TV-reception antennas for VHF are constructed with cross wires supported at their centers, as shown in Figure \({\rm{24}}{\rm{.27}}\). The ideal length for the cross wires is one-half the wavelength to be received, with the more expensive antennas having one for each channel. Suppose you measure the lengths of the wires for particular channels and find them to be \({\rm{1}}{\rm{.94}}\) and \({\rm{0}}{\rm{.753\;m}}\) long, respectively. What are the frequencies for these channels?

Wavelength: The distance between identical points (adjacent crests) in adjacent cycles of a waveform signal propagated in space or along a wire is defined as the wavelength. This length is typically specified in wireless systems in meters (m), centimeters (cm), or millimeters (mm).

Considering the given information,

The length of wire 1 is${\mathrm{L}}_1=1.94\mathrm{m}$

The length of wire 2 is ${\mathrm{L}}_2=0.753\mathrm{m}$

Apply the formula,

The relationship between frequency and wavelength is as follows:

$\mathrm{f}=\frac{\mathrm{c}}{\mathrm{\lambda }}$

The wavelength is since the optimal length of the cross wire is half the

wavelength.

$\lambda =2\mathrm{L}$

The frequency for wire 1's channel is,

$$\begin{gathered} {\mathrm{f}}_1=\frac{\mathrm{c}}{2{\mathrm{L}}_1} \\ =\frac{3\times {10}^8\mathrm{m}/\mathrm{s}}{2\times 1.94\mathrm{m}} \\ =7.732\times {10}^7\mathrm{Hz}\left(\frac{1\mathrm{MHz}}{{10}^6\mathrm{Hz}}\right) \\ =77.32\mathrm{MHz} \end{gathered}$$

The frequency for wire 2's channel is,

$$\begin{gathered} {\mathrm{f}}_2=\frac{\mathrm{c}}{2{\mathrm{L}}_2} \\ =\frac{3\times {10}^8\mathrm{m}/\mathrm{s}}{2\times 0.753\mathrm{m}} \\ =1.992\times {10}^8\mathrm{Hz}\left(\frac{1\mathrm{MHz}}{{10}^6\mathrm{Hz}}\right) \\ =199.2\mathrm{MHz} \end{gathered}$$

Therefore, the frequency of the channel for the length of wire $1.94\mathrm{m}$is $77.32\mathrm{MHz}$and the frequency of the channel for the length of wire$0.753\mathrm{m}$ is $1.992\mathrm{MHz}$.