Chapter 22, Problem 30P

To determine

The wavelength for two TV channels.

Answer to Problem 30P

The wavelength of channel $\left(2\right)$ is $5.56m$ .

The wavelength of channel $\left(69\right)$ is $0.372m$ .

Explanation of Solution

Given info:

Frequency of channel $\left(2\right)$ , $f_2=54.0MHz=54.0\times {10}^{6}Hz$

Frequency of channel $\left(69\right)$ , $f_{69}=806MHz=806\times {10}^{6}Hz$

Factor, $c=3.00\times {10}^{8}m/s$

Formula used:

The relation between frequency and wavelength: $f=\frac{c}{\lambda }$

Where,

  $f$ is frequency

  $\lambda$ is wavelength

  $c$ is a factor

Calculation:

Rearrange the relation between frequency and wavelength: $f=\frac{c}{\lambda }$ . So, $\lambda =\frac{c}{f}$

Calculation for wavelength of channel $\left(2\right)$

Substitute the given values in formula, ${\lambda }_2=\frac{c}{f_2}$

  ${\lambda }_2=\frac{3\times {10}^{8}m/s}{54.0\times {10}^{6}Hz}=5.56m$

Calculation for wavelength of channel $\left(69\right)$

Substitute the given values in formula, ${\lambda }_{69}=\frac{c}{f_{69}}$

  ${\lambda }_{69}=\frac{3\times {10}^{8}m/s}{806\times {10}^{6}Hz}=0.372m$

Conclusion: The wavelength of channel $\left(2\right)$ is $5.56m$ and wavelength of channel $\left(69\right)$ is $0.372m$