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March 26, 2017

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We want to rotate the direction of polarization of a beam of polarized light through 90 degrees by sending the beam through one or more polarizing sheets.(a) What is the mini-mum number of sheets required? (b) What is the minimum number of sheets required if the transmitted intensity is to be more than 60% of the original intensity?

  • Physics - ,

    Since when can polarizing sheets rotate a linearly polarized light beam?

  • Physics - ,

    I don't really know. I did ln(1/100)=-4.605 and ln(cos(90)^2)=-1.605 and 4.605/1.605=2.869. The answer for part (a) should be 2 but 2.869 is closer to 3 so am I doing anything wrong?

  • Physics - ,

    (a)If we place one sheet with its polarization direction 90⁰ to the polarized light then the transmitted intensity will be zero (since I=I₀•cos²α = I₀•cos²90⁰=0).
    If we use 2 sheets, the 1st sheet rotates by angle α (0< α<90⁰) and the 2nd sheet rotate by the angle (90⁰-α)
    I=I₀•cos²α•cos²(90⁰-α)= I₀•cos²α•sin²α.
    Thus, we need two sheets.
    (b) I=I₀•cos²ⁿ(90⁰/n)
    n=2 I₂=I₀•cos⁴(90⁰/2)=0.25•I₀
    n=3 I₃=I₀•cos⁶(90⁰/3)=0.422•I₀
    n=4 I₄=I₀•cos⁸(90⁰/4)=0.53•I₀
    n=5 I₅=I₀•cos¹⁰(90⁰/5)=0.6054•I₀
    I ո ≥0.6•I₀
    Thus, the minimum value n=5 sheets.

  • Physics - ,

    I still don't understand part (a). I know that cos^2(90)=0 but how did you get to the answer of 2 sheets?

  • Physics - ,

    2 because....that's the minimum no. of angles of that can ad up to 90

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