickness of $d$ of such a medium corresponds to an equivalent thickness of $nd$ in air (vacuum), where $n$ is the refractive index. However, that works when considering the phase of the field. In the scenario described ...

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The questions seems to boil down to "what is so quantum about a single photon?" In the absence of interactions, a single photon is described by the same mathematics as a classical electromagnetic (EM) field. (Helmholtz equation for the electric field for instance when one considers the propagating field without charges and currents.) The only difference is ...

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The properties of the wall do have a moderate efect on the interference pattern. However for most opaque walls the simple physical optics model sort of ignoring the wall seems to work reasonably well. Precise mathematics based on Maxwell equations for light are complex and details change significantly with say horizontal or vertical polarisation for a ...

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Hei so, the short answer is: you need a Positive lens with focal length about equal to the distance lens-screen; like the Virtual Reality goggles. I will try to add some detail and links, today is impossible because my sole source of wisdom, Wikipedia, is DOWN. Formula, I will say something like: the focal lenght should be $f= b / (1 - \frac{a}{d})$ ; ...

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The factor of three enters from the flux equation during the derivation. Not a full answer, but this was too large for a comment. See here equation 5.

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Yes. Fresnel equation as written as valid for some fixed wavelength since the index of refraction $n$ is usually wavelength-dependent. If your source is not monochromatic, you need to compute $R_\lambda$ for each wavelength (using the appropriate $n$ is each case), multiply each $R_\lambda$ by its proportion in the incident beam, and sum over all $\lambda$ (...

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This is a real effect. See these ray trace simulation. Focal length of the lens is defined with parallel input rays. The focal point of the beam changes as the divergence changes (and with distance of light source to the lens - not shown).

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The lens type you need is called an ultra wide angle lens. the "widest" lens of this sort is a "fisheye" lens with an acceptance angle of 180 degrees, but these create significant amounts of image distortion. There are "near-fisheye" lenses with smaller acceptance angles and almost no fisheye distortion; one of these will make the subject in the image appear ...

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The power stays the same (other than loss at the lens), but the intensity (power per unit area) will increase when you focus down. In order to calculate your spot size (beam waist), you need to use the equations for a Gaussian beam (assuming you have a Gaussian beam, but it's a reasonably good approximation for most laser beams). There's a good intro with ...

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Maybe not the perfect answer for your question, but one that is inspired by the practicality: Because the conversion efficiency (not just the converted power directly) increases with increasing fundamental intensity, you will get more SH with higher intensity (as long as you stay under the damage threshold of your crystal/potential coatings of the crystal ...

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The other way to invert an LCD so the digit segments and the background are "flipped" is to have extra LCD elements built into the background which can be switched on and off independently of the digit segments. When you want backlit segments against a black background, you de-energize the segments and energize the background. When you want backlit ...

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In a typical LCD panel with two linear polarizers, rotating either linear polarizer by 90 degrees will invert the colors. For instance, if the linear polarizers are originally cross-oriented (as in the Wikipedia article) so that applying a bias across the liquid crystal prevents the system from transmitting light, rotating one of the polarizers so that the ...

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One difficulty with OAPs and toroidal mirrors is that they are significantly harder to align than spherical mirrors, and they're also more sensitive to beam pointing. Without knowing more about your experiment I can't really comment on which trade-offs might work best for your specific application. I don't have much experience working with 400nm light, but ...

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You get reflection at every interface between mismatched refractive indices, it is only a matter of percentage. If the surface is smooth enough and there is not too much light coming in fom above the surface to swamp out the image, a reflection will be seen.

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1) No, reflection, emission and absorption are all distinct processes. Reflection happens when an incident wave hits an interface (like a typical air-glass interface). Due to this discontinuity in the propagating medium part of the wave is transmitted and part is reflected. So no incident wave implies no reflected wave. 2) The color of an object depends on ...

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In a comment you said: The intensity of light emitted after passing through first polarizer is always I/2. That's correct if the incoming light is unpolarised, but in this exercise we are told that the incoming light has vertical polarisation. So to get the intensity after the 1st polariser we need to use the cosine formula. But we are also told that ...

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Picture is worth a thousand words: Btw, different colors here does NOT represent different wavelengths - just different light rays.

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"Is there a way to create an optical transistor that runs using a cheap laser and made using mirrors and lenses?" Yes and no. It depends on the definition of "cheap", and on how fast the "transistor" needs to be. Mirrors and lenses per se won't do anything but steer and concentrate light. You need a nonlinear medium, as @EmilioPisanty pointed out. There ...

