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Edit: Reading other people's answers, I forgot to mention I assumed a flat mirror. Excellent question, but the answer is no. The reason is because the object (in the strict optics meaning) in the case of the photograph is actually on the paper whereas in the case of the mirror it is still at the same place, far behind: the rays of light coming from it are ...

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The other answers have addressed rather well what happens in a vacuum. However, the situation is rather different in a medium, and there are indeed certain fundamental constraints which mean you absolutely shouldn't go above a certain intensity in material media. The name of the game is Kerr lensing. The Kerr effect is the most basic of nonlinear optical ...

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There is a limit, though I'm not sure exactly where that limit will be. We know there must be a limit because if you concentrate enough energy into a small volume you get a black hole. However this limit is far, far above anything we can achieve at (for example) the NIF. Photons are bosons so there is no limit to the number that can pack into the same ...

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This is what I think the first bit of the calculation does. Suppose you start with a spherical eye with a hole in it (e.g. the pupil in the human eye): The radius of the eye is $ER$ and the radius of the hole is $AR$, and with the length $DA$ these form a right angled triangle. Pythagoras' theorem tells us: $$DA^2 + AR^2 = ER^2$$ so: DA = ... 5 With a flat mirror, it would be worse. The distance from eye to object is the distance without the mirror plus twice the distance from eye to mirror. With a convex mirror, it is even worse. With a concave mirror, the answer is it depends on the curvature, the severity of myopia, and the distance to the mirror. 4 You make a good point which requires us to be more careful about what Fermat's Principle says and how the proof proceeds. The upshot of what I'm going to say is The statement of the Law of Reflection must include an appropriate constraint. Here's what I mean in detail. First, let's give a precise statement of Fermat's Principle: Fermat's ... 4 Assuming the mirror has no curvature then things would not look clearer. You have to think about the image as it is being projected, our eyes doesn't perceive reflected image as being at the point of reflection but rather perceives the image as being a set distance from our eyes. To clarify if you imagine your eyes couldn't look at anything but the mirror ... 3 The vacuum is polarizable. The polarization can be with respect to electric charge or color charge. In the presence of an electric field, virtual electron-positron pairs briefly exist (created from virtual photons of sufficient energy). The virtual pairs act as dipoles and orient with respect to the field. For example, near a proton, the virtual electron ... 3 The "color" of a body is not a property of the body nor of the light it reflects or emits, but rather of the human eye and brain that receive and process the light. Of course, this is based on a property of the light, for this is how the eye receives the information that there is some object in there in the first place. This property is the spectrum of the ... 3 It depends on what you mean by the energy of light. A light wave has an associated energy flux, that is how much energy passes through a surface at right angles to the light ray. This energy flux is a constant and doesn't change when a light wave passes through materials of different refractive index. However, since the velocity of light is reduced in a ... 3 It depends on the position of the sun. A rainbow does not exist at a particular location in the sky. Its relative position depends on the position of the observer and the sun. All raindrops refract sunlight in the same way, but only the light from some raindrops reach the observer's eye. This light is what constitutes the rainbow for that observer. The bow ... 3 Here's the logic (well a particular rendition): Recall that n is defined as the ratio of the speed of light c in vacuum to the speed of light v in the given medium; \begin{align} n = \frac{c}{v} \end{align} Note that in a linear medium, Maxwell's equations are exactly the same as in vacuum, except \mu_0 and \epsilon_0 are replaced by \mu and ... 2 The acoustic wave in the material causes a variation in refractive index and the amount of light scattered at a particular angle (its similar to Bragg scattering) depends on the intensity of the modulation. The scattered light does carry both frequencies in the form of blue or red shifting. It shifts by mf where f is the driving frequency and m is the order. ... 2 The diffraction at the single slit will form an sinc pattern. The image below (taken from this hyperphysics article on single slit diffraction) shows what this will look like. The first dark fringe is just outside the large central lobe and the next bright fringe is the second largest lobe just above the dark fringe. Good luck! 2 In principle yes. But to make a good \text{Zn}_3\text{N}_2 homojunction LED you need the capability to incorporating both p-type and n-type dopants (normally oxide materials are naturally n-type) which might not be possible. From what I have read, this material has been proposed as a way of making p-type ZnO (which is naturally n-type) by a post growth ... 2 I'm not certain what "backed by a convergent lens" means in this context. A divergent lens by itself cannot form a real image, since a divergent lens has a negative focal distance. Use the thin lens equation:\frac{1}{d_o} + \frac{1}{d_i} = \frac{1}{f}$$Since f < 0 and d_o > 0 by convention, (\text{positive}) + \frac{1}{d_i} = ... 2 I don't think it would be a good idea. Solar power plants are effective only because they direct light beams to a small area, which result in high heat rate concentration. To make the water evaporate "significantly" faster You would need huge amount of mirrors or else the heat rate concentration would be too small to make a difference. Answering to the ... 2 Not necessarily. It depends on where the image is formed from the mirror. Depending on the radius of curvature (assuming spherical curvature) the image ould form anywhere, but the person would want it to form on their retina. You can calculate this using the mirror equation \frac{1}{d_{0}}+\frac{1}{d_{i}}=\frac{1}{f} Where d_{0} is the distance (from ... 2 There's no way that you can add a vector pointing along the x axis, and have it cancel a vector pointing in the y direction. So the amplitude of the light is zero nowhere, and there are no intensity fringes. Instead, the polarization of the wave changes at the screen. As you study the light along the direction that you expect to see fringes, you would ... 2 I hope you know that intensity (I) of light at any point on the screen due to interference in the Young's Double Slit experiment can be given as$$A^2=I=a_1^2+a_2^2+2a_1a_2\cos{\phi} where $a_1, a_2$ are the amplitudes of the light waves with constant phase difference of $\phi$, $A$ is the amplitude of the resultant displaement at the point on ...

