Take the 2-minute tour ×
Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It's 100% free, no registration required.

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)?

share|improve this question
add comment

1 Answer

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 $$ \mathrm{FSR}(\nu) = c \cdot \mathrm{FSR}(\bar{\nu}) = \frac{c}{2\pi} \mathrm{FSR}(k) $$

share|improve this answer
    
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
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.