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 130 m. What is the magnitude of the phase difference at the detector?

Answer 1

`X=165 \textdegree`

Step-by-step explanation:

From the question we are told that:

Wavelength
`\lambda =360m`

Angle
`\theta=90 \textdegree`

Distance variation
`d=130m`

Generally the equation for phase difference X is mathematically given by

`X=\theta+(d)/(\lambda)*100`

`X=90+(150)/(360)*100`

`X=165 \textdegree`

Therefore Phase difference X is given as

`X=165 \textdegree`