Please help with all parts, thanks in advance! Unpolarized light of intensity I_

Question

Please help with all parts, thanks in advance!

Unpolarized light of intensity I_0 = 750 W/m^2 is incident upon three polarizers. The axis of the first polarizer is vertical. The axis of the second polarizer is rotated at an angle theta = 15degree from the vertical. The axis of the third polarizer is horizontal. Randomized Variables I_0 = 750 W/m^2 theta = 15 degree Part (a) What is the intensity of the light in W/m^2 after it passes through the first polarizer? Part (b) What is the intensity of the light in W/m^2 after it passes through the second polarizer? Part (c) What is the intensity in W/m^2 of the light after it passes through the third polarizer? Part (d) Write an equation for the intensity of light after it has passed through all three polarizers in terms of the intensity of unpolarized light entering the first polarizer I_0 and the angle of the second polarizer relative to the first, given that the first and third polarizers are crossed (90 degree between them). Use trigonometric identities to simplify and give the results in terms of a single trigonometric function of phi = 2theta. Part (e) What is the smallest angle theta_2 degrees, from the vertical that the second polarizer could be rotated such that the intensity of light passing through all three polarizers is I_3 = 24 W/m^2? Theta_2 =

Explanation / Answer

part(a)

Intensity of light after passing through first polarizer I1 = I0/2 = 750/2 = 375 W/m^2

part(b)

Intensity of light after passing through second polarizer I2 = I1*(costheta)^2

I2 = 375*(cos15)^2 = 350 W/m^2

part(c)

Intensity of light after passing through third polarizer I3 = I2*(cos(90-theta))^2

I3 = 350*(cos(90-15))^2 = 23.4 W/m^2

part(d)

given I3' = 24

I3' = I2'*(cos(90-theta))^2

I2' = I1'*(costheta)^2

I1' = Io/2

I3' = Io/2'*(costheta)^2*(cos(90-theta))^2

I3' = Io/2*(costheta)^2*(sintheta)^2

I3' = Io/2*(sin2theta)^2/4

24 = 750/2*(sin2theta)^2/4

theta = 15.2 degrees

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