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ElectroMagnetic Waves

  1. Oct 3, 2006 #1
    An electromagnatic wave is traveling in vacuum with frequency 5.7 x 1014 Hz. The wave has average total energy density of 4.6 x 10-6 J/m3.


    (e) How much energy does a 1.9 m2 flat surface (perpendicular to the wave propagation direction) receive in 9 s?

    E = 23598 J, how do i get that number?


    4.6*10^-6 = ATC??
     
  2. jcsd
  3. Oct 3, 2006 #2
    I1 / I0 = .871572

    (b) Calculate the electric field amplitude E1 of the light after it has passed through the first polarizer. Express your answer as a fraction of the electric field amplitude E0 of the initial beam.

    E1 / E0 = .933


    I = c*e0*E2

    I cant get the answer either thanks
     
  4. Oct 3, 2006 #3

    Hootenanny

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    How far will the wave have travelled in 9 seconds? What is the volume of the cuboid traced by the surface and the distance travelled by the wave?
     
  5. Oct 3, 2006 #4
    the wave would have traveled 17.1 Meters in 9 seconds is that right? (1.9*9) = 17.1
     
  6. Oct 3, 2006 #5

    Hootenanny

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    Are you sure about that? How fast does an EM wave travel?
     
  7. Oct 3, 2006 #6
    3*10^8 m/s

    So 3*10^8 * 1.9 = 5.7E8 meters in a second so 5.13E9 meters in 9 seconds

    Ok Got it thanks
     
    Last edited: Oct 3, 2006
  8. Oct 3, 2006 #7
    i have no idea on this one, i cant get the ratio at all
     
  9. Oct 3, 2006 #8

    Hootenanny

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    Careful, your dealing with a volume there not a length, never the less your answer should be correct.

    [tex]\hline[/tex]
    For your next question it may be useful to note that;

    [tex]\frac{I}{I_{0}} = \cos^2\theta[/tex]

    and

    [tex]E = E_{0}\cos\theta[/tex]

    Although it is useful to remember that The ratio of the intensities is equal to the square of the rms of the Electric field.
     
  10. Oct 3, 2006 #9
    Thanks alot, one last question


    (c) The second polarizer is set at various angles within the range q 2 = 0 to 90°. Calculate the intensity of the light after it has passed through the second polarizer for the following values of q 2. Express each answer as a fraction of I1.



    At q 2 = 22°: I2 / I1 =




    So i am guessin i need to find final intensity which would be


    S = S0 ((cos theta)^2)^2

    is that right?
     
  11. Oct 4, 2006 #10
    do u have to calculate two diffrent numbers and then add them?
     
  12. Oct 4, 2006 #11
    help meeeeeee
     
  13. Oct 5, 2006 #12

    Hootenanny

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    Note my original Equation above;

    [tex]\frac{I}{I_{0}} = \cos^2\theta[/tex]
     
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