Unpolarized light whose intensity is 1.76 W/m2 is incident on the polarizer in t

Question

Unpolarized light whose intensity is 1.76 W/m2 is incident on the polarizer in the drawing. (a) What is the intensity of the light leaving the polarizer? (b) If the analyzer is set at an angle of = 43.0 with respect to the polarizer, what is the intensity of the light that reaches the photocell?

Unpolarized light whose intensity is 1.76 W/m2 is incident on the polarizer in the drawing. (a) What is the intensity of the light leaving the polarizer? (b) If the analyzer is set at an angle of = 43.0 with respect to the polarizer, what is the intensity of the light that reaches the photocell?

Explanation / Answer

Given that, I0 = 1.76 W/m2

(a)

Unpolarized light always loses half its intensity on passing through an ideal polarizer

I1 = 1/2(I0) = 1.76/2 = 0.88 W/m2

Hence, I1 = 0.88 W/m2

(b) I2 = I1 Cos2(theta) = 0.88 Cos2(43)

I2 = 0.88 Cos2(43) W/m2

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