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E and B Fields of Monochromatic Plane Waves

  1. May 26, 2015 #1
    1. The problem statement, all variables and given/known data

    560 nm light is collimated and passes as a parallel beam in a direction perpendicular to the ##x+y+z=0## plane. It is polarized parallel to the ##(y-z)## plane. Treating it as a plane wave, what are the real E and B fields?

    Intensity of the beam is ##1 \ mW/cm^2##. Make your expressions for the fields as numerical as possible.

    2. Relevant equations

    ##E(r, t) = E_0 \ cos (k.r - \omega t) \hat{n}##

    ##B(r,t) = \frac{1}{c} E_0 \ cos (k.r - \omega t) (\hat{k} \times \hat{n})##

    Where ##k## is the propagation/wave vector, and ##\hat{n}## is the direction of popularization.

    ##k=\frac{\omega}{\lambda f}=\frac{\omega}{c}= \frac{2 \pi}{\lambda}##

    Using the following relationship, I used the given intensity to work out the amplitude:

    ##I=\frac{1}{2} c \epsilon_0 E_0^2 \implies E_0 = \sqrt{\frac{2I}{c \epsilon_0}} = 7.685 \times 10^{-10} \ V/m##


    3. The attempt at a solution

    So, to find ##k.r## and ##k \times \hat{n}##, I need the equation for a plane perpendicular to the ##x+y+z=0## plane. But how can I find this equation when there are infinitely many such vectors? :confused:

    For instance the wave vector is ##k= \omega/c \ \hat{k}##, where ##\hat{k}## is a unit vector in the direction of the vector perpendicular to the ##x+y+z=0## plane. I also need that in order to work out the cross product ##\hat{k} \times \hat{n}## for the B field.

    As for the polarization ##\hat{n}## (parallel to y-z plane), would the unit vector be ##\hat{n} = \frac{\hat{y}+\hat{z}}{\sqrt{2}}##? Or does the plus sign need to be changed to a minus? Any explanation would be greatly appreciated.
     
  2. jcsd
  3. May 28, 2015 #2

    blue_leaf77

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    No, there is only one unit vector perpendicular to a given plane. Given ##\lambda## and the unit vector associated with the propagation, ##k## will be uniquely defined.
    After finding such unit vector, use the fact that ##k## must be perpendicular to ##n##.
     
  4. May 28, 2015 #3
    Yes, we need to have ##\hat{n} . k = 0##. But what is the actual expression for this unit vector (that is perpendicular to ##x+y+z=0##)?

    (I believe it should be of the form ##\hat{k}=(a\hat{x}+b\hat{y}+c\hat{z})/\sqrt{3}##, I then need to write ##k=\frac{2 \pi}{\lambda} \hat{k}##).

    And what expression do you use for ##\hat{n}##? Is it ##(\hat{y} + \hat{z})/\sqrt{2}## or ##(\hat{y} - \hat{z})/\sqrt{2}## ? :confused:
     
  5. May 28, 2015 #4

    blue_leaf77

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    Note that you can view the expression ## x+y+z=0## as a dot product between two vectors which are perpendicular to each other.
    Another alternative will be apparent if you are more familiar with the use of gradient ##\nabla## to find a normal vector w.r.t. to a surface.
    Let's first solve the vector ##k##, after that it should be easy to find the correct ##\hat{n}##.
     
    Last edited: May 28, 2015
  6. May 28, 2015 #5
    How should I use the gradient ##\nabla## here? What are we differentiating?
     
  7. May 28, 2015 #6

    blue_leaf77

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    It depends on which one is easier to you for use, if you think you can find the required unit vector for ##k## by viewing ##x+y+z=0## as a dot product as I said above then take this way.
    As for the gradient, read this http://mathworld.wolfram.com/NormalVector.html.
     
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