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Is there a way to create an optical transistor that runs using a cheap laser and made using mirrors and lenses? No. Classical electromagnetism is a linear theory, which means that it always obeys the superposition principle, and if you only use linear optical elements (such as mirrors and lenses) you stay within that box. The nontrivial logical behaviour of ...

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Without knowing more about how much access you have to the actual laser, what type of laser you're using, or what type of measurement you're doing, it's difficult to suggest alternative solutions to pulse pickers, but there are several options to change the repetition rate of a pulsed laser (also depending on the technology used for pulsing, but based on ...

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The moon looks bigger when it is near the horizon than when it is higher in the sky. Many observers also report that the horizon moon appears to be closer than the zenith moon. Of course, the moon is of constant size and distance from the Earth, so, these appearences are illusory. Yet there is little agreement among perceptual psychologists about the cause ...

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temporal coherence is a measure of the correlation of the phase of the light wave at different points along the direction of propagation .It tells us how monochromatic a source is Spatial coherence is the beam of light specifically a laser to a tight spot and also allow a laser beam to stay narrow over a large distance.

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In physics one should have clearly defined terms. Black body radiation is a mathematical model where a function can describe the effects of radiation for an object that radiates off , after absorbing, any radiation that falls on it: "Blackbody radiation" or "cavity radiation" refers to an object or system which absorbs all radiation incident upon it and ...

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Monochromatic simply means that the light is of constant frequency. The frequency only affects the value of the refractive index for dispersive media. It has no affect on the formulae for the Fresnel coefficients. If your monochromatic light is unpolarized, then you need to average over all possible polarizations to properly estimate the reflectance.

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As other answers have pointed out, if one surface of the lens is made reflective by for instance coating it with a thin reflective film, for light incident from the coated side it will behave as a convex mirror, and for light incident from the other side it will behave as a concave mirror. Assuming the lens is sufficiently thin, we can also calculate how "...

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As others have pointed out, it will behave as a lens and a mirror. I wanted to point out a particular version of this with interesting applications: the cat's eye retroreflector: The cat's eye retroreflector is a transparent sphere that is silvered on one side, much like the eye of a cat with its reflective tapetum lucidum at the back. This structure has ...

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It will behave as a concave mirror with a convex lens in front of it, or just as a convex mirror if the other side is reflecting (with a lens behind it that doesn't really do anything because it's blocked). You would have to do ray tracing to see more precise behavior for the former case.

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Yes. Convex mirrors are typically made by depositing a metal coating on the curved surface of a convex piece of glass. But the focal lengths are incomparable.

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You are asked about the superposition of two sets of plane wavefronts. I have redrawn part of the diagram which shows some wavefronts at an instant in time, ie like a photograph. $k_1$ and $k_2$ are two rays with wavefronts (eg crests) perpendicular to the rays and one wavelength apart shown in the same colour as the corresponding rays. What ...

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Yes, though the fiber will be in the multi-mode regime for that wavelength, and it's possible that the glass will exhibit significant losses if you want to run this wavelength of light over substantial lengths of your fiber (probably in the multi-kilometer regime or higher). Other than that, there's nothing stopping this light from propagating down your ...

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The angular separation between the two lines of sight from the two rails on a train track gets smaller (and approaches zero) when viewing the rails at an increasing distance (angle in radians = track separation/viewing distance). This means that the image on the retina of your eye also gets smaller.

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Laser light is typically polarized, though there can be exceptions where there isn't quite enough coherence and the light comes out as unpolarized. However, the fact that you're using a given laser source as a local oscillator in a homodyne setup essentially means that you trust it to be one of the most stable, coherent sources in your experiment. As such, ...

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The human eye is close to being fully diffraction-limited, at least for photopic (cone-based) vision at the center of the visual field (i.e. for images wholly within the fovea), though it's not quite there for most people. According to Yanoff and Duker (Ophthalmology, 3rd ed. (Mosby, 2009), p. 57): The 20/20 (6/6) Snellen line represents the ability to ...

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To evaluate this integral, first separate it into its radial and angular dependences, by setting $\lambda = r e^{i\theta}$ and $\alpha=s e^{i\varphi}$, which produce \begin{align} I & := \frac{1}{\pi^2}\int_{-\infty}^{\infty}L_n(\lvert\lambda\rvert^2)e^{\lambda^*\alpha-\lambda\alpha^*-\frac{1}{2}\lvert\lambda\rvert^2}\mathrm d^2\lambda % \\& = % \...

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