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13 June, 12:09

The average intensity of light emerging from a polarizing sheet is 0.708 W/m2, and that of the horizontally polarized light incident on the sheet is 0.960 W/m2. Determine the angle that the transmission axis of the polarizing sheet makes with the horizontal.

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  1. 13 June, 13:03
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    Angle θ = 30.82°

    Explanation:

    From Malus's law, since the intensity of a wave is proportional to its amplitude squared, the intensity I of the transmitted wave is related to the incident wave by; I = I_o cos²θ

    where;

    I_o is the intensity of the polarized wave before passing through the filter.

    In this question,

    I is 0.708 W/m²

    While I_o is 0.960 W/m²

    Thus, plugging in these values into the equation, we have;

    0.708 W/m² = 0.960 W/m² •cos²θ

    Thus, cos²θ = 0.708 W/m²/0.960 W/m²

    cos²θ = 0.7375

    Cos θ = √0.7375

    Cos θ = 0.8588

    θ = Cos^ (-1) 0.8588

    θ = 30.82°
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