# If most types of practical lasers produce a large number of half-wavelengths, then how are they practical for use at all?

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.

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.

If most types of practical lasers produce such a large number of half-wavelengths, then how are they practical for use at all, since we usually require single wavelengths for applications?

• there is no such a thing as single wavelength oscillation, all oscillators quantum and otherwise have a finite bandwidth, so does a laser and so does an LC resonator, because ultimately if for no other reason there is no way to extract energy without actually extracting and thereby having some amount of dissipation in the resonator. Dec 20 '20 at 20:41

The paragraph you cited only states that the cavity length $$L$$ is an integer multiple of half the wavelength $$\tfrac{\lambda}{2}$$. This condition is usually fulfilled for many wavelengths $$\lambda_i$$ with different intergers $$m_i$$. This is why usually there are other mode-selective elements inside the laser cavity. This could for example be a birefringent element, which rotates the polarization of the light dependent on its wavelength. Light of the wrong polarization can then be filtered out.