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Final+sol

# Final+sol - 1 You have available a source of unpolarized...

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1. You have available a source of unpolarized light (intensity I o ) and a number of "perfect" linear polarizers (each of them transmitting without losses all the light polarized parallel to its "transmission axis" and blocking totally all light polarized perpendicular to this axis.) (A) Using a sketch of the light polarization and the polarizer orientation , derive and show the simple relation for the light intensity I( θ ) relative to I o transmitted by two polarizers as a function of the angle θ between their transmission axes. I 0 I 1 I 2 ! I 1 = 1 2 I 0 , and I 2 = 1 2 I 0 cos 2 θ

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In considering the electric field components, we can obtain the relations between intensities before and after each polarizer. E ! " E E ! E | | I o = 1 2 ε o c {(E ) 2 + (E ) 2 } I 1 = 1 2 ε o c (E ) 2 = 1 2 I o I 2 = 1 2 ε o c (E cos θ ) 2 = 1 2 I o cos 2 θ (B) Imagine two polarizers are set with their transmission axes "crossed" at 0° and 90°. What is the transmitted light intensity? For θ = 90°, I 2 = 0.
(C) Five additional polarizers are placed between the two "crossed" polarizers with their transmission axes set at 60° , 45° , 75° , 15° , and 30° in that order, with the 60° polarizer adjacent to the 0° polarizer and so on (all angles are measured from the axis of the first 0° polarizer). What is the intensity relative to the incident one, transmitted by these seven polarizers? The orientation angles of a series of polarizers are θ o = 0°, θ 1 = 60°, θ 2 = 45°, θ 3 = 75°, θ 4 = 15°, θ 5 = 30°, and θ 6 = 90°. I 1 = 1 2 I o I 2 = I 1 cos 2 ( θ 1 θ o ) = 1 2 I o cos 2 ( 60 – 0) I 3 = I 2 cos 2 ( θ 2 θ 1 ) = 1 2 I o cos 2 (60) cos 2 (15) I 4 = I 3 cos 2 ( θ 3 θ 2 ) = 1 2 I o cos 2 (60) cos 2 (15) cos 2 (30) I 5 = I 4 cos 2 ( θ 4 θ 3 ) = 1 2 I o cos 2 (60) cos 2 (15) cos 2 (30) cos 2 (60) I 6 = I 5 cos 2 ( θ 5 θ 4 ) = 1 2 I o cos 2 (60) cos 2 (15) cos 2 (30) cos 2 (60) cos 2 (15) I 7 = I 6 cos 2 ( θ 6 θ 5 )= 1 2 I o cos 2 (60) cos 2 (15) cos 2 (30) cos 2 (60) cos 2 (15)

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