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Sequence of Polaroids

  1. Jun 7, 2010 #1
    1. The problem statement, all variables and given/known data
    An unpolarized beam of light is incident on a sequence of polarizers. Assume that the angle between the axis of the first polarizer and the last is 90 (deg.).


    2. Relevant equations



    3. The attempt at a solution
    1. If there are three polarizers in the sequence what should be the angles between the axes of the polarizers to maximize the light transmitted? What is the maximum amount of light transmitted?

    Let x be the angle between the first and second polaroid axes. Hence the final intensity is given by:

    [tex]I_f = (1/2)I_0Cos^2[x]Cos^2[90-x][/tex]

    Differentiating and determining critical points, it can be shown x = (1/4)n(Pi). So if the angle between the first and last (third) polaroid is 90 (deg.), then a maximum occurs at x = Pi/4 and the final maximum intensity is I_f = (1/8)*I_0 (W/m^2). {Correct me if I am wrong...}

    2. If there are 8 polarizers in the sequence what should be the angles between the axes of the polarizers to maximize the light transmitted? What is the maximum amount of light transmitted?

    First, I did a little trial and error to see if a maximum intensity would occur for four polarizers if each polaroid axes was an angular equidistance x from each other. It can be shown that when x = 30 (deg.) {I created a table and computed products of each cosine function of the intensity equation}, the intensity obtains a maximum value. So this suggested that for eight polarizers, a maximum intensity would be obtained for when each polaroid axes was an angular equidistance from each other. Keeping in mind that the angle between the first and last (eighth) polaroid axes is 90 (deg.), this means, (letting x be the common angular separation between each polaroid axis) 7x = 90 (deg.), or x = ~12.857 (deg.). The final intensity for this common angle is I_f = 812*I_0 (mW/m^2).

    3. (My question) What is the minimum number of polarizers required in order for the maximum final light intensity to be .485*I_0 ?

    Okay so this is where I am stumped. It seems that if there are three polarizers, then the maximum final intensity is less than .485*I_0 so this would suggest that in order to obtain this final maximum intensity, there would have to be less than three polarizers. However, the next viable obtain (two polarizers), the final intensity is always 0 (W/m^2) so long as the angle between the first and last polarizers is 90 (deg.). I guess this might suggest I am doing parts (1) and (2) wrong, but I really don't see any other way than a numerical solution (which would require a tremendous amount of calculations, even using excel....) Can anyone explain this to me?

    Thank you

    - Sam
     
  2. jcsd
  3. Jun 7, 2010 #2

    vela

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    With two polaroids, you get no light. With three, you get some. So adding another filter counterintuitively increases the amount of light coming through. So instead of three, how much light could come through if you had four?
     
  4. Jun 8, 2010 #3
    Well the maximum light intensity comes through when there exists an angular equidistance of x = 30 (deg.) between each polaroid axes and I_f = (27/128)*I_0 (W/m^2).
     
  5. Jun 8, 2010 #4

    vela

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    How does that compare to the amount of light that comes through when you have just three polaroids?
     
  6. Jun 8, 2010 #5
    Its less than. I understand that as you add more polaroids, the light that passes through is less intense. What is your point?
     
  7. Jun 8, 2010 #6

    vela

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    It is? If you have three polaroids, only 1/8 (or 16/128) comes through. With four, you have 27/128. Seems like more to me.
     
  8. Jun 8, 2010 #7
    I'm sorry. I am an idiot. :)

    Thank you, vela.
     
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