As the filter is rotated, cos2θvaries from a minimum of 0 to a maximum of 1, so the transmitted intensity varies from a minimum of IfImin()=−1210to a maximum of max00011(1)) .22IfIfI=+−=+The ratio of Imaxto Iminis IIffmax.=+−11Setting the ratio equal to 5.0 and solving for f, we get f= 0.67. 41. Let I0be the intensity of the incident beam and fbe the fraction that is polarized. Thus, the intensity of the polarized portion is f I0. After transmission, this portion contributes f I0cos2to the intensity of the transmitted beam. Here
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