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Lets say your receiver is a copper plate, if you leave it in sunshine in the conditions you describe you get certain temperature, lets say 50 c, lets say you calculate the area of your dish an it is 1 square meter, now you place the piece of copper close to the focal point, and you can see a concentrated bright dot, and you measure the area of that dot, lets ...


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You are quite right: Einstein is using highly idealised concepts in a thought experiment. In general ALL interactions between light and matter take nonzero time. But Einstein is justified in this approach because: As stated by Nikos M in his comment: "Einstein's observation does not need the light to be reflected, one can very well get the same ...


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One way to imagine computing the reflection delay is to assume that the reflection process takes place within the first skin depth of the material. Taking numbers straight from the Wikipedia article, a copper conductor reflecting a 100 MHz (radio) wave would do so within its outermost 6.6 μm. In vacuum, light crosses 6.6 μm in about 20 femtoseconds. Skin ...


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A rough estimate can be obtained as follows. At the extremely high intensity radiation the temperature will be determined by radiative equilibrium. Now, in the case of purely radiative equilibrium, an object that receives the solar radiation that reaches the Earth would reach the a similar temperature as the lunar surface. At the equator, the maximum ...


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So taking your questions one at a time 1) The concave mirror will form a virtual image if the object is placed closer to the mirror than the focal point of the mirror. The formula for the position of the final image is $s'=\frac{sf}{s-f}$ where s is the object-mirror distance and f is the focal length of the mirror. You can see that if $s<f$ this will be ...


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You can reach the end of a rainbow, if you make one using a hose pipe on a sunny day. It is possible to locate the ends to within a few cm


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Why can't we reach the ends of rainbow? The WIkipedia article you linked to contains the explanation The rainbow is not located at a specific distance, but comes from an optical illusion caused by any water droplets viewed from a certain angle relative to a light source. Thus, a rainbow is not an object and cannot be physically approached. Indeed, ...


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You are totally right about photons and their reflection and absorption on bright or dark surfaces if you illuminate these surfaces with visible light. The photons in the range of visible light will be reflected or absorbed (and re-emitted with a longer wavelength) and that is the reason we see these surfaces as bright or as dark. And the dark surface has a ...


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For a parabolic mirror, the answer is "all rays go through the focus". For a spherical mirror, you can see the answer most easily by looking at the limit as the spherical mirror is almost a hemisphere: Clearly the rays that start further off axis are focused closer to the mirror.


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Unfortunately, as interesting an idea as this is, and as creative as you must be for thinking of it, it's not an actual possibility as far as I'm concerned. A one-way mirror works much in the same way that a metallic screen door works. It allows you to see from the inside of your house, outward. However, this is due to the fact that there is far more ...


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The lens works because it takes all the sunlight falling on its area, $A_1$, and focuses it onto a small spot $A_2$. The intensity in the spot is the intensity of the sunlight multiplied by $A_1/A_2$. Exactly the same applies to a mirror. So provided your mirror has the same cross sectional area as the lens, and provided it can focus the light as ...


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The rise in temperature would not be linear. As you are heating a piece of material, the energy loss due to radiation will increase (and so will the energy loss due to conduction and convection in air, but we are not going to discuss those here). Radiation losses can be estimated by a simple formula for black body radiators called the Stefan-Boltzmann law: ...


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Yes there is a formula for this No. Of images = 360/A - 1 Hope this will be helpful


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In a classical electrodynamics picture there is an exponentially dampened wave penetrating the medium with the smaller index of refraction. See e.g. http://en.wikipedia.org/wiki/Total_internal_reflection. This wave can indeed be detected. The easiest way to see this is by bringing a third optical surface close to the interface. There will be non-zero ...


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Reflection,refraction and transmission of light are macroscopic manifestation of a phenomenon called scattering.In this incoming photons are absorbed and either the quantum energy level of an atom is raised (as in case of resonance absorption) or the outer electron cloud is set into motion(this is responsible for light around us).Almost instantaneously ...


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Let me go a little further than @mark-h's answer: The behavior of light at an interface is described by the electro-magnetic field solution to the Helmhotlz equation. It gives the reflected and transmitted electric and magnetic components as a function of the refractive indices of the incident and exiting media. From those solutions we can derive the ...


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The short answer is that, if steeped in water, the optical power decreases, possibly a great deal if your lens is a crown glass (like N-BK7) with a refractive index of 1.5 or so. It doesn't matter whether the lens is concave or convex: the power is scaled down (made less negative if the lens is diverging) by the same scaling constant: ...



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