Question

Two radio waves are used in the operation of a cellular telephone. To receive a call, the phone detects the wave emitted at one frequency by the transmitting station or base unit. To send your message to the base unit, your phone emits its own wave at a different frequency. The difference between these two frequencies is fixed for all channels of cell phone operation. Suppose the wavelength of the wave emitted by the base unit is 0.34394 m and the wavelength of the wave emitted by the phone is 0.36140 m. Using a value of 2.9979 108 m/s for the speed of light, determine the difference between the two frequencies used in the operation of a cell phone.

Answer 1

Answer:

$4.2111\cdot 10^7 Hz$

Explanation:

The frequency of an electromagnetic wave is given by:

$f=\frac{c}{\lambda}$

where

$c=2.9979\cdot 10^8 m/s$ is the speed of light

$\lambda$ is the wavelength of the wave

For the first wave,

$\lambda=0.34394 m$

so the corresponding frequency is

$f=\frac{2.9979\cdot 10^8 m/s}{0.34394 m}=8.71635\cdot 10^8 Hz$

For the second wave,

$\lambda=0.36140 m$

so the corresponding frequency is

$f=\frac{2.9979\cdot 10^8 m/s}{0.36140 m}=8.29524\cdot 10^8 Hz$

So the difference between the two frequencies is

$\Delta f=8.71635\cdot 10^8 Hz-8.29524\cdot 10^8 Hz=4.2111\cdot 10^7 Hz$