Absorption and Gain for Lasers

[PH^S!C5]

Homework Statement

If a round trip gain in a 56.8 cm long laser is 5.16%, what is the net gain coefficient (g-\alpha).

g is the small signal gain coefficient
\alpha is the absorption coefficient
L-length of cavity

Homework Equations

The Round Trip Power Gain: Gr= R1*R2* exp[(g-\alpha)*2L]
where I have to solve for (g-\alpha)

The Attempt at a Solution

I have set R1=R2=1 assuming high reflectivity but I am not sure if this is alright.
First, solve as follows: ln(Gr)/2L = (g-\alpha)
Then, substitute the values for L and Gr:
ln(0.0516)/(2*56.8 cm) = -0.0260936057 = (g-\alpha)

However, I am not sure what this represents in terms of laser net gain coefficient and if I have correctly solved this (assuming R1=R2=1 and where Gr=0.0516 instead of 5.16% form.
I would appreciate some revision and explanation. I have read the textbooks from Milonni (Laser Physics) and Silfvast (Laser Fundamentals) but they don't really explain the meaning of this in detail. I have searched on google and this ebook helped a bit but I really can't say I understand completely... Any help is appreciated! THANKS!

Google ebook link: http://books.google.ca/books?id=DlW...EUQ6AEwAw#v=onepage&q=round trip gain&f=false

 

[TSny]

Would the round trip power gain be Gr = 1 + .0516 = 1.0516? (Instead of .0516)

 

[PH^S!C5]

Truthfully, I am not certain. I have read over the textbook again and the only possible explanation for adding a one is to have a net gain per round trip increase through the amplifier?
I have tried your suggestion and I obtained a value of 4.43x10^-4 cm^-1... at least it's not negative. Do you know if this is correct? Why? Thanks again!

 

[TSny]

I'm not real certain. But "gain" is often defined as the ratio of the output power to the input power. See http://en.wikipedia.org/wiki/Amplifier.

So, in your case, I would think the round trip gain would be the power after one round trip divided by the power at the start of the round trip. So, if $P_o$ is the power at the start of the round trip and if the power increases by 5 % in one round trip, then the power after one round trip would be $P_o + .05P_o = (1+.05)P_o = 1.05 P_o$. So the gain would be $\frac{1.05P_o}{P_o} = 1.05$