Waves - A two-source interference problem

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The discussion focuses on calculating the angle to the first point of minimum signal (m=0) for radio waves transmitted from two synchronized towers. The relevant formula for destructive interference is dsin(theta) = (m + 1/2)λ, where d is the distance between the towers, and λ is the wavelength. Given the frequency of 1000 kilohertz corresponds to a wavelength of 300 meters, the calculation involves substituting these values into the formula. Participants note the similarity to Young's double slit experiment, emphasizing the need for the destructive interference approach. The conversation concludes with acknowledgment of the method and its relation to established physics concepts.
fatkat444
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Consider a road that runs parallel to the line connecting a pair of radio towers that transmit the same station (assume that their transmissions are synchronized), which has an AM frequency of 1000 kilohertz. If the road is 5 kilometers from the towers and the towers are separated by 400 meters, find the angle to the first point of minimum signal (m=0). Hint: A frequency of 1000 kilohertz corresponds to a wavelength of 300 meters for radio waves.
Express your answer in radians, to two significant figures.




Because it's constructive interference i think we use dsin(theta) = m x lambda



So that would be: 400 x sin(theta) = 0 x 300

Is this correct? And if so how would you go about working it out?
 
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I am guessing you have studied Youngs double slit experiment. This is very similar to that, but you have to use formula for destructive interference, since they have asked for the minima.

Consider the two towers as sources and the road as the screen in youngs DS Expt...
 
Raze2dust said:
I am guessing you have studied Youngs double slit experiment. This is very similar to that, but you have to use formula for destructive interference, since they have asked for the minima.

Consider the two towers as sources and the road as the screen in youngs DS Expt...

OK, Thanks alot.
 
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