InterviewSolution
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InterviewSolution
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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» (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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