In experiment 1, unpolarized light falls on the polarizer inthe figure below. Th

Question

In experiment 1, unpolarized light falls on the polarizer inthe figure below. The angle of the analyzer is =69.4°. 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 additionalrotation. (Use a positive number to indicate an increase in theangle. Use a negative number to indicate a decrease in theangle.)
° I have no idea how to work through this problem. Anyhelp would be greatly appreciated. (Please show the steps onhow to work through this problem!) Will rate lifesaver! Thanks.

Explanation / Answer

Given angle of analyzer = 69.4o,initial intensity = I0    From Mallu's law In experiment 1,    From Mallu's law In experiment 1,    the intensity on the photocell I1 = (I0/2)*cos2
In experiment 2, the intensity on the photocell I2 =I*cos2' =(I0/2)*cos2
    cos2' =(1/2)*cos269.4 = 0.06189
     cos' = 0.2487
          ' =75.6o
additional angle = 75.6 - 69.4 = +6.2o

the intensity on the photocell I2 =I*cos2' =(I0/2)*cos2
    cos2' =(1/2)*cos269.4 = 0.06189
     cos' = 0.2487
          ' =75.6o
additional angle = 75.6 - 69.4 = +6.2o

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