# Half-wavelengths of the electric field fit in the cavity length $L$: $m \frac{\lambda}{2n} = L$

I am currently studying Laser Systems Engineering by Keith Kasunic. Chapter 1.2.1 Temporal Coherence says the following:

Axial (or longitudinal) modes are determined by the geometrical fit (or resonance) of a given wavelength in the laser cavity. That is, if the two mirrors that define a laser cavity are nearly planar and perfectly reflecting – an ideal assumption, given that one mirror will be designed not to be so that light can escape the cavity as output power – then Fig. 1.12 shows that an integer number $$m = 1, 2, 3$$, etc., of half-wavelengths of the electric field fit in the cavity length $$L$$: $$m \dfrac{\lambda}{2n} = L \ \ \ \ \ \text{[m]} \tag{1.4}$$ where the refractive index $$n$$ of the gain medium is included to account for the reduction in wavelength in comparison with its free-space ($$n = 1$$) value. With the exception of a specific type of semiconductor laser known as a vertical-cavity surface-emitting laser (VCSEL) with a cavity length $$L \approx \lambda$$, the number of half-wavelengths is large in practice. For example, for a HeNe laser emitting at $$\lambda = 633$$ nm with $$L = 100$$ mm and $$n \approx 1$$, $$m = 2L/\lambda = 2 \times 0.1 \ \text{m} / 633 \ \text{nm} = 315,955$$ half-wavelengths.

I understand the cavity length $$m \dfrac{\lambda}{2n} = L$$, as a standalone, but what is meant by "half-wavelengths" of the electric field, and how is this represented mathematically, in terms of electromagnetism and electromagnetic waves? And, relatedly, how does the $$m$$ values, as shown in figure 1.12, influence the electromagnetic waves themselves, beyond the simple equation of the cavity length $$m \dfrac{\lambda}{2n} = L$$ (that is, where in the mathematics of electromagnetic waves does the $$m \dfrac{\lambda}{2n} = L$$ come into play)?

• It's very simple, a half-wavelength is a distance equal to half the wavelength. If the wavelength is 632 nm, then a half-wavelength is 316 nm. Dec 14 '20 at 6:07
• @ThePhoton But that doesn't answer my questions with regards to the context of electromagnetic waves. I'm looking for more detail within that context. Dec 14 '20 at 6:39
• There's nothing special about EM waves in explaining this term. If 1 meter is 100 cm, then a half meter is 50 cm. If one wavelength is 632 nm, then a half wavelength is 316 nm. It's represented mathematically by $\lambda/2$, just like you used it in your post. Dec 14 '20 at 6:41
• They're being over-specific. The wavelength would be the same if you measured the magnetic field. The diagram shows a case where 6 wavelengths (i.e. 12 half-wavelengths) fit in the cavity (labeled m=12) and a case where 5 wavelengths (10 half-wavelengths) fit labeled m=10, and a case where 4 wavelengths (8 half-wavelengths) fit, labeled m=8. In each case the number m is equal to the number of half-wavelengths in the cavity. Dec 14 '20 at 6:47
• Presumably n is the index of refraction of the material in the cavity and $\lambda$ is the vacuum wavelength, so that the actual wavelength in the material is $\lambda/n$. Dec 14 '20 at 6:48