Explanation:
The frequency range of the TV channel, [tex]f_1=51\ Mhz\ to\ f_2=57\ Mhz[/tex]
We need to find the corresponding range of wavelengths. The relation between the frequency and the wavelength is given by :
[tex]\lambda=\dfrac{c}{f}[/tex]
For minimum frequency
[tex]\lambda_{min}=\dfrac{c}{f_1}[/tex]
[tex]\lambda_{min}=\dfrac{3\times 10^8}{51\times 10^6}[/tex]
[tex]\lambda_{min}=5.88\ m[/tex]
For maximum frequency,
[tex]\lambda_{max}=\dfrac{c}{f_2}[/tex]
[tex]\lambda_{max}=\dfrac{3\times 10^8}{57\times 10^6}[/tex]
[tex]\lambda_{max}=5.26\ m[/tex]
So, the corresponding wavelength range is from 5.88 m to 5.26 m. Hence, this is the required solution.
