Question

In a certain experiment, a radio transmitter emits sinusoidal electromagnetic waves of fre- quency 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?

Answer 1

To solve the problem it is necessary to take into account the concepts related to frequency depending on the wavelength and the speed of light.

By definition we know that the frequency is equivalent to,

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

where,

c= Speed of light

$\lambda = Wavelength$

While the wavelength is equal to,

$\lambda = \frac{2L}{n}$

Where,

L = Length

n = Number of antinodes/nodes

PART A) For the first part we have that our wavelength is 110MHz, therefore

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

$\lambda = \frac{3*10^8}{11*10^6}$

$\lambda = 1.36m$

Therefore the distance between the nodal planes is 1.36m

PART B) For this part we need to find the Length through the number of nodes (8) and the wavelength, that is,

$\lambda'=\frac{2L}{n}$

$L = \frac{\lambda n}{2}$

$L = \frac{8*2.72}{2}$

$L = 10.90m$

Therefore the length of the cavity is 10.90m