Question

Waves from a radio station have a wavelength of 220 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

Answer 1

Answer:

$x_{min} = 55$

Given:

wavelength, $\lambda$ = 220 m

distance from reciever = 20 km

Solution:

To calculate the path difference in case of the destructive interference:

path difference, $\Delta x = (m + \frac{1}{2})\times \lambda$           (1)

where m = 0, 1, 2, 3.....

here, m = 0

Therefore, eqn (1) is reduced to:

 $\Delta x = (0 + \frac{1}{2})\times\lambda= \frac{\lambda }{2} = \frac{220}{2} = 110 m$

Now, the minimum distance $x_{min}$ from mountain to reciever to generate destructive interference at the reciever:

$x_{min} = \frac{\Delta x}{2} = \frac{110}{2}$ = 55 m