For sure the frequency of the conducted wave (after absorption intoI will consider an antenna receiving signal from a propagating electromagnetic field, though the reverse, an antenna) emitting a propagating electromagnetic field will matchjust be the frequency ofreverse.
When a propagating EM field impinges upon an antenna the radiated waveelectric field (before absorption intowhich is perpendicular to the antennapropagation direction). The speed of light for puts forces on the conducted wave is within a factor of order unity fromelectrons in the speed of light forantenna and drives them at the radiated wave (probably something like 2/3 depending onsame frequency as the propertiesfrequency of the receiver)incident EM field. This means that the electrons oscillate at the same frequency as the incident field.
So this means after absorptionNow, the conductor wave will have a similar wavelength (maybe slightly longer) thanantenna, plus the radiated wave. The receiver electronics havecable plugged into the lowest impedanceantenna, act as a waveguide for a conductednew guided wave with. This waveguide has a wavelength matched tocertain frequency/wavelength dependent impedance. It turns out, that offor some standard antennas, the antenna size. Becauseimpedance exhibits resonances (local minima) at wavelengths corresponding to approximately $$ \lambda_{\text{guided}} = 2L/n $$ where $\lambda_{\text{guided}}$ is the frequencywavelength of the conductorguided wave, $L$ is related to the wavelengthlength of the conducted waveantenna, and because$n$ is a positive integer.* Given other types of resonant cavities we're familiar with in physics, it is not so surprising that there is a simple relationship between the frequencylength of the conductedantenna and the resonant wavelengths for the guided wave. This relationship is equal toexplained well in the frequencyother answers.
However, the above (and many of the radiated wave, and becauseother answers) doesn't address the frequencyheart the OPs question which is "why does the wavelength of the radiatedpropagating wave isin $z$ related to the wavelength of the radiatedcaptured wave, we find there in $x$?" The answer is a relationship betweenthat these two wavelengths are related by the size ofcorresponding frequencies. We have
\begin{align} f_{\text{propagating}} =& c_{\text{air}}/\lambda_{\text{propagating}}\\ f_{\text{guided}} =& c_{\text{waveguide}}/\lambda_{\text{guided}}.\\ \end{align} But, recall above, that I explained that, because the antenna andelectrons are directly driven by the wavelengthpropagating wave, that $$ f_{\text{guided}} = f_{\text{propagating}}, $$ so from this we can quickly conclude that \begin{align} \lambda_{\text{guided}} = \frac{c_{\text{waveguide}}}{c_{\text{air}}} \lambda_{\text{propagating}} = r \lambda_{\text{propagating}}. \end{align} And we have resonances when $$ \lambda_{\text{propagating}} = \frac{1}{r} \frac{2L}{n}. $$ Based on properties of the radiated waveantenna/receiver we probably have $r\approx 0.6$, so a factor of order unity.
Note: AboveSo we can see that
- the resonant wavelengths of the guided wave are related to the length of the antenna because the antenna is like a cavity for the guided wave
- the frequencies of the guided and propagating waves are equal because the guided wave is directly driven by the propagating wave
- the wavelength of the propagating/guided wave is directly related to the frequency of the propagating/guided wave by the corresponding speed of light
- Therefore the resonant wavelengths of the propagating wave are simply related to the lengths of the antenna.
I callthink it is an interesting observation the electromagneticOP makes about the direction of energy propagation being possibly redirected when the propagating wave is captured by the antenna and the receiver device the "conducted" wave. It would perhaps be more appropriateBut, of course, this is exactly what a waveguide does. More generally, when converting a propagating wave to call it the "guided" or "captured"a guided wave we need to satisfy certain phase/momentum matching conditions. At frequencies where anIn the case of e.g. a dipole antenna being used itthis is likely we can't analyzebest satisfied when the circuit using a simple lumped circuit DC analysisantenna is parallel with the electric field. Instead we think
*Very possibly I'm off by a factor of two or something here or have $n$ in the antenna and cables as serving as waveguides for electromagnetic waves travelling throughoutwrong spot in the circuitsequation. If so let me know in the comments and I'll correct it.