Question

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B: Scrateh papers are not allowed during the exam. An ed and used as seratch but will not be considered for grading y sheet of the booklet can be Four polarizing sheets are stacked as follows: the first is at 10.0 clockwise the vertical, the second is at an angle of 10.0° counterclockwise (CCw) with the W) with cal, the third is at 40.0° CW with the vertical, and the fourth is horizontal. The incident light n intensity lo 500 W/m2 Calculate 14, the transmitted intensity by the system if the incident light lo is unpolarized? the incident light lo is vertically polarized? For the remainder of the problem assume that 10-500 W/㎡ and is vertically polarized. Calculate the transmitted intensity if the second polarizer is removed. What would be the transmitted intensity if the second and the third polarizer are 以- exchanged? If the second polarizer is removed, what should be the angle (with the vertical) of the third polarizer so that maximum intensity can be transmitted through the system? Calculate this maximum intensity Light is incident on air-water interface at the Brewster angle θΒ of that interface an reflects toward the original system of 4 polarizers. Calculate the fraction of the incident light u is transmitted by the system. cindex of

Answer 1

Intensity of incident light is I_0 = 500 w/m^2 \r\n from malus law I_1 = I_0/2 = 500/2 = 250 w/m^2 again from malus law I_2 = I_1 cos^2 theta where theta = 10 degree + 10 degree I_2 = 250 cos^2 (20) I_2 = 220.75 w/m^2 \r\n now I_3 = I_2 cos^2 theta where theta = 10 degree + 40 degree = 50 degree I_2 = 220.75 cos^2 (50 degree) I_3 = 91.20 w/m^2 now I_4 = I_3 cos^2 theta where theta = 90 - 40 = 50 degree I_4 = 91.20 cos^2 (50 degree) I_4 = 37.68 w/m^2 \r\n I_1 = I_0 cos^2 theta where theta = 10 degree I_1 = 500 cos^2 10 degree = 484.92 w/m^2 now I_2 = I_1 cos^2 theta where theta = 10 degree + 10 degree = 20 degree I_2 = 484.92 cos 62 (20 degree) I_2 = 428.195 w/m^2 now, I_3 = I_2 cos^2 theta