Chapter 27, Problem 67A

Linsey’s favorite radio station broadcasts at a frequency f.

a. Write an equation for the wavelength of these radio waves.

b. What is the ratio of ceiling height h to wavelength $\lambda$ for this wave?

c. Calculate the answers for parts (a) and (b) above for a frequency of 97.3 MHz and a ceiling height of 2.4 meters,

To determine

The equation for the wavelength.

Explanation of Solution

Introduction:

The radio waves are the electromagnetic waves which travel with the speed of the light.

The speed of the wave is represented as the product of the wavelength and frequency of the light.

  $v_{wave}=\lambda \times f$

Therefore,

  $\lambda =\frac{v_{wave}}{f}$

Conclusion:

The equation for the wavelength is $\lambda =\frac{v_{wave}}{f}$ .

To determine

The ratio of ceiling height to the wavelength.

Explanation of Solution

Introduction:

The speed of the wave is given as

  $v_{wave}=\lambda \times f$

  $\begin{array}{l}{v}_{wave}=\lambda \times f\\ \frac{\text{h}}{\text{λ}}\text{=}\frac{f\times h}{v_{wave}}\end{array}$

To determine

The wavelength and ratio of ceiling height and wavelength.

Answer to Problem 67A

The wavelength of the wave is 3.1m.

The ratio of ceiling height to the wavelength is $\text{0}\text{.8}$ .

Explanation of Solution

Given:

Frequency of wave, $f=97.3\times 10^{6}m$

The ceiling height, $h=2.4\text{ m}$

Formula used:

  $v_{wave}=\lambda \times f$

  $\begin{array}{l}\lambda =\frac{v_{wave}}{f}\\ \lambda =\frac{c}{f}\end{array}$

Calculation:

Radio waves travel with the speed of light, so the wavelength of the wave is

  $\begin{array}{l}\lambda =\frac{c}{f}\\ \lambda =\frac{3\times 10^8}{97.3\times 10^6}\\ \lambda =3.1m\end{array}$

And the ratio of ceiling height to the wavelength is

  $\begin{array}{l}\frac{h}{\lambda }=\frac{2.4}{3.1}\\ =0.8\end{array}$

Conclusion:

The wavelength is 3.1 meters and their ratio is $\text{0}\text{.8}$ .