Question

(6%) Problem 6: Unpolarized light of intensity lo-1350 W/m2 s incident upon three polarizers. The axis of the first polarizer is vertical. The axis of the second polarizer is rotated at an angle θ-15° from the vertical. The axis of the third polarizer is horizontal Randomized Variables o 1350 W/m2 015° Otheexpertta.com @theexpertta.com-tracking id: 8075-83-54-49-94F2-19889. In accordance with Expert TA's Terms of Service. copying this information to any solutions sharing website is strictly forbidden. Doing so may result in termination of your Expert TA Account 20% Part (a) What is the intensity of the light in W/m2 after it passes through the first polarizer? Grade Summary II629.78 0% 100% tan acos sinh cotanh Attempts remaining: 5 cotan atan cos asin % per attempt) detailed view 10% 0 END 10% o Degrees Radians BACKSPACE DELI CLEAR Submit I give up! Feedback: 0% deduction per feedback. Hints: 20% Part (b) What is the intensity of the light in W/m2 after it passes through the second polarizer? 20% Part (c) What is the intensity in W/m2 of the light after it passes through the third polarizer? 20% 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 Io and the angle of the second polarizer relative to the first, given that the first and third polarizers are crossed (90° between them. Use trigonometric identities to simplify and give the results in terms of a single trigonometric function of φ-2θ 쇼 20% Part (e) What is the smallest angle θ2 in degrees from the vertical that the second polarizer could be rotated such that the intensity of light passing through all three polarizers is 13'-78 W/m2?

Answer 1

a)

I = Io / 2 = 675 W/m^2

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b)

Intensity after 2nd polariser

I2 = 675* ( cos15) ^2 = 629.78 W/m^2.

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c)

Intensity after 3rd polariser

I3 = 629.78* cos ( 90 - 15)^2

I3 = 42.19 W/m^2

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d)

I = ( io/ 2) (cos x) ^2 ( cos ( 90 - x)) ^2

I = (io/2) (cos x) ^2 ( sin x) ^2

I = (io/2) ( sin x cos x) ^2

I = (io/2) (2 sin x cos x) ^2/ 4

I = io sin 2x / 8

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e)

78 = 1350 sin 2 x/ 8

x = 13.765

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