Certainly, you can. See e.g. Article The electric field standing wave effect in infrared transfle...

Article The Electric Field Standing Wave Effect in Infrared Transmis...

Hi MIA. You've asked a very interesting question. TM, above, has referred you to some articles that suggest good answers. I'm going to try to give a relatively short answer that might also be helpful. A lossless cavity has a set of resonant modes that satisfy all the boundary conditions of the cavity as you know. If truly lossless, these modes have very narrow resonances, i.e., the Q of the modes is very high -- infinitely high in the idealized lossless case. (Consider an evacuated cavity formed by superconducting walls.) But most cavities have loss, for example due to dissipation in the walls of the cavity or absorption within the cavity itself. You've anticipated this in your question. For these cavities, the modes are broader, their width being an inverse measure of the time that a resonant excitation in the cavity takes to ring down. In the case of Fabry-Perot (FP) resonators, Q can be very high with good coatings and little internal loss. These modes don't respond to any excitations except those very, very close to resonance. But any pair of reflectors can form a FP cavity. With low reflectivity, the modes will be very broad. But the calculation is the same as for a FP cavity with high Q -- with an imaginary component to the index. The loss can be small or can even be large -- the math is the same. Wide modes will respond to excitations that are significantly off-resonance. Now consider the case of a FP cavity with loss but also with a source of gain, e.g., a semiconductor laser. Below transparency, the cavity modes are wide but as the laser is pumped closer to transparency, the modes narrow. At transparency, the modes are still relatively wide because there is typically lots of loss out the ends of the cavity and some internal loss. But continue to pump the gain higher and the modes narrow as the gain compensates for mirror and internal losses. Eventually, gain is high enough that the net losses in the cavity approach (but do not reach) zero. The modes of a semiconductor laser in this state can be extremely narrow. The arithmetic of a FP cavity with gain is simple as I've suggested and is beautifully treated by McCall and Platzman in the warm-up intro to "An Optimized pi/2 Distributed Feedback Laser," JQE, vol 21, Dec 1985, pp. 1899-1904. With this article, you can calculate the continuous spectrum of a laser. And with the intuition that you'll get from working through TM's articles, McCall and Platzman and some of your own calculations, you be able to figure out the loss in a guitar string just by listening to it slowly quiet down after being plucked. Best of luck on this important question.