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I assume you ask if there is also reflection when refraction occurs. The answer is yes. It happen when light (or any kind of wave) passes from a medium 1 to another medium 2, those medium being characterized with different index (i.e. wave velocity). At the interface between the two medium, in the case of light wave, the electric field has to obey certain ...


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Everything is about electromagnetic waves. At the border of medium one of the components of electric field have to be continuus and other component of electric displacement field. Such calculations leads to Fresnel equations


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The diagram you want to use looks something like this: Depending on how much attenuation there is in the membrane, you need to consider the potential of multiple reflections (or not). I actually analyzed this problem in some depth - considering not only the intensity of reflections on the different surfaces, but also multiple reflections and even the ...


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There are several options to increase the signal size: 1) increase the length of the path (look at the slab from side to side instead of from the front - you may have seen a "green edge" on a glass coffee table - that's because you see light coming through a much thicker slab of glass) 2) Use multiple sheets of glass on top of each other. Be careful about ...


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There is a fallacy in your question which might be the source of your confusion. It is not all light that gets focused to a point, it is only rays parallel to the lens axis! For the case of two (point source) stars, if the rays of one star are parallel to the lens axis, the rays of the other star will not be (not co-linear), therefore its image will appear ...


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Have a look at Floris' answer to Calculating light's lateral shift in a glass slab. If you shine your laser through the glass plate at an angle $\theta$ then the beam will be deflected by a distance $x$ given by: $$ x =d\sin\theta\left(1-\frac{\sqrt{1-\sin^2\theta}}{\sqrt{n^2-\sin^2\theta}}\right) $$ where $d$ is the unknown thickness of the plate and ...


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Your diagram shows parallel light beams originating from infinity. Light entering the eye in the real world isn't all parallel. In the real world, all the light that bounces off a single point will hit the sensor (eye-cone, digital sensor, etc) at a single point, assuming that point is in focus. However, light from different points will strike the sensor ...


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Truth be told: geometric optics is wrong. It is sufficient for small angles and not too complicated setups, but is still an approximation. The first thing: Normal lenses do have not a focal point, but something which resembles more a line, the so-called caustic. The correct shape for getting a perfect point on the line crossing the middle of the lens (I ...


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A convex lens does not focus all the rays to a single point, only the axis parallel ones. It focuses all the rays emanating from a given point to a corresponding point on the other side. That point is the image of the original point. The standard diagram shows that a lens sends all axis parallel incoming rays to the focal point. Parallel rays can be thought ...


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...if a lens bends all incoming rays of light to intersect at the focal point? Shouldn't this produce a single dot of light...? (In your diagram, the source image is at infinity. I will continue the analysis along that idea.) It is true that all rays parallel to the axis focus to that single dot. Not all rays, however, are parallel to the axis: ...


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You are seeing only one color from each droplet. Even though each droplet reflects the whole spectrum, only one color gets to your eye. The rest of the light from a single droplet is sent somewhere else than your eye (maybe into the eye of someone standing near you). So if you see a thick blue band in a rainbow, the blue band is formed by many reflections ...


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This Wikipedia article gives a LOT of detail regarding rainbow formation https://en.wikipedia.org/wiki/Rainbow To make a long story short, each wavelength (color) reflects off of the back of a raindrop at a slightly different angle, of approximately 42 degrees. To see the rainbow, the sun has to be at your back, to allow the reflected light to enter your ...


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Interesting question because the diagram is correct but I too have never seen this in the lab. It turns out that the dispersion of the material and its thickness has to be very large to observe this. Here is a ray trace simulation: Bar geometry, 7.5cm x 1cm, Glass SF11 (Highly dispersive with Abbe number ~25), Ray wavelengths 380nm (blue) and 780nm ...


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On each ray, the red and the violet are strictly on top of each other (together, of course, with all the other colours of the rainbow, to make white light). At the prism, they get displaced by different amounts so each red ray is displaced from the violet ray it sat on top of before the prism. Image source In the middle of the beam, this doesn't matter, ...


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The reflexion co-efficient equation you have given is part of the Fresnel Equations. There are two ways wherein this equation varies with frequency: ultimately both down to the variation of the refractive indices $n_1$ and $n_2$ with frequency. Clearly the equation you have written can vary with frequency if $n_1$ and $n_2$ do so: also $\theta_2$ does ...


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The frequency (wavelength) dependence comes in through the refractive index, and it is specific to the medium. In the lab, the manufacturer of your optics will generally provide this dependence, as in e.g. this resource. Calculating this theoretically is generally a hard problem, though.


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Possibly in the slightly negative energy region of a Casimir mirror pair However, the effect is really small


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There may be scope in this question to mention the faster than light (FTL) capabilities of Tachyons (should they exist). If (in the language of Speacial Relativity) we apply the first postulate of SR to tachyons and there is an instantaneous transmission of information in one inertial frame then it would be true to say this occurs in all inertial frames. I ...



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