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 130 MHz . One microwave generator is set to emit microwaves with a wavelength of 1.250cm. If the second generator emits the longer wavelength, what is that wavelength? Express your answer to four significant figures and include the appropriate units.

Answer 1

Answer:

The wavelength of second generator is found to be 1.257 cm.

Explanation:

Let,

fb = beat frequency

f1 = frequency of given wave

f2 = frequency of second generator

λ1 = wavelength of given generator = 1.250 cm = 0.0125 m

λ2 = wavelength of second generator

since,

fb = |f1 - f2|

therefore,

fb = c/λ1 - c/λ2

1/λ2 = 1/λ1 - fb/c  

1/λ2 = 1/0.0125 m - (130,000,000 Hz)/(300,000,000 m/s) + 1/0.0125 m

λ2 = 1/79.566

λ2 = 0.01257 m = 1.257 cm

Answer 2

Answer:

The wavelength is 1.30 cm.

Explanation:

Given that,

Beat frequency = 130 MHz

Wavelength = 1.250 cm

We need to calculate the frequency

Using formula of frequency

$f=\dfrac{v}{\lambda}$

Put the value into the formula

$f=\dfrac{3\times10^{8}}{1.250\times10^{-2}}$

$f=2.4\times10^{10}\ Hz$

$f=24000\ MHz$

We need to calculate the frequency of beat produced by the two waves

The second generator emits the longer wavelength, so the frequency of second generator is less than that of the first generator

Using formula of beat frequency

$f_{b}=f_{1}-f_{2}$

Put the value into the formula

$130=24000-f_{2}$

$f_{2}=130-24000$

$f_{1}=23870\ MHz$

We need to calculate the wavelength

Using formula of wavelength

$\lambda=\dfrac{v}{f_{2}}$

$\lambda=\dfrac{3\times10^{8}}{2.3\times10^{10}}$

$\lambda=1.30\ cm$

Hence, The wavelength is 1.30 cm.