ch33-p041 - 41. Let I0 be the intensity of the incident...

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As the filter is rotated, cos 2 θ varies from a minimum of 0 to a maximum of 1, so the transmitted intensity varies from a minimum of If I min () =− 1 2 1 0 to a maximum of max 0 0 0 11 (1 ) ) . 22 I f I fI =+− =+ The ratio of I max to I min is I I f f max . = + 1 1 Setting the ratio equal to 5.0 and solving for f , we get f = 0.67. 41. Let I 0 be the intensity of the incident beam and f be the fraction that is polarized. Thus, the intensity of the polarized portion is f I 0 . After transmission, this portion contributes f I 0 cos 2 to the intensity of the transmitted beam. Here
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