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Wavlnumber in media

  1. Aug 26, 2014 #1
    1. The problem statement, all variables and given/known data
    Glass has a permeability of ε=2.0 and relative permeability μ=1.
    Electromagnetic radiation of angular frequency ω=2.8x109 s-1. travels through the glass. Calculate
    a) refractive index
    b) speed of radiation
    c) frequency of radiation
    d) wavenumber of radiation

    a, b and c are straightforward but d I have questions for


    2. Relevant equations
    n=√ε
    n=c/v
    f=ω/2∏
    v=ω/k
    ω=2∏f
    k=n2∏f/v



    3. The attempt at a solution

    a) refractive index, n=√ε = √2

    b) speed of radiation travelling through glass,
    n = c/v
    so v = c/n
    v = 3.0x108 ms-1 / √2
    v = 2.12x108 ms-1

    c) frequency of radiation, f = ω/2∏
    f = 2.8x109 s-1 / 2∏
    f = 4.46x106 Hz

    d)wavenumber of radiation
    now v = ω/k - where k is the wavenumber
    so make k the subject
    k = ω/v
    k = 2.8x109 s-1 / 2.12x108 ms-1
    k = 13.2

    BUT

    ω = 2∏f
    and
    k=n2∏f/v

    so k = (2x2∏x 4.46x106) / 2.12x108 ms-1
    k = 0.132

    So which is right, please?
     
  2. jcsd
  3. Aug 26, 2014 #2

    Simon Bridge

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    The first one.

    Note: ##y(x,t)=A\sin(kx-\omega t + \delta)## ... which means that ##k=2\pi/\lambda## ... i.e. inverse wavelength in the medium.

    If v is the speed in the medium, then ##v=f\lambda \implies c=nf\lambda## is the speed in a vacuum.

    So ##k=2n\pi f/c## ... recognize it?
     
  4. Aug 26, 2014 #3

    collinsmark

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    Don't you mean relative permittivity? :biggrin:

    Try that one again. Keep an eye on the exponent. You're a couple orders of magnitude off there.
     
  5. Aug 26, 2014 #4
    Yes, I see.
    And yes, typo's.

    Ta.
     
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