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Optical systems not involving magnetic fields are symmetric. So, if the display passes light in one direction, it will pass light in the other. Putting a mirror at the back of the TFT and lighting it from the front is therefore equivalent, expect that some light will be attenuated on the way in as pointed out by @CarlWitthoft in the comments. As a ...

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A couple things: this site is not specifically for engineering questions, so there's a small chance this may be closed by mods. Second, many commercial high-power LED's are listed not by luminous flux, but by nominal diode power consumption; a 1 watt LED actually emits far less than 1 watt of radiant power, so you'll need to think about what radiant flux ...

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You have two good fundamental answers, but a slightly different take on your answer is that yes, this is done all the time in optical systems to compensate for various distortions, aberrations and errors. A good example is an achromatic doublet, where a convex (converging) lens of one material is put in direct contact with a concave (diverging) lens along a ...

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Any discussion of the type of image a lens can form (real or virtual) must include information about the type of object that is being used. A divergent lens, by itself, can form only a virtual image of a real object. if we pre-condition the light from the object by passing it through a converging lens, then the resulting intermediate image can be a real or ...

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The Casimir effect is used as experimental proof of the existence of the vacuum virtual exchanges. The typical example is of two uncharged metallic plates in a vacuum, placed a few micrometers apart. In a classical description, the lack of an external field also means that there is no field between the plates, and no force would be measured between them. ...

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Keeping in mind that a surface cannot have a refractive index, but rather the bulk material which has that surface, what shape is the material you're measuring, and do you get to modify it? If, for example, it's a planar solid (front and back faces parallel), point the laser at an angle and measure the lateral displacement of the output beam. Given that ...

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It does not matter where the mirror is kept for you to see the entire image. A mirror with half the length of the man should be sufficient irrespective of the position. The only thing you do have to make sure is that when the mirror is brought closer, it must not be moved vertically. This can be understood from the following diagram:- The rectangle is ...

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Simply, at Brewster's angle, the $\pi$ - component of incident electromagnetic wave always transmits 100%. What is left is the $\sigma$ - component. The $\sigma$ - component of EM wave (which is already lying in xz - plane) striking on the second plate with respect to that plate is equal its amplitude times cosine of the angle the glass plate B will rotate. ...

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I've heard that clear things such as the gaseous components of the atmosphere or water reflect no light, and that is why they are clear Yes, "clear things" reflect light in very small percentage. But, they do reflect certain percentage of light falling on them. If they don't reflect that small percentage of light, you would have not been able to see ...

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There is something you should think about and then hopefully things become clearer for you: The athmosphere is only "clear" in a rather narrow region of the electromagnetic spectrum (see the nice graphic on the wikpedia page). For most wavelengths, our air is unclear and little of the incident light reaches earth's surface. Another useful example here is ...

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