Question

Waves from a radio station have a wavelength of 369 m. They travel by two paths to a home receiver 20.0 km from the transmitter. One path is a direct path, and the second is by reflection from a mountain directly behind the home receiver. What is the minimum distance from the mountain to the receiver that produces destructive interference at the receiver? (Assume that no phase change occurs on reflection from the mountain.)

Answer 1

Answer:

92.25m

Explanation:

In order to solve the exercise, it is necessary to apply the concept of construtive interference due to a path difference.

The formula is given by,

$\delta = (m+\frac{1}{2})\frac{\lambda}{n}$

where,

n is the index of refraction of the medium in which the wave is traveling

$\lambda =$ wavelenght

$\delta =$ is the path difference

m = integer (0,1,2,3...)

Since in this case we are dealing with an atmospheric environment, where air is predominant, we approximate n to 1.

And since we need the reflected wave,

$\delta = 2x$

Where x is the distance in one direction without return.

The distance must correspond to the minimum therefore m = 0, so

$\delta = (m+\frac{1}{2})\frac{\lambda}{n}$

$\delta = ({0+\frac{1}{2})\frac{369}{1}$

$\delta = 184.5m$

Then the minimum distance is:

$x= \frac{delta}{2}$

$x = \frac{184.6}{2}$

$x = 92.25m$

Therefore the minimum distance from the mountain to the receiver that produces destructive interference at the receiver is 92.25m