# Laser flux power density and Raman scattered radiation

I am currently studying the textbook Infrared and Raman Spectroscopy, 2nd edition, by Peter Larkin. In a section entitled The Raman Scattering Process, the author says the following:

The intensity of the Raman scattered radiation $$I_R$$ is given by:

$$I_R \propto \nu^4 I_o N \left( \dfrac{\partial{\alpha}}{\partial{Q}} \right)^2$$

where $$I_o$$ is the incident laser intensity, $$N$$ is the number of scattering molecules in a given state, $$\nu$$ is the frequency of the exciting laser, $$\alpha$$ is the polarizability of the molecules, and $$Q$$ is the vibrational amplitude.

The above expression indicates that the Raman signal has several important parameters for Raman spectroscopy. First, since the signal is concentration dependent, quantitation is possible. Secondly, using shorter wavelength excitation or increasing the laser flux power density can increase the Raman intensity. Lastly, only molecular vibrations that cause a change in polarizability are Raman active. Here the change in the polarizability with respect to a change in the vibrational amplitude, $$Q$$, is greater than zero.

$$(\partial \alpha / \partial Q) \not= 0$$

The Raman intensity is proportional to the square of the above quantity.

What is "laser flux power density", and what about $$I_R \propto \nu^4 I_o N \left( \dfrac{\partial{\alpha}}{\partial{Q}} \right)^2$$ indicates that increasing the laser flux power density can increase the Raman intensity?

I would greatly appreciate it if people would please take the time to explain this.

Well, sorry for the short answer, but "laser flux power density" is just $$I_0$$, which is the power per unit surface area (which is why it is called flux power density) and the Raman intensity is proportional to $$I_R$$, so by this equation you can see that when $$I_0$$ increases, so does $$I_R$$