In experiment 1, unpolarized light falls on the polarizer in thefigure below. Th

Question

In experiment 1, unpolarized light falls on the polarizer in thefigure below. The angle of the analyzer is =66.2°. In experiment 2, theunpolarized light is replaced by light of the same intensity, butthe light is polarized along the direction of the polarizer'stransmission axis. By how many additional degrees must theanalyzer be rotated so that the light falling on the photocell willhave the same intensity as it did in experiment 1? Explain whether is increased or deceased by this additional numberof degrees. (Use a positive number to indicate an increase in theangle. Use a negative number to indicate a decrease in theangle.)

Explanation / Answer

Given that    Angle = 66.2o Let us take initial intensity be I For experiment 1    The intensity on the photocell is I1 = (I/2)*cos2 for experiment 2 The intensity on the photocell is    I2 =I*cos2' I1= I2 I*cos2' =(I/2)*cos2 cos2' = (1/2)*cos2            =(1/2)*cos266.2o          Find '. so, the additional angle = ' - 66.2 o Substitute the values.

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