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Metals refractive index is always complex number (and not only for metals). Imagine part shows the extinction coefficient $k$ - absorption in a material. Real and imagine part isn't connected. P.S. For engineering calculations real part sometimes is less than 1. Theoretically even for Fresnel reflection in dielectric we must use full formula with complex ...

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The problem with this question (although your question is still a natural one for those thinking about light to ask) is that it mixes the ideal and the real. You describe an ideal situation with your mirrors, but then ask for what would happen in real life. No actual mirror has reflection coefficient of 1 (which would represent 100% reflectivity) and so any ...

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(This type of question has been asked by 4 users but in those questions they either gave an example of a wooden box or a room and they got answers that the light is absorbed by the wood or the walls of the room. But in my question its the case of mirrors.) In this case, the light would be absorbed by de "viewer". You would need some type of device ...

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When your book says energy it should say radiant intensity. I didn't read Lambert's Photometria myself, but multiple sources say that this is how Lambert defined his law. A lambertian surface follows Lambert's cosine law, so for this surface we have: $$I_\theta=I_n \cos\theta$$ Radiance's definition can be written as: $$L=\frac{\partial I}{\partial ... -2 Internal reflections from the facets of the object 0 The central idea is that you can translate probabilities from a single-particle picture into fractions of particles in a many-particle picture (assuming no interaction). Consider T for a moment for a single particle. Let's just, for ease of writing, say T=.9. So if you send in a single particle from the far left, 90% of the probability density ... 1 If the bottle and liquid are made of dielectric material, then the interfaces between different mediums reflect light, they don't absorb it (i.e. dissipate it as heat in the glass). This is probably a good approximation for your bottle. As a first approximation, once you have worked out your incidence angles with Snell's laww, you need to use the Fresnel ... 0 Most likely what you are seeing is a thin film of oil floating on the surface of the water that comes either from natural underwater sources, runoff from the shore, or from ships. The oil breaks the surface tension of the water and reduces traction forces from the wind - thus ripple amplitudes are reduced or entirely diminished. This phenomena has been ... 0 I've also seen this, and wondered what it is! I thought they may have been some sort of surface currents as they looked like "rivers" to me. However, I've just done some searching and found this and this. So the answer is biogenic slicks, oily substances exuded by algae washed off from the shore! These don't dissolve well with water and form a thin film over ... 0 Thinking about it, maybe what selects the special direction is that the creases in the water that have wavefronts perpendicular to the line of sight cover each other up while the ones with wavefronts parallel to the line of sight don't. This may mean that as diffraction off small features causes the large features in the angle distribution (as it's a ... 0 If the image would be formed in front of the mirror it would not be a virtual image anymore, it would mean that your eye would have the impression to see it in front of the mirror. As soon as the ray is reflected, the eye which interprets light as travelling in straight line necessary sees the object as coming from behind the mirror and that is why the ... 3 It is theoretically predicted that superconducting layers might be able to act as reflectors through the so called Heisenberg-Coulomb effect. Out of these, you could of course form a cavity able to contain a gravitational wave in principle. This effect has, to my knowledge, not yet been experimentally tested, although several tests have been proposed, see, ... 1 The main question is how to contain the gravitational wave, and also how to make a cavity large enough to contain the gravitational wave. Also, reflecting the gravitational wave would be a big problem. So, all in all, no. 0 The photons would be able to bounce around indefinitely. However, if this were really done, the photon would quickly be absorbed by the particles of the mirror 0 It would stay inside of a perfect mirror. Of course, perfect mirrors don't exist in reality. It is covered in this video. 0 The response time of the human eye is something like 60 Hz. This means that you need to draw the full line in 1/60th of a second. Based on your drawing (which isn't particularly clear) you are trying to draw a line using roughly 90° of mirror rotation. Put this all together and you need the mirror to be rotating at an angular speed of 900 rotations ... 1 There are several ways to approach this problem. If we can estimate the power density achieved in W/m^2, then the temperature that can be reached follows from the Stefan-Boltzmann law. First method: 1) Take the total power collected, and see the size it got focused down to. You state the area of the mirror array is 0.6 m^2 (roughly), and with power ... 0 Simple geometry tells you that the size of the image s, for a given distance from mirror to screen d, is a function of angle according to$$\tan\frac{\theta}{2} = \frac{s}{2d} For example, an image of 1 meter at a distance of 10 meters needs the mirror to sweep through an angle of $2 \cdot \tan^{-1}(0.05)\approx5.7°$

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