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Help with impedance matching problem

  1. Oct 8, 2005 #1
    Ok, need some help with the end part of this, heres the problem...

    The relation between the impedance Z and the refractive index n of a dielectric is given by Z=1/n. Light travelling in free space enters a glass lens which has a refractive index of 1.5 for free space wavelength of 5.5E-7 m. Show that the reflections of this wavelength are avoided by a coating of refractive index 1.22 and thickness 1.12E-7 m.

    Alright, since this is just a impedance mathching problem, you know that Z2=sqrt(Z1*Z3). In this case, Z1 is just 1 since its air and Z3=1/1.5. Solving for this you get Z2=1/1.22 where n=1.22.

    The next part is where I have a problem. You know from the boundary conditions that Z1/Z2=(A2-B2)/(A1-B1) where A1 is the amplitude of the light in Z1, B1 is the amplitude of the reflected wave in Z1 and the same for A2 and B2. If you go back a few steps you also know that k1/k2=(A2-B2)/(A1-B1) where k1 and k2 are the wave number (2PI/wavelength). So, you should be able to set Z1/Z2=k1/k2 and then plug in what you know and solve for the wavelength of the wave when its in Z2. Once you get that you know that the thickness of the coating should be l=wavelenth/4 since this is a given in a impedance matching problem.

    However, when i solve out everything, I dont get 1.12E-7. I find that I get 1.67E-7 instead and the only way to get 1.12E-7 is if Z1/Z2=k2/k1. I've checked my algebra many times I really don't see a problem in it. Anyone have any idea what I could be doing wrong?
  2. jcsd
  3. Oct 8, 2005 #2


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    Homework Helper

    Wow, you're doing quite alot of things here, and I'm not sure how to follow. The theoretical condition for no reflection from an impedence point of view is for the input impedance into the coating to be equal to the intrinsic impedence of EM waves in air. The waves will thus "feel" like there has been no change in medium at all.

    Let medium 1 be air with intrinsic impedence [tex]\sqrt{\frac{\mu_0}{\epsilon_0}}[/tex], medium 2 be the coating, 3 be the glass and propagation in the positive z direction. Assume lossless dielectrics. Now from the equation:

    [tex] \eta_i_n = \eta_2\frac{\eta_3 + j\,\eta_2tan\beta z}{\eta_2 + j\,\eta_3tan\beta z}[/tex]

    \eta_i_n = \eta_1

    \eta_1 = \eta_2\frac{\eta_3 + j\,\eta_2tan\beta z}{\eta_2 + j\,\eta_3tan\beta z}[/tex]

    For them to be equal, we have two solutions, but we are more interested in the one where the imaginary side approaches infinity. Then

    [tex]\eta_1 = \frac{\eta_2^2}{\eta_3}[/tex]

    For the imaginary to approach infinity, we must have
    [tex]\tan\beta z \rightarrow \infty[/tex] or [tex]\beta z = \pi/2, 3\pi/4,...[/tex]

    Solving simply for z, or d, as the propagation constant beta is fixed for a specific wave length and medium (medium 2 - coating).

    [tex]d = \frac{\lambda_2}{4}[/tex]

    If you calculate the wavelength of 5.5E-7 m in the coating or 5.5E-7/1.22, and plug it into that equation, you get your answer.

    For the goodness of GOD! Finally I've fixed the latex.
    Last edited: Oct 8, 2005
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