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Distance travelled by wave at whichamplitude decreases

  1. Apr 18, 2010 #1
    okay so the question involves a boundary between 2 dielectrics.
    one with n1=1.5 the other n2=1.4
    incident angle of wave I= 80deg
    critical angle i found to be C= 69deg
    given wavelength in air is 700nm

    1) prove component of wave vector parallel to boundary is = ky = k2*(n1/n2)*sin(I)
    ...............this i have done
    2)Prove that if the wave incident angle I>C then kx in the second dielectric is purely imaginary............this i have done

    3)If I=80deg calculate the depth in the second dielectric at which the amplitude of the electric field has fallen by a factor of exp(-5) from its surface value
    .........THIS I HAVE NOT DONE

    i was told in lectures that if a wave has an absorption coeff = (ni*w)/c then the, the absorption length is = c/(ni*w) but i dont know what to do from here
  2. jcsd
  3. Apr 18, 2010 #2
    Re: distance travelled by wave at which amplitude decreases

    ok so ive tried this:

    =Eo*exp[i(wt-krz)]exp[-kiz)] - shows amp falls by exp[-1]

    =Eo*exp[i(wt-krz)]exp[-5kiz)] - shows amp falls by exp[-5]

    therefore absorption coeff = 5ki
    absorption length = 1/5ki = c/5niw

    ki = niw/c
    w = 2pi*c/n2L...........L = wavelength = 700nm

    so absorption length = n2L/ni10pi = 3.1194x10-8/ni

    but how do i find ni and is this right anyway?
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