Answer :
- The distance from source A at which there is constructive interference between points A and B is 3.50 meters.
- There will be two separate interference fringes between points A and B, with the first fringe occurring at 3.00 meters and the second fringe occurring at 6.00 meters.
The distance from source A at which there is constructive interference between points A and B can be found by setting the phase difference between the waves from sources A and B to a multiple of 2π. The phase difference between the waves at a given point is given by:
To determine the distance from source A at which there is constructive interference between points A and B, we can use the formula for the constructive interference of two waves:
d = (m + 1/2) *λ / 2
where d is the distance from source A, m is an integer, lambda is the wavelength of the radio waves, and the term (m + 1/2) is the phase difference between the two waves.
Plugging in the values given in the problem, we get:
d = (m + 1/2) * 6.00 meters / 2
= (m + 0.5) * 3.00 meters
Since we want constructive interference, the phase difference must be an integer multiple of 2π, which means the value of m must be an even integer. Therefore, we can set m = 2 to get:
d = (2 + 0.5) * 3.00 meters
= 3.50 meters
So the distance from source A at which there is constructive interference between points A and B is 3.50 meters.
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