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4 November, 00:31

Plane polarized light with intensity I0 is incident on a polarizer. What angle should the principle axis make with respenct to the incident polarization to get a transmission intensity that is 0.464 I0?

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  1. 4 November, 03:17
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    Q = 47.06 degrees

    Explanation:

    Given:

    - The transmitted intensity I = 0.464 I_o

    - Incident Intensity I = I_o

    Find:

    What angle should the principle axis make with respect to the incident polarization

    Solution:

    - The relation of transmitted Intensity I to to the incident intensity I_o on a plane paper with its principle axis is given by:

    I = I_o * cos^2 (Q)

    - Where Q is the angle between the Incident polarized Light and its angle with the principle axis. Hence, Using the relation given above:

    Q = cos ^-1 (sqrt (I / I_o))

    - Plug the values in:

    Q = cos^-1 (sqrt (0.464))

    Q = cos^-1 (0.6811754546)

    Q = 47.06 degrees
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