In a certain experiment, a radio transmitter emits sinusoidal electromagnetic waves of frequency 110.0 MHz in opposite directions inside a narrow cavity with reflectors at both ends, causing a standing-wave pattern to occur. (a) How far apart are the nodal planes of the magnetic field? (b) If the standing-wave pattern is determined to be in its eighth harmonic, how long is the cavity?

The electromagnetic wave travels with the speed of light and as the frequency f of the wave is related to the wavelength and the speed as$\mathbf{c}\mathbf{=}\mathbf{f\lambda }$Thus,$\mathbf{\lambda }\mathbf{=}\frac{\mathbf{c}}{\mathbf{f}}$

The frequency is given by f =$110\times {10}^6\mathrm{Hz}$ The frequency of the electromagnetic waves doesn't change with changing the medium while the wavelength changes as the speed of the light changes by the medium. The wavelength is

${\mathrm{\lambda }}_{\mathrm{air}}=\frac{\mathrm{c}}{\mathrm{f}}=\frac{3\times {10}^8\mathrm{m}/\mathrm{s}}{110\times {10}^6\mathrm{Hz}}=2.73\mathrm{m}$

The nodal planes of the magnetic field are apart with distance. Plug the values for to get l

$\mathrm{l}=\frac{\mathrm{\lambda }}{2}=\frac{2.73}{2}=1.36\mathrm{m}$

Therefore,${\mathrm{\lambda }}_{\mathrm{air}}=2.73\mathrm{m}$

For eight antinodes planes of the electric field, use equation$\mathrm{x}=\frac{8}{2}\mathrm{\lambda }$ to get the distance between adjacent nodal planes x which represents the length of the cavity. Substitute the values we have,

$\mathrm{x}=\frac{8}{2}(2.73\mathrm{m})=10.92\mathrm{m}$

Therefore, the distance between adjacent nodal planes x is 10.92m