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I believe my answer to be simple but I will explain as I cannot figure it out.

I am writing a simulation for an optical experiment. I have a very small square plate target of dimentions $W \times W$. I want to set up a laser linewidth that has a center wavelength of $0.1W$ and a half-width-half-maximum of $0.002W$. I am assuming that the bandwidth of the laser follows a Cauchy distribution, which has the following form:

\begin{equation} P(\lambda) = \frac{1}{\pi}\frac{\gamma}{(\lambda-\lambda_0)^2+\gamma^2} \end{equation}

where $\lambda_0$ is the center wavelength, which is equal to $0.1W$, and $\gamma$ is the half-width-half-maximum, equal to $0.002W$.

So my goal is to try to leave actual dimensions out of this and keep everything as ratios in relation to $W$. If I plug in my parameters I get:

\begin{align} P(\lambda) &= \frac{1}{\pi}\frac{0.002W}{(\lambda-0.1W)^2+(0.002)^2W^2} \\ & \frac{1}{\pi W}\frac{0.002}{(\frac{\lambda}{W}-0.1)^2+(0.002)^2} \end{align}

So as you can see, by virtue of the $1/(\pi W)$ term, I am left with an equation that forces me to use a value for $W$, instead of a ratio of $W$. I don't think this is correct because the value for $W$ will be on the order of micrometers, which will make $P(\lambda)$ blow up. Not to mention that the actual function part is never less than 1 because the input parameters are less than 1 and get smaller when squared.

Does anyone have any thoughts or suggestions?

EDIT: I didn't make it clear that I want to use the linewidth of the laser to make a pdf of how probable a certain wavelength is to illuminate the target. That is the motivation for this construction.

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    What do you mean by "$W$ will make $P(\lambda)$ blow up"? No matter how you look at the tails, the width of the distribution is roughly on the order of $W$, it is only natural that the height of the distribution is on the order of $1/W$, which is pretty big but finite and totally fine. We call $W$ the scale parameter exactly for this. – 2017-01-09
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    I guess I was just under the assumption that the values for a pdf will always be less than 1, since, when integrated, the value will equal 1. But I guess when integrated from -infinity to infinity, everything will work out okay. – 2017-01-09

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This seems to be a fairly common misconception. Actually, there is no reason for the probability density $f(x)$ to be less than one everywhere. It's the cumulative distribution function $$\mathbb{P}(X \le x) = \int_{-\infty}^x f(t) \, \mathrm{d}t$$ that should always be less than $1$.

In fact, the p.d.f. does not even need to be bounded at all (although yours is). The most common example of a p.d.f. that isn't bounded might be the chi-square distribution (with 1 dof) with density $$f(x) = \frac{1}{\sqrt{2\pi x e^x}}, \; \; x > 0,$$ which describes the square of a standard normally distributed variable.

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    There is no such thing as a "cumulative density". The correct term is "cumulative distribution function". The word "cumulative" contradicts the word "density": something cumulative is _extensive_ and density is _intesive_. $\qquad$ – 2017-01-09