Question

You are listening to the FM radio in your car. As you come to a stop at a traffic light, you notice that the radio signal is fuzzy. By pulling up a short distance, you can make the reception clear again. In this problem, we work through a simple model of what is happening.

Our model is that the radio waves are taking two paths to your radio antenna:

• the direct route from the transmitter

• an indirect route via reflection off a building

Because the two paths have different lengths, they can constructively or destructively interfere. Assume that the transmitter is very far away and that the building is at an a45-degree angle from the path to the transmitter.(Intro 1figure)

Point A in the figure is where you originally stopped, and point B is where the station is completely clear again. Finally, assume that the signal is at its worst at point A and its clearest at point B.

Part AWhat is the distance d between points A and B? Express your answer in wavelengths, as a fraction.


Part BYour FM station has a frequency of 100 megahertz. The speed of light is about 3.00×10⁸ meters per second. What is the distance d between points A and B? Express your answer in meters to two significant figures.

Answer 1

Concepts and reason

The concept of constructive and destructive interference is required to solve the problem.

Initially, in the first part determine the path lengths for constructive and destructive interference. Then, determine the expression for the distance between point A and point B by subtracting the path lengths for destructive and constructive interference.

In the next part, re-write the wavelength in terms of frequency and speed of the wave using relation $f = \frac{c}{\lambda }$. Substitute the values in the re-arranged expression of path difference to calculate the distance between point A and B.

Fundamentals

The path length for constructive interference is given as,

${x_{\rm{c}}} = n\lambda$

Here, ${x_{\rm{c}}}$is the constructive path length, $\lambda$ is the wavelength of light and n is the order of fringe.

The path length for destructive interference is given as,

${x_{\rm{d}}} = \left( {n + \frac{1}{2}} \right)\lambda$

Here, ${x_{\rm{d}}}$is the destructive path length, $\lambda$ is the wavelength of light and n is the order of fringe.

The wavelength of is determined by using the relation,

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

Here, $\lambda$is the wavelength, c is the speed of light and f is the frequency of the signal.

(A)

Consider the waves at point A interfere destructively and the waves at point B interfere constructively.

The waves at point A interfere destructively. The path length for point A is given as,

${x_{\rm{A}}} = \left( {n + \frac{1}{2}} \right)\lambda$

The waves at point B interfere constructively. The path length for point B is given as,

${x_{\rm{B}}} = n\lambda$

The distance between point A and point B is given as,

$\Delta x = {x_{\rm{A}}} - {x_{\rm{B}}}$

Substitute $\left( {n + \frac{1}{2}} \right)\lambda$ for ${x_{\rm{A}}}$ and $n\lambda$ for ${x_{\rm{B}}}$in the equation $\Delta x = {x_{\rm{A}}} - {x_{\rm{B}}}$.

$\begin{array}{c}\\\Delta x = \left( {n + \frac{1}{2}} \right)\lambda - n\lambda \\\\ = \frac{\lambda }{2}\\\end{array}$

(B)

Calculate the distance between points A and B.

The distance between points A and B is given as,

$\Delta x = \frac{\lambda }{2}$

Substitute $\frac{c}{f}$ for $\lambda$ in the equation $\Delta x = \frac{\lambda }{2}$.

$\Delta x = \frac{c}{{2f}}$

Substitute 100 MHz for f and $3.00 \times {10^8}{\rm{ m/s}}$ for c in the equation $\Delta x = \frac{c}{{2f}}$.

$\begin{array}{c}\\\Delta x = \frac{{3.00 \times {{10}^8}{\rm{ m/s}}}}{{2\left( {100{\rm{ MHz}}\left( {\frac{{{{10}^6}{\rm{ Hz}}}}{{1{\rm{ MHz}}}}} \right)} \right)}}\\\\ = 1.5{\rm{ m}}\\\end{array}$

Ans: Part A

The distance between the points A and B in terms of wavelength is $\frac{\lambda }{2}$.

Part B

The distance between the points A and B is 1.5 m.