Answer
a. From $\lambda=2.78m$ to $\lambda=3.41m$
b. From $\lambda=180m$ to $\lambda=561m$
Work Step by Step
Use equation 22–4, $c=\lambda f$.
a. Calculate the range of FM radio wavelengths. The highest frequency has the shortest wavelength.
$$\lambda=\frac{c}{f}=\frac{3.00\times10^8m/s }{1.08\times10^8Hz}$$
$$\lambda=2.78m$$
$$\lambda=\frac{c}{f}=\frac{3.00\times10^8m/s }{8.8\times10^7Hz}$$
$$\lambda=3.41m$$
b. Calculate the range of AM radio wavelengths. The highest frequency has the shortest wavelength.
$$\lambda=\frac{c}{f}=\frac{3.00\times10^8m/s }{1.7\times10^6Hz}$$
$$\lambda=180m$$
$$\lambda=\frac{c}{f}=\frac{3.00\times10^8m/s }{5.35\times10^5Hz}$$
$$\lambda=561m$$
