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3

If you increase the length of the fibre, you will get more reasonable results. The losses -in my mind- need to be distributed along the entire length of the fiber . Currently, you seem to model a very small portion of the fiber, and as such you have a very small area that need to transfer a seemingly small amount of energy. However, if that energy is going ...


3

If a model does not reflect reality, the model is not accurate enough for that situation. Consider some of the minutia in the workings of light. First an alternate model for the fiber: As light passes through, some of it passes straight through, some is "absorbed." (quotes because I suggest including reflection and refraction in here as well as ...


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When you project an image on a reflective medium, you inherit the projection distance. So what you are seing is a plane that is hovering behind the glass. In essence your eyes see is the whole path to the target as if it never had reflected at all. So no need to focus that close.


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Your assumption on temperature increase is flawed You can't convert watts to temperature directly unless the object is perfectly insulated and the application of heat has a limited time. The equation from your link would result in a temperature increase directly proportional to time, increasing forever. Temperature is a measure of "heat content," ...


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After the exchange of comments, (more specifically) that the glass is transparent, my "guess" is the following: You state that the signal power loss is in the order of 1 W. Assuming the measurement system before and after can accurately measure the energy difference, this power drop can be attributed to the following things: increasing the energy ...


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