Optical Fibres: Spectral Width Calculation

[Master1022]

Homework Statement

A pulsed GaAlAs laser operates at 850 nm wavelength. Under pulsed conditions there are several lasing modes present separated in frequency by $\Delta f = c/2nl$ where typically $l$ = 300 $\micro m$ (the length of the laser cavity), and $n$ = 3.5 is the refractive index of the laser cavity. By assuming that three modes are lasing, calculate the spectral width $\Delta \lambda$ of the emitted light.

Relevant Equations

$v = f \lambda$

Hi,

I was working on this problem that I think should be quite simple, but I cannot seem to get the correct answer.

Question:
A pulsed laser operates at 850 nm wavelength. Under pulsed conditions there are several lasing modes present separated in frequency by $\Delta f = c/2nl$ where typically $l$ = 300 $\mu m$ (the length of the laser cavity), and $n$ = 3.5 is the refractive index of the laser cavity. By assuming that three modes are lasing, calculate the spectral width $\Delta \lambda$ of the emitted light.

Attempt:
Three modes means that there are two frequency spacings, $\Delta f$, between the lower and upper frequencies. Thus:
2 \Delta f = \frac{c}{nl} = \frac{3 \times 10^{8}}{3.5 \cdot 300 \times 10^{-6}} = 2.867... \times 10^{11}
Therefore, I thought that:
\text{Spectral width} = \Delta \lambda = \frac{c/n}{2 \Delta f} = 3 \times 10^{-4} \text{m}

However, the answer is 0.6 nm which is quite a bit smaller than what I have... I do not really understand what approach I should be using instead.

Any help would be greatly appreciated.

 

[BvU]

I thought that

Thinking is one thing. Writing an equation is something else --- and in this case, better

What is $\ {d\lambda\over df}\$ if $\ \lambda = {c/n\over f}$ ?

$\$

 

[Master1022]

Thanks for the reply @BvU

Thinking is one thing. Writing an equation is something else --- and in this case, better

What is $\ {d\lambda\over df}\$ if $\ \lambda = {c/n\over f}$ ?
$\$

So $\frac{d\lambda}{df} = -\frac{c/n}{f^2}$

I am then slightly confused how to proceed from there. Would I calculate $f$ from the 850 nm we are given? If so, then I could do:
d\lambda = \frac{c/n}{(c/n\lambda)^2} \cdot df = \frac{\lambda^2}{c/n} \cdot 2 \Delta f
= \frac{\lambda^2}{c/n} \cdot \frac{c}{nl} = \frac{\lambda^2}{l}
but I am still making an error as that does not give me the right answer.

 

[BvU]

If so, then I could do

I mislead you to think $f = c/(n\lambda)\ \$ (not on purpose, I'm afraid )

The 850 nm is in vacuo, so $\ f\$ should be $\ f = c/\lambda$

Numerically I still don't get the 0.6 nm but 0.68 nm for $2\Delta f$.

$\$

 

[Master1022]

I mislead you to think $f = c/(n\lambda)\ \$ (not on purpose, I'm afraid )

The 850 nm is in vacuo, so $\ f\$ should be $\ f = c/\lambda$

Numerically I still don't get the 0.6 nm but 0.68 nm for $2\Delta f$.

$\$

Thank you very much once again for your reply @BvU !

Using that, then I get:
\lambda = \frac{c/n}{f} \rightarrow \frac{d\lambda}{df} = -\frac{c/n}{f^2}
= \frac{-c/n}{\left( \frac{c}{\lambda} \right)^2 } = \frac{-c/n}{c^2 / \lambda^2} = \frac{\lambda^2}{nc}
Then we can multiply by $2\Delta f$ to get $\frac{\lambda^2}{n^2 l}$. However, I think I need to have just a factor of $n$ in the denominator (instead of $n^2$) to get the same answer as you. Could you see how I can reconcile this difference

[EDIT]: Nevermind, I have just realized that if I start with $\lambda = \frac{c}{f}$, then I get the same answer as you