# Tag Info

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Try this: Make a paper cut-out of the letter F (or any easy-to-cut-out letter which is different from its horizontally and vertically flipped images). Shade or mark one side of the cut-out to distinguish it from the other. Get a second piece of paper and write the same letter on it several times. Hold the cut-out in front of you while standing in front of ...

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Vsauce answered this question exactly in one episode. I suggest checking it out, it explains the different ways the eyes see the mirror, and gets right to the point.

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Instead of scattering, think of it as diffuse reflection. The bidirectional reflectance distribution function (BRDF) describes optical surface properties. It's application is as well in computer graphics, as in-depth ray tracing simulations. It depends on angle of incident light $\vec \omega_i$ (2 dimensions) and angle of observation $\vec \omega_r$, also 2 ...

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The best way to understand this phase shift is to solve and study solutions of the Helmholtz equation near the boundary between two dielectric mediums. You don't quite have to solve the full Maxwell equations: the assumption that the light field can be modelled by one scalar field (approximately equal to one transverse component of the electric field) rather ...

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To calculate the transmitted and scattered wave state at any angle from a plane interface, you need to resolve the incident field into s- and p-linear polarised wave amplitudes and multiply the two complex amplitudes by the amplitude Fresnel transmission and reflexion co-efficients. By "amplitude Fresnel co-efficient I mean the ratio of complex wave ...

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If we look at an actual, finite, laser incident on an interface, it is not a plane wave. It has finite extent, so it is a superposition of different plane waves which all have different angles of incidences. In optics, we can model the total field by adding all the incident plane waves and all the reflected plane waves together. Because of the different ...

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A boundary is defined as the line or surface between two different types of materials. For any given boundary, that reflects and/or transmits, we can give it transmission $T$ and reflection $R$ constants to find the amount of reflected or transmitted stuff(like light). $$\text{Stuff} \cdot R = \text{Reflected Stuff}$$ \text{Stuff} \cdot T = ...

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If we assume that mirrors will leak some energy, then is it possible to put objects such as a photo multiplier tube (in combination with a mirror) and adjust it in such a way that only the amount of energy lost by reflection of the mirror is recovered and sent back to the other mirror. This cannot be done even as a thought experiment. Photomultipliers ...

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You have separate Fresnel equations for s- and p-polarized light. The two polarizations reflect/refract separately. You can reconstitute them on the other side to recover the new polarization vector if you want.

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Your approach is correct, but you really need to draw even a crude sketch. The lamp is $x$ m from the mirror; the image is $5$ m from the lamp, which puts it $(5+x)$ from the mirror. These are the d values for your magnification formula, which is correct...

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I like to view the sign convention as, the positive direction is the direction the light is moving. (This means the incident and reflected light will have opposite positive directions.) Example: An object is on the left, and the concave mirror is on the right. They are relatively far away $(d_o>f)$. Light rays travel to the right from the object to the ...

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You are asking about retroreflectors. Some background: In general (implying 3D) and with the assumption that no functional refractive elements are allowed, a hollow corner cube reflector would be the tool of choice. Nevertheless equally useful glass corner cubes exist. Both types are colloquially known as triple-prisms. So that's where your first answer "3 ...

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I think the way to do it is to realize that the ray direction of the incoming wavefront has 3 components, and we can invert the ray direction in a 2-dimensional plane using 2 orthogonal mirrors (this is just a consequence of the angles of a triangle adding up to 180). So from $(x,y,z)$, we can invert them 2 at a time: $(x,y,z)$ $(-x,-y,z)$ (mirrors in the ...

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I suggest at looking at the rays from the edges of the object, like this: where the thin lines show where the image appears to the observer at A.

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Your query is valid: How is voltage a 'wave' that reflects and creates standing waves? Well, the answer is quite simple when you stop and think about it. All signals travelling across transmission lines are merely electromagnetic waves. Now these lines are commonly driven by voltage sources, hence they are 'voltage waves'. This makes perfect sense: If a ...

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