Question

TV channel 2 broadcasts in the frequency range 51 to 57 MHz. What is the corresponding range of wavelengths? (Let us denote the minimum and maximum wavelengths by λmin and λmax, respectively.)

Answer 1

Step-by-step explanation:

The frequency range of the TV channel,
`f_1=51\ Mhz\ to\ f_2=57\ Mhz`

We need to find the corresponding range of wavelengths. The relation between the frequency and the wavelength is given by :

`\lambda=(c)/(f)`

For minimum frequency

`\lambda_(min)=(c)/(f_1)`

`\lambda_(min)=(3* 10^8)/(51* 10^6)`

`\lambda_(min)=5.88\ m`

For maximum frequency,

`\lambda_(max)=(c)/(f_2)`

`\lambda_(max)=(3* 10^8)/(57* 10^6)`

`\lambda_(max)=5.26\ m`

So, the corresponding wavelength range is from 5.88 m to 5.26 m. Hence, this is the required solution.