Question

Two microwave signals of nearly equal wavelengths can generate a beat frequency if both are directed onto the same microwave detector. In an experiment, the beat frequency is 140 MHz . One microwave generator is set to emit microwaves with a wavelength of 1.500cm. If the second generator emits the longer wavelength, what is that wavelength?

Answer 1

Answer:

1.5106 cm

Explanation:

The beat frequency is equal to the absolute value of the difference between the frequencies of the two signals:

$f_B = |f_1 - f_2|$

using the wave equation, we can re-write each frequency as

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

where c is the speed of light and $\lambda$ is the wavelength. Therefore,

$f_B = |\frac{c}{\lambda_1}-\frac{c}{\lambda_2}|$

where:

$f_B = 140 MHz = 140\cdot 10^6 Hz$ is the beat frequency

$\lambda_1 = 1.50 cm = 0.015 m$ is the wavelength of the first generator

$\lambda_2$ is the wavelength of the second generator

We also know that the second generator emits the longer wavelength, so we already know that the term inside the module is positive. Therefore, we can now solve for $\lambda_2$:

$f_B = c(\frac{1}{\lambda_1}-\frac{1}{\lambda_2})\\\lambda_2=(\frac{1}{\lambda_1}-\frac{f_B}{c})^{-1}=(\frac{1}{0.015}-\frac{140\cdot 10^6}{3\cdot 10^8})^{-1}=0.015106 m = 1.5106 cm$