Waves - A two-source interference problem

In summary, the problem involves finding the angle to the first point of minimum signal (m=0) for a road that runs parallel to a pair of synchronized radio towers with an AM frequency of 1000 kilohertz, separated by 400 meters. Using the formula for destructive interference, dsin(theta) = m x lambda, and considering the two towers as sources and the road as the screen in Young's double slit experiment, the answer can be expressed in radians, to two significant figures.
  • #1
fatkat444
4
0
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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  • #2
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...
 
  • #3
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.
 

FAQ: Waves - A two-source interference problem

What is the concept of two-source interference in waves?

The concept of two-source interference in waves refers to the phenomenon where two waves with the same frequency and amplitude intersect and create a new wave with a different amplitude. This is caused by the constructive and destructive interference of the two waves.

How do you calculate the resulting amplitude in a two-source interference problem?

The resulting amplitude in a two-source interference problem can be calculated by adding the amplitudes of the two individual waves and taking into account the phase difference between them. The resulting amplitude can be greater or smaller than the individual amplitudes, depending on the phase difference.

What is the difference between constructive and destructive interference in two-source interference?

Constructive interference occurs when the two waves have a phase difference of 0 degrees or a multiple of 360 degrees, resulting in a larger amplitude in the new wave. Destructive interference, on the other hand, occurs when the two waves have a phase difference of 180 degrees or a multiple of 360 degrees, resulting in a smaller amplitude or cancellation of the new wave.

How does the distance between the two sources affect the interference pattern in waves?

The distance between the two sources plays a crucial role in the interference pattern of waves. As the distance increases, the interference pattern becomes more spread out, resulting in a decrease in the amplitude of the new wave. In contrast, as the distance decreases, the interference pattern becomes more concentrated, resulting in an increase in the amplitude of the new wave.

Can two waves with different frequencies interfere with each other?

No, two waves with different frequencies cannot interfere with each other in a two-source interference problem. In order for interference to occur, the two waves must have the same frequency and amplitude. If the frequencies are different, the waves will not have a constant phase difference and will not create an interference pattern.

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