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    Intensity of light passing through series of polarizers

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    Unpolarized light is passing through three successive polarizers placed one after the other. Polarizing axis of Polarizer 1 is vertically arranged i.e. polarizing axis is making 0 degree angle with vertical axis. Next polarizers, Polarizer 2 and Polarizer 3 are placed in such a way that their polarizing axis are making 30 and 90 degree angle with vertical axis respectively. (a) Find the intensity of light coming out of 3rd polarizer. And (b) show what will happen if second polarizer is removed. (intensity of incident unpolarized light as 1 unit)

    © BrainMass Inc. brainmass.com September 27, 2022, 11:08 am ad1c9bdddf
    https://brainmass.com/physics/polarisation/intensity-light-passing-through-series-polarizers-590942

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    A. When polarised light is passing through the polariser the intensity of light passes through the polariser is given by
    I = Io * Cos^2(theta) ; Where, Io - intensity of incident light and theta - angle between polarization direction of incident light and axis of polarizer.
    Now in our case incident light is unpolarized light beam which can be taken as a mixture of linear polarizations uniformly for all possible angles (0 degree to 360 degree circularly). Now the average value of cos^2(theta) over 360 degree is 1/2. Hence, intensity of light passing through polariser 1 will be,
    I1 = I0 * 1/2 = 1 * 1/2 = 0.5 = 50 % and this light will be vertically polarised.
    Now angle for second polariser is 30 degree and incident intensity is 0.5.
    So I2 = I1*Cos^2(theta) = 0.5 * Cos^2(30) = 0.5 * (sqrt(3)/2)^2 = 0.5 * 3/4 = 0.375 =37.5 %
    Means 0.375 unit (or 37.5 %) intensity of light is passing through polarizer 2 and falling on polarizer 3.
    Angle between polarizer 2 and polarizer 3 will be our next 'theta' as light coming out of p-2 is having polarizing direction of 30 degree with respect to vertical axis and p-3 is at 90: degree to the vertical. So new theta = 60 degree. So intensity of light passing through p-2 is
    I3 = I2 * Cos^2(60) = 0.375 * (1/2)^2 = 0.375/4 = 0.09375 = 9.375 %
    Thus 0.09375 unit (or 9.375 % of incident light ) intensity of light will come out from the polarizer 3.
    B. Now, if we remove a second polariser (p-2) then intensity of light falling on p-3 will be 0.5 unit and it will have vertical polarization direction so angle ' theta ' will be 90 degree. So intensity of light coming out of p-3 will be ,
    I = 0.5 * Cos^2(90) = 0.5 * 0 =0.
    So in this case no light will pass through crossed polarised polarizers.

    Reference:
    http://en.wikipedia.org/wiki/Polarizer

    This content was COPIED from BrainMass.com - View the original, and get the already-completed solution here!

    © BrainMass Inc. brainmass.com September 27, 2022, 11:08 am ad1c9bdddf>
    https://brainmass.com/physics/polarisation/intensity-light-passing-through-series-polarizers-590942

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