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In spatial sampling of a wave, a detector records a wave at positions $x_n=np$ separated by pitch $p>0$. Angular wavenumbers differing by $2\pi m/p$, with integers $m$, give identical samples of a complex Fourier mode:This is spatial aliasing. To distinguish arbitrary sinusoids within a prescribed band, require $|k|<\pi /p$, equivalently $|{\nu }_x|<1/(2p)$, the Nyquist spatial frequency. Thus $p<{\lambda }_{\min }/2$ for the shortest allowed wavelength. At equality a sine can vanish at every sample, so phase information can be lost. The Nyquist–Shannon sampling theorem states the precise reconstruction conditions for a bandlimited function; a finite scan introduces further limitations.

The synthetic two-tone snapshot iswhere both ${\nu }_j$ count cycles per metre. The component wavelengths are $0.50\mathrm{m}$ and $0.125\mathrm{m}$. With $p=0.05\mathrm{m}$, both lie below the Nyquist spatial frequency of $10$ cycles per metre. With $p=0.10\mathrm{m}$, the $8$-cycle component aliases to the $2$-cycle component: at every $x_n=n/10$, $\cos (2\pi \cdot 8x_n)=\cos (2\pi \cdot 2x_n)$. The sampled sum is therefore indistinguishable from $1.5\cos (4\pi x)$, although the continuous curves differ between samples. Drawing lines between sample points cannot resolve this ambiguity.

The wavelength $\lambda$ is the spatial period of a monochromatic wave, measured at a fixed time along the wavevector. For a plane wave with a sinusoid as its profile, it is the distance between consecutive wave crests, or between any consecutive points with the same wave phase modulo $2\pi$:Here $\mathbf{k}$ uses angular wavenumber, $f$ is ordinary frequency, and $v_p$ is phase speed. A distance between arbitrary zero crossings need not be a wavelength: consecutive zeros of a sinusoid are only $\lambda /2$ apart, with opposite slopes. In a standing wave, consecutive nodes of a standing wave are also $\lambda /2$ apart; $\lambda$ refers to the constituent travelling waves, even though the nodes of a standing wave do not travel.

Across a stationary interface between stationary linear media, the incident, reflected and transmitted monochromatic waves have the same frequency. The boundary conditions must hold at every time: terms oscillating with different angular frequencies cannot cancel throughout the entire time record. Each medium's dispersion relation then determines its own angular wavenumber and wavelength, soFor electromagnetic radiation in a transparent medium with refractive index $n$, $v_p=c_0/n$ and $\lambda ={\lambda }_0/n$, where $c_0$ is the vacuum speed of light. A synthetic example with ${\lambda }_0=600\mathrm{nm}$ and $n=1.5$ gives $\lambda =400\mathrm{nm}$ at the same frequency. This conservation argument assumes time-independent linear material response and a stationary interface in the chosen frame. A moving interface can produce a Doppler effect; time-dependent or nonlinear response can change or generate frequencies.^[ref]

A broad frequency spectrum generally supplies many component wavelengths, rather than one wavelength for an entire pulse. Even two tones have both their individual wavelengths and a different overall repetition distance: the $2$- and $8$-cycle-per-metre sum in spatial sampling of a wave repeats after $0.50\mathrm{m}$, but contains the finer $0.125\mathrm{m}$ structure. A narrow wave packet can have an approximate carrier wavelength; the wave envelope propagates approximately at group velocity, determined by the slope of the dispersion relation. Substituting that group velocity into $\lambda =v_p/f$ generally gives the wrong carrier wavelength. Where the medium varies slowly, local wavelength gives a more useful description.