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Fig. shows the plot of the function y(x) representing a fraction of the total power of thermal radiation falling within the spectral interval from 0 to x. Here x = lambda//lambda_(m)(lambda_(m) is the wavelength corresponding to the maximum of spectral radiation density). Using this plot, find: (a) the wavelength which divides the radiation spectrum into two equal (in terms of energy) parts at the temperature 3700K, (b) the fraction of the total radiation power falling within the visible range of the spectrum (0.40-0.76 mu m) at the temperature 5000K, (c) how manu times the power radiated at wavelengths exceeding 0.76mu m will increase if the temperature rises form 3000 to 5000K. |
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Answer» Solution :(a) From the CURVE of the FUNCTION `y(x)` we see that `y = 0.5` when `x = 1.41` Thus `lambda = 1.41 xx (0.29)/(3700)cm = 1.105mum` (b) At `5000K` `lambda = (0.29)/(5)xx10^(-6)m = 0.58mu m` So the VISIBLE range `(0.40` to `0.70)mu m` corresponds to a range `(0.69` to `1.31)` of `x`. From the curve `y(0.69) = 0.07` `y(1.31) = 0.44` so the fraction is `0.37` (c ) The value of `x` corresponding to `0.76` are `x_(1) = 0.76//(0.29)/(0.3) = 0.786` at `3000K` `x_(2) = 0.76//(0.29)/(0.5) = 1.31` at `5000K` The requisite fraction is then `((P_(2))/(P_(1))) = underset("ratio of t otal POWER")underset(uarr)(((T_(2))/(T_(1)))^(4))xxunderset("ratio of the fra ction of required wavel eng ths i n the radiated power")underset(uarr)((1-y_(2))/(1-y_(1)))` `= ((5)/(3))^(4) xx(1-0.44)/(1-0.12) = 4.91` |
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