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Photon Polarization Question

  1. Oct 10, 2015 #1
    1. The problem statement, all variables and given/known data
    Linearly polarized light of wavelength 5890 A is incident normally on a birefringent crystal that has its optic axis parallel to the face of the crystal, along the x axis. If the incident light is polarized at an angle of 45° to the x and y axes, what is the probability that the photons exiting a crystal of thickness 100.0 microns will be right-circularly polarized? The index of refraction for light of this wavelength polarized along y (perpendicular to the optic axis) is 1.66 and the index of refraction for light polarized along x (parallel to the optic axis is 1.49.

    2. Relevant equations
    I know that |x'> = cosΦ|x> + sinΦ|y> and that |y'>= -sinΦ|x> + cosΦ|y>.
    I also know that the right-polarized state is given by |R> = 1/√2 |x> + i/√2 |y>.

    3. The attempt at a solution
    When the light first enters, it is in the state |x'> = 1/√2|x> + 1/√2|y>. So, when it is refracted, it has a new angle with respect to the x axis. This new angle given by Snell's Law is, sinα = √2 / 1.49. So, the particle's state with respect to x is |x''> = cosα|x> + sinα|y>. The probability of this happening is |<x''|x'>|2 which is (cosα/√2 + sinα/√2)2 ~.7988.

    Since its a birefringent crystal, some particles have also been scattered with respect to the y axis, where its angle is given by sinβ = √2 / 1.66. So, these particles' state with respect to y is |y''> -sinβ|x> + cosβ|y>. The probability of this happening is |<y''|y'>|2 which is (-sinβ/√2 + cosβ/√2)2 ~0.054.

    When it reaches the other side of the crystal, the probability that it exits in the state |R> is given by |<R|x''>|2. This gives ( (cosα)/√2 + (i sinα)/√2 )*( (cosα)/√2 - (i sinα/√2 ) = ½.

    For the other stream that was in the |y''> state, the probability it exits in the state |R> is given by |<R|y''>|2. This gives ( (-sinβ)/√2 + (i cosβ)/√2 )*( (-sinβ)/√2 - (i cosβ)/√2 ) = ½.

    The total probability then becomes the first probability of going into the x'' state multiplied by the probability of going from the x'' state to the R state, which is ~ (.7988)*.5 = .3994 added to the probability of going into the y'' state multiplied by the probability of going from the y'' state to the R state, which is ~ (.054)*.5 = .027. The sum of these probabilities is ~ 0.4267 or roughly 43%. However, when I checked the answer at the back of the book, it said 0.12 or 12%.

    I didn't use the thickness of the crystal or the wavelength of the light, but I'm not sure whether they are required in solving the problem. Any help would be appreciated. Thanks.
     
  2. jcsd
  3. Oct 10, 2015 #2

    blue_leaf77

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    Homework Helper

    No, you don't use Snell's law in that way. The angles involved in your calculation are the angles subtended by the field's vector with the fast and slow axes of the crystal. It's also said indirectly that the incoming photons is perpendicular to the crystal axes, so Snell's law is not really needed here.
    The crystal's indices and thickness is used to calculate the phase shift accumulated by the ##|x\rangle## and ##|y\rangle## components of the beam. That is, the outgoing photons is in general no longer linearly polarized as it was before the crystal. Remember that the accumulated phase of a light beam having wave number ##k## over a distance ##d## is expressed in the quantity ##e^{ikd}##, now this quantity will be different for ##|x\rangle## and ##|y\rangle## components as the crystal has birefringence nature.
     
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