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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 θ 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.

Respuesta :

Answer:

θ = 0.384 rad = 22 degrees

Explanation:

Given:

- The wavelength of the source λ = 300 meters

- The distance between towers d = 400 m

- The order of minimum signal m = 0

Find:

find the angle θ to the first point of minimum signal (m=0)

Solution:

- The given problem is analogous to Young's double slit experiment where the split through which the waves are initiated are towers with slit separation d = 400 m ( i.e distance between towers ).

- For the minimum signal, we are asked for destructive interference for which the formula is given as:

                                  d*sin(θ) = ( m + 0.5 )*λ

For m = 0,

                                  sin(θ) = 0.5*λ / d

                                  sin(θ) = 0.5*300 / 400

                                  θ = arc sin ( 3/8)

                                  θ = 0.384 rad = 22 degrees

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