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I don't quite understand why you need an optical fiber - one can cut sheet steel with infrared radiation (https://www.coherent.com/Applications/index.cfm?fuseaction=forms.AppLevel2&AppLevel2ID=62 ). Furthermore, efficient heating can be even possible in a situation that is inverse with respect to what you describe: when the IR radiation beam is ...


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As your using the optical fiber for transmission no heat is wasted there. Then we have, $$P_{radiated}= \epsilon k A T ^4$$ Where $\epsilon$ is the shininess of the surface,k Boltzmann constant ,A is area of surface and T is temperature of the surface. And P is power. Now, $H(heat)=P.t(time)$ so that $H$ can be plugged on to the equation $H=mC\Delta T$ to ...


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I think I understand your confusion. The answer would be: You've been misguided. There is no special link between heat and infrared radiation, except for the fact that most bodies radiate most of their heat in the infrared spectrum because they don't have enough energy (heat) to radiate at a higher frequency. See the graphs in this thread. So one could ...


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The answer is simple: your camera CCD is sensitive to infrared light, while your eye is not. The real question here is: why does your camera render the infrared light as visible light on the screen? The only answer I can hazard on that is that it allows for more visible photographs when they're taken in low (visible) light.


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A room with area $60\rm\,m^2$ is about $8\rm\,m$ on a side; call it $x= 10\rm\,m$ to make arithmetic easier. The time it takes for a reflection to cross such a room is $$ t = \frac x c = \rm \frac{10\,m}{ 3\times10^8 m/s} = \frac 1 3 \times 10^{-7} s = 30\,ns, $$ corresponding to a reflection frequency of $$ f = \frac 1 t = \rm \frac 1{30}\,GHz = 30\,MHz. ...



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