Calculating the number of quanta emitted as Cherenkov radiation in a wavelength interval dl? This website gives an equation for it http://math.ucr.edu/home/baez/physics/Relativity/SpeedOfLight/cherenkov.html
It appears as plaintext but this is what I believe it is parsed:
$n = \dfrac{dl  2πα  L \sin^2θ}{l^2}$
And this is the text surrounding and about the equation:

For most media blue light predominates over longer wavelengths of light, because the number of quanta emitted as Cherenkov radiation in a wavelength interval dl at wavelength l over a path length L is given by:
where α is the fine structure constant, equal to about 1/137.  Notice that the refractive index, and therefore the angle θ, changes with wavelength l as demonstrated when a prism produces a spectrum from white light.  This suppresses the rate at the small wavelengths of ultraviolet and beyond.

Other websites list this formula but appear to have copied and pasted the surrounding text around it.
It's not an equation in the given form but I assume from the text around it that an $n = $ could be added for the number of quanta.
Three questions: Have I interpreted this equation correctly?
If so, why didn't they cancel the common factor of $l$.
And can someone link or show a derivation of this equation or whatever the correct equation is?
 A: I'm by no means an expert on this, but since I found this question interesting I searched in the infinite vastness of the internet for a hint and came up with three links:


*

*Number 1. (search for the word "celebrated") 

*Number 2. eq. 1.10 on p.6 (warning: it's a.) a
pdf and b.) it's in german) 

*Number 3. eq. 2.5 on p.7 (same
warning)



Have I interpreted this equation correctly?

Basically yes. You should write it as follows:
$$ N = \dfrac{  2πα  Z^2 L\sin^2θ d\lambda}{\lambda^2}$$
(so there's no confusion between the length L and the wavelength lambda)
In the links they write $\frac{dN}{dx}$, so when you write it the way you did you have already integrated over dx. But you still have to integrate over the range of wavelenghts of emitted photons, this is why you can't cancel the $d\lambda$-factor. 
Since $\cos(\theta)=\frac{1}{n \beta}$, you can write the $sin(\theta)^2$ as:
$\left(1-\frac{1}{\beta^2 n^2}\right)$
which then yields 
$$\frac{dN}{dx} = \dfrac{  2πα  Z^2 d\lambda }{\lambda^2}\left(1-\frac{1}{\beta^2 n^2}\right)$$ as given in the second two links. 
edit: I just realized that I completely ignored that there's no Z in your formula. So Z is the charge of the particle traveling through the medium. When you want to calculate the number of photons emitted by an electron, then Z would obviously be 1.
