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Homework Help: Infinite number of polarisers (Malus's law)

  1. Apr 18, 2012 #1
    My problem:
    "Imagine you have two identical perfect linear polarisers and a source of natural light. Place them one behind the other and position their transmission axes at 0° and 90° respectively. If 1000 W.m-2 of randomly polarised light is incident.

    Determine an expression for the maximum possible intensity of light that can be transmitted through if you insert ‘n’ polarisers between them."

    I understand that I need to be using, Malus's law which is...
    [tex] I_θ = I_0 Cos(θ_n)^2[/tex]

    My attempt:

    I know that the first and last polariser transmission axes must be at 0° and 90°, I know how to write the expression out... eg. [tex] I_θ = I_0 Cos(θ_1)^2 Cos(θ_2)^2......... Cos(θ_∞)^2[/tex] But I am just not sure how to generalize the expression.
  2. jcsd
  3. Apr 18, 2012 #2


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    Homework Helper
    Gold Member

    I'll try to get you started.

    I don't think the problem is actually asking about an infinite number of polarizers. It's just asking for n polarizers (plus the two given polarizers on each end), where n is some arbitrary natural number (and can be finite).

    Here are some hints that I promised to get you started.
    • The light's intensity after the first polarizer is [itex] \frac{1}{2}I_0 [/itex], where [itex] I_0 [/itex] is the original intensity. That's because the original light is unpolarized.
    • Now let's start adding polarizers in between the two given ones. The problem statement said that the transmission axes should be such that the maximum possible intensity gets through the system. That means that you need to adjust the transmission axes of the polarizers in the middle such that the total of 90o of rotation is divided up evenly between them. In other words, if you add 1 polarizer, its angle should be 45o. If you add two polarizers, the angles should be at 30o and 60o, and so on.
    • Let's start with 1 polarizer (n = 1). Remember, the intensity after the first polarizer is [itex] \frac{1}{2}I_0 [/itex]. That intensity is then reduced by a factor of [itex] \cos^2(45^{\circ}) [/itex] by the middle polarizer. Finally, the intensity is reduced by another factor of [itex] \cos^2(90^{\circ} - 45^{\circ}) = \cos^2(45^{\circ})[/itex] by the last polarizer. So the final intensity in this case is [itex] I = \frac{1}{2}I_0\cos^2(45^{\circ})\cos^2(45^{\circ}) = \frac{1}{2}I_0 \cos^4(45^{\circ}) [/itex]

    Now try it with n = 2. Then n = 3. Do you see a pattern forming?
    Last edited: Apr 18, 2012
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