# Fast Q-switching of flash pumped Nd:YLF laser

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1. Jul 23, 2016

### ppedro

Hi friends!

I solved the problem 8.3 of Problems in Lasers Physics'' book (by Cerullo, Longhi, Nisoli, Stagira and Svelto) but I think there's a mistake on the solution presented in this book in page 196. This is a problem book with problems and solutions that follows closely the laser physics theory as presented in Principles of Laser Physics'' book (by Svelto), referring many times to the equations derived in it.

My question has to do with the book's solution way of calculating $A_{b}$ which is the laser beam area inside the crystal (at its center, in the middle of the confocal resonant cavity). This laser beam is, as far as I understand, approximated to be of cylindrical form inside the crystal, and thus it has volume $V=A_{b}l$, where $l$ is the length of the crystal.

Alright, the book's solution says the area of the beam at the center of the resonator is given by
$$A_{b}=\frac{\pi w_{b}^{2}}{2}$$
where $w_{b}$ is the beam spot size at the center. This value of $w_{b}$ is said to be given by $w_{b}=\left(\frac{L\lambda}{2\pi}\right)^{1/2}$ for the case of a confocal resonator, in accordance with the second equation in Eq.5.5.11 of the Principles of Laser Physics'' book, which I agree.

However, I don't see the reasoning in the division by 2, because, by definition, $w_{b}$ is the radius of the cross section'' (the spot'') of the gaussian beam, as can be seen throughout Principles of Laser Physics'' starting from all the definitions in section 4.7 Gaussian beams'', and which is, for example, represented in Fig.5.9.a. To me this would just be $A_{b}=\pi w_{b}^{2}$. Am I wrong? This might seem minor, but to me it's important because it makes a difference on a request to reevaluate my exam correction.

2. Jul 28, 2016

### Greg Bernhardt

Thanks for the post! This is an automated courtesy bump. Sorry you aren't generating responses at the moment. Do you have any further information, come to any new conclusions or is it possible to reword the post?

3. Jul 30, 2016

### ppedro

Thanks. I don't have any further information. I think I asked it pretty clearly, but I admit it's a pretty technical question.

4. Jul 30, 2016

### Fred Wright

I don't have a copy of your reference but usually the normalized intensity profile for a gaussian beam is defined to be:
$I(r)= exp(-2\frac {r^2} {w^2})$
If you define the area of the "spot" to be at FWHM then,
$\frac{1} {2} = exp(-2\frac {r^2} {w^2})$
$r^2 = \frac {w^2} {2}ln(2), A = \frac {\pi} {2} w^2 ln(2)$
It could be that the authors of your reference are more restrictive in the area by setting,
$I=e^{-1}I_{max}$ (because of fast switching?) which would give the answer they quoted. Please check your authors' definition of the gaussian beam profile and where they determine the area of the "spot" on the profile.

5. Aug 6, 2016

### ppedro

Thanks for the input. Yes, that definition gives the computed answer. However, the textbook does not strictly define the area of the spot nor gives a reason for that choice in this particular problem. There are several possible definitions and my answer is the one using $I=e^{-2}I_{max}$.