Engineering Optical Fibres: Spectral Width Calculation

AI Thread Summary
The discussion revolves around calculating the spectral width of light emitted from a pulsed laser operating at 850 nm. The user initially miscalculates the spectral width, arriving at a value significantly larger than the expected 0.6 nm. Key points include the correct application of frequency spacing and the relationship between wavelength and frequency. The user clarifies the correct formula for frequency, realizing that using the vacuum wavelength leads to the correct calculations. Ultimately, the user reconciles their approach and arrives at the expected answer by correctly applying the relationship between wavelength and frequency.
Master1022
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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.
 
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Master1022 said:
I thought that
Thinking is one thing. Writing an equation is something else --- and in this case, better :smile:

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

##\ ##
 
Thanks for the reply @BvU

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

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.
 
Master1022 said:
If so, then I could do
I mislead you to think ##f = c/(n\lambda)\ \ ## (not on purpose, I'm afraid o:) )

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##.

##\ ##
 
BvU said:
I mislead you to think ##f = c/(n\lambda)\ \ ## (not on purpose, I'm afraid o:) )

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
 

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