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In the exercise we are given that the spectrum of a light source consists of two spectral lines, which both have wavelengths around $500 \text{ nm}$ and the separation between them - given in wavenumbers - is close to $0.1 \text{ cm}^{-1}$.

One part asks us to calculate the distance between the mirrors if we want the free spectral range of our Fabry-Perot etalon to satisfy $FSR(\text{in } \overline \nu) = 2.5(\overline \nu_2 - \overline \nu_1)$.

Now, I know that the free spectral range ($\theta = 0$, $n_{air} \approx 1$) is given by

$$FSR = \frac{c}{2d}$$

where $d$ is the distance between the mirrors and we have

$$\Delta\overline \nu = \frac{c}{\lambda_2} - \frac{c}{\lambda_1} = \frac{c}{2\pi}\Delta k $$

So shouldn't then

$$d = \frac{c}{2 FSR} = \frac{c}{2 \cdot 2.5c/(2\pi) \Delta k} = \frac{\pi}{2.5\Delta k} \approx 12.566 \text{ cm}$$

According to the solution this is wrong: They get $FSR = 0.25 \text{ cm}^{-1}$ and $d = \frac{1}{2 \cdot FSR} = 20 \text{ mm}$

I seem to be missing something. If they calculate the free spectral range in terms of the wavenumber, shouldn't they convert it to frequency by

$$FSR(k) \to FSR(\nu) = \frac{c}{2\pi} FSR(k)$$

instead of

$$FSR(k) \to FSR(\nu) = c \cdot FSR(k)$$

Does anybody see what's going on here (maybe the solution is simply wrong, but I'd like to know for sure)?

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up vote 1 down vote accepted

The magnitude that is denoted as $\bar{\nu}$ is not the frequency, it is the wavenumber defined as follows: $$ \bar{\nu} = \frac{1}{\lambda} $$ The wavenumber used in spectroscopy $\bar{\nu}$ and "usual" wavenumber $k$ are different. The corect conversion of FSR is $$ \mathrm{FSR}(\nu) = c \cdot \mathrm{FSR}(\bar{\nu}) = \frac{c}{2\pi} \mathrm{FSR}(k) $$

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Ah, this clears everything up. Using one single term for two things which have the same units and come up in the same contexts is just plane vicious! :-( physicists... – Sam Jan 1 '12 at 15:27
Thank you very much, by the way. =) – Sam Jan 2 '12 at 7:58

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