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I am a newbie to the world of lasers and was working my way through some basic problems, when I encountered this one:

Optical Electronics, A.K. Ghatak and K. Thyagarajan (Cambridge University Press, 1989). First example on p. 210.

Now , I understand that we have to find the gain coefficient of the Ruby Laser , but I am unable to make heads or tails of the solution given :

  • Primarily, how did $g(\omega_0) = 1/\Delta\omega$ come about?
  • Also, if someone can guide me through the solution, it would be a great help.

Disclaimer: This is not a homework question . I am preparing for a physics exam and was solving these questions / examples from the book recommended by my instructor.

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Searching that book for "lineshape function" will return this page, which explains what that means. Essentially, the atoms in the gain medium are usually able to respond to frequencies $\omega$ which are close to, but not necessarily exactly equal to, the central frequency $\omega_0$. The response is strongest at $\omega_0$ and then it tapers off over some interval, which we call the bandwidth and usually denote $\Delta\omega$. The manner of this tapering-off is described by the gain medium's lineshape function $g(\omega)$, which measures the strength of the coupling to radiation of frequency $\omega$.

This function is always real and nonnegative ($g(\omega)\geq0\,\forall\omega$), and it is usually normalized via $$\int g(\omega)\,\mathrm d\omega=1\tag1$$ or via $$\tilde g(\omega_0)=1.\tag2$$ The property you asked about, $g(\omega_0)\approx 1/\Delta \omega$ obviously applies to the normalization $(1)$. This states that, all else being equal, if you increase the frequency span at which $g$ is relevant, and simultaneously keep the area under $g$ constant, its height must decrease in inverse proportion to the increase in the bandwidth.

That is by itself a very fuzzy statement: it is very widely applicable, but it fails to provide any sort of constant on the numerator of $1/\Delta\omega$. There will always be some constant in there, but in practice (unless you choose some crazy $g$) this constant will be relatively close to $1$. As the book mentions, if you have some specific example of $g$ that you want to investigate, then you can do a more in-depth calculation and produce a more precise statement. However, in terms of dimensional analysis, functional dependence, and order-of-magnitude estimates, the fuzzy argument works just fine.

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  • $\begingroup$ Hi , @EmilioPisanty , thank you for the explanation . What , I am unable to understand is , how did the book calculate $g(\omega_0)$ ? What is $\Delta\omega$ ? Would be grateful for a reply ... $\endgroup$
    – pranav
    Feb 24, 2015 at 1:21
  • $\begingroup$ That is sort of already in the answer. Unless you have a more specific question there's very little more I can say. You should read up on the concept of [bandwidth](). Also, keep in mind that the statement $g(\omega_0)\approx 1/\Delta \omega$ is less of a "calculation" than a general property and an order-of-magnitude estimation, as I explained in the answer. $\endgroup$ Feb 24, 2015 at 11:04
  • $\begingroup$ Hi , @EmilioPisanty - you mean that $g(\omega_0)$ is a pre-determined property for an active material , right ? $\endgroup$
    – pranav
    Feb 24, 2015 at 11:44
  • $\begingroup$ Hi @EmilioPisanty , also would be grateful if you can have a look at physics.stackexchange.com/questions/166774/… -- sorry if you find it slightly rude :) $\endgroup$
    – pranav
    Feb 24, 2015 at 11:47
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    $\begingroup$ Yes, $g$ is an intrinsic property of the gain medium. However, its behaviour is not entirely arbitrary. In particular, if you fix the shape of $g$, then its height ($\propto g(\omega_0)$), its width ($=\Delta\omega$) and its area ($=\int g(\omega)\,\mathrm d\omega$) are not all independent. $\endgroup$ Feb 24, 2015 at 11:52

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