Waves from a radio station have a wavelength of 308 m. They travel by two paths

Question

Waves from a radio station have a wavelength of 308 m. They travel by two paths to a home receiver 20.6 km from the transmitter. One path is a direct path, and the other 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.)

I'm finding it hard to conceptualize this- our professor gave us the equations for detructive interference:

dsin(theta)=(m +1/2)(lambda)

or

d(y/L)=(m +1/2)(lambda)

If you could include a diagram that would be very helpful!Thanks!

Explanation / Answer

An extra full wavelength would add up and produce Constructive Interference.

For Destructive Interference one path needs to be 1/2 wavelength longer.

But since the mountain to receiver would add 2 X the distance to the mountain it looks like the mountain would have to be 1/4 wavelength from the receiver

So, the minimum distance required from the mountain to the receiver = 308 / 4 = 77 m

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