I'm trying to follow this worked example from my lecturer.

Here's the question:


and here's the answer to part 1:

enter image description here

When I was attempting this without looking at the answer, I did correctly identify the formula to be used, but I got an answer about seven orders of magnitude off. It seems that the lecturer's answer comes from ignoring the speed of light in the formula. It seems there's no other way to get there; the square root term produces a complex number with magnitudes $10^1 + 10^{-1}i$, so you need the frequency derived from the wavelength of order $10^{-9}$ which has a magnitude of $10^7$ to get the result shown in the official mark scheme. Obviously if you then divide the frequency by the speed of light that cuts the magnitude down to $10^1$ leaving you with a complex number that has nowhere near the required magnitudes of $10^7$ or $10^6$.

Sure enough, once I exclude the speed of light from the calculation, I get his result exactly. This isn't a mistake, since he does it on a number of other occasions when using the same formula.

Is this some kind of convention (i.e. a unique system of units -- I have heard that theoretical physicists will often work in unit systems where $c = 1$, but this was never explicitly mentioned) or is there some underlying physics that I don't know?


1 Answer 1


You don't need the speed of light: $$ k_x=\frac{\omega}{c}\sqrt{\frac{\epsilon_d \epsilon_m}{\epsilon_d+\epsilon_m}}=\frac{2\pi}{\lambda}\sqrt{\frac{\epsilon_d \epsilon_m}{\epsilon_d+\epsilon_m}}, $$ since $$ \omega=2\pi\nu \,\,\,\text{and}\,\,\,c=\lambda\nu. $$ Or you could simply say: $$ k_x=k_{\rm Vak}\cdot n=\frac{2\pi}{\lambda}\cdot\sqrt{\frac{\epsilon_d \epsilon_m}{\epsilon_d+\epsilon_m}}. $$ How did you use the speed of light if you just know $\omega$ thrue $\lambda$, which would take c away (if you know what I mean).

  • $\begingroup$ Oh, I thought that $\omega = 2\pi f = \frac{2\pi}{\lambda}$? $\endgroup$
    – lordnoob
    Commented Jan 4, 2021 at 13:36
  • 4
    $\begingroup$ @lordnoob that would imply $f=1/\lambda$, or $\lambda f=1$. I'm sure you can see the units don't work out, and the correct equation is $\lambda f=c$. There's your missing factor of $c$. $\endgroup$ Commented Jan 4, 2021 at 14:12
  • $\begingroup$ The frequency is the amount of turns per time which is equivalent to 1 turn over the period $T$ (Time for one turn): $f=\nu=\frac{1}{T}$. $\endgroup$
    – Roger
    Commented Jan 4, 2021 at 14:35

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