Question

Sources A and B emit long-range radio waves of wavelength 360 m, with the phase of the emission from A ahead of that from source B by 90°. The distance rA from A to a detector is greater than the corresponding distance rB from B by 150 m. What is the magnitude of the phase difference at the detector?

Answer 1

Answer:  The phase difference is 60° with source B ahead

Explanation: Considering the waves A and B emited with an initial phase difference of 90°  and also we must take into account the distance to the detector, in this case the distace rB is shorter 150 m than rA.

Considering the wavelength of 360 m a phase difference of 90° is equal to 90 m in distance,  so source A is ahead this path but the distance to the detector is larger so the difference that both can reach the detector is: 150m -90m = 60 m it is equivalent to 60° in phase each other wave of arriving to the detector. The wave B is ahead.

Answer 2

The magnitude of the phase difference at the detector is equal to -1.04 radians.

Given the following data:

Wavelength = 360 m.

Angle = 90°.

Distance variation = 150 m.

How to calculate the phase difference at the detector.

First of all, we would determine the phase difference that these sources at a distance of 150 meters create by emitting a long-range radio waves of wavelength 360 meters;

$\phi = \frac{150}{360} \times 360$

Phase difference = 150°.

Next, we would find the difference in the angles from the source:

X = 90 - 150

X = -60°.

In radians, we have:

X = -60 × π/180

X = -1.04 radians.

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