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A mixture of plane polarised and unpolarised light falls normally on a polarising sheet. On rotating the polarising sheet about the direction of the incident beam, the transmitted intensity varies by a factor 4. Find the ratio of the intensities `I_(P)` and `I_(0)` respectively of the polarized and unpolarised components in the incident beam. Next the axis of polarising sheet is fixed at an angle of `45^(@)` with the direction when the transmitted intensity is maximum. Then obtain the total intensity of the transmitted beam in terms of `I_(0)`. `[(3)/(2),(5I_(0))/(4)]`A. `(2)/(1)`B. `(3)/(2)`C. `(4)/(3)`D. `(4)/(1)` |
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Answer» Correct Answer - 2 Intensity of polarised light `I__(p) = I_(p) cos ^(2) theta//2` `theta I_(p) ={ {:(,"max" =I_(p)),(,"min"=0):}` But intensity of unpolarised light remains const. `therefore I_(0) = I_(0)//2` `therefore I_("max")= I_(p) + I_(0)//2` `I_("min") = I_(0)//2` but problem `I_(p) +(I_(0))/(2) = 4(I_(0))/(2)` `therefore (I_(p))/(I_(0)) =(3)/(2)` |
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