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Two linear polarizers in series

  1. May 6, 2016 #1
    1. The problem statement, all variables and given/known data
    The optical power of a HeNe -laser is ##P_0 = 5.0mW## and the wavelength ##\lambda = 633nm##. The emitted light is linearly polarized. As the laser beam travels through two in-series -polarizers, the power detected behind the second polarizer ##P_2 = 1mW## . If the first polarizer is removed, the detected power drops to zero.

    At what angle with respect to the direction of polarization of the laser is the polarization axis of the first polarizer?

    2. Relevant equations
    The power of a harmonic wave:
    \begin{equation}
    P = \frac{1}{2}\mu \omega ^2 A^2 v = \frac{1}{2} \mu (2\pi f)^2 A^2 v = \frac{1}{2} \mu (2\pi \frac{v}{\lambda})^2 A^2 v,
    \end{equation}
    where ##\mu## is the linear density of the medium of propagation, ##\omega## is the angular frequency, ##A## is the amplitude of the wave and ##v## its velocity.

    Sine of a double angle:
    \begin{equation}
    sin(2\theta) = 2sin(\theta)cos(\theta)
    \end{equation}

    3. The attempt at a solution

    This question was on today's exam, and I'm pretty sure I messed up. I would like some feedback if at all possible.

    What I did was assume that only the projection of the electric field component of the wave onto the polarization axis of the first (and second) polarizer passes through, meaning the electric field component of the wave after the first polarizer
    \begin{align*}
    E_1
    &= E_0 cos(\theta),
    \end{align*}
    where ##\theta## is the angle we want to find out. As the light passes through the second polarizer, the effect is similar:
    \begin{align*}
    E_2
    &= E_1 cos(\phi)\\
    &= E_0 cos(\theta)cos(\phi),
    \end{align*}
    where ##\phi## is the angle between the axes of polarization of the first and the second polarizers. Since we know that no light passes through the second polarizer if the first one is removed, we know that the projection of the electric field component of the wave onto the axis of polarization of the second polarizer must be zero, and thus they must be at a right angle with respect to each other, meaning that
    \begin{align*}
    \phi + \theta = 90^{\circ} &\iff \phi = 90^{\circ}-\theta.
    \end{align*}
    Therefore
    \begin{align*}
    E_2
    &= E_0 cos(\theta)cos(90^{\circ}-\theta)\\
    &= E_0 cos(\theta)sin(\theta)|| \text{ Using equation (2)}\\
    &= \frac{1}{2}E_0 sin(2\theta)
    \end{align*}
    The next thing I did was to look at equation (1) and say that the intensity (and power) is proportional (equal) to the square of the amplitude of the electric field component of the wave. I did this assumption, because I didn't have the linear density available to me (what would that be in this context, even?). Regardless, if this is true, then
    \begin{align*}
    P_2
    &= E_2^2\\
    &= (\frac{1}{2}E_0 sin(2\theta))^2\\
    &= \frac{1}{4}E_0^2 sin^2(2\theta)\\
    &= \frac{1}{4}P_0 sin^2(2\theta)
    \end{align*}
    Solving this for ##sin^2(2\theta)## yields
    \begin{align*}
    sin^2(2\theta)
    &= \frac{4P_2}{P_0}
    = \frac{4}{5} \text{ (Since } P_0 = 5mW \text{ and } P_2 = 1mW \text{)}\\
    \iff\\
    sin(2\theta)
    &= \pm \sqrt{\frac{4}{5}}
    = \pm \frac{2}{\sqrt{5}}\\
    \iff\\
    2\theta &= arcsin(\pm \frac{2}{\sqrt{5}}) = \pm 63.43^{\circ}
    \end{align*}
    and the desired angle is therefore
    \begin{align*}
    \theta &\approx \pm 31.72^{\circ}
    \end{align*}
    My hunch is this is wrong, mainly because some people I talked to got a different value for ##\theta## (something like ##53^{\circ}##). My guess is that I made a mistake when I made the assumption about the power being equal to the amplitude of the electric field, but I don't really know what else I could have done there...
     
  2. jcsd
  3. May 6, 2016 #2

    gneill

    User Avatar

    Staff: Mentor

    Have you considered that there might be two valid angles and that they might be complementary angles?
     
  4. May 6, 2016 #3
    I can see how we could come to a different solution if we used the symmetry of the cosine,
    \begin{equation}
    cos(x) = cos (-x)
    \end{equation}
    Since the axis of polarization can be though of as extending onto both sides of the origin if we slapped it onto a unit circle, it doesn't really matter if we rotate the axis of polarization ##-x## degrees or ##\pi - x## degrees. However in this case it would give me something like ##150^{\circ}##, not ##50something^{\circ}##. Why would the complementary angle be a valid solution, instead of the supplementary angle?
     
  5. May 6, 2016 #4

    blue_leaf77

    User Avatar
    Science Advisor
    Homework Helper

    Yours is also right. This is due to the nature of ##y=\sin 2\theta##. If you plot this function, you will find that there are two values of ##\theta## less than 90 degrees which have the same ##y##.
     
  6. May 6, 2016 #5

    gneill

    User Avatar

    Staff: Mentor

    You have two polarizers where the second one has its orientation fixed by the given conditions to be at 90° to the polarization of the incoming light. The first polarizer will drop the intensity of the incoming light but also re-orient its polarization to match that of the angle of the polarizer. When the resulting light encounters the second polarizer the effective angle of the light with respect to the polarizer is 90° - Φ, where Φ is the rotation angle of the first polarizer with respect to the incoming light.

    Consider that the power is proportional to the intensity, and the Law of Malus for intensity is

    ##I = I_o cos^2(\phi)##

    For the two polarizers then,

    ##I = I_o cos^2(\phi) cos^2(90° - \phi)##

    You can interchange the order of the two angles without changing the result. So there is an angle and its complement that will satisfy the equation. I'll bet the angle that your classmate found was 90 - 31.72 = 58.28 degrees.
     
  7. May 6, 2016 #6
    Hmm, you're right. I guess this is the result of me not really ever taking an extensive trig course and relying too much on my calculator. Since our sine has ##2\theta## instead of just ##\theta##, you can fit two angles inside ##90^{\circ}## that satisfy the equation. I keep forgetting about the other possible angles...

    Thanks for the feedback.
     
  8. May 6, 2016 #7
    That does sound familiar. Blue_leaf77 above also noted that my answer could be valid, although they gave a different justification for it. I do see what you mean, though, so thank you. I will mark this as solved.
     
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