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

Answer:

$X=165 \textdegree$

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

 $X=\theta+\frac{d}{\lambda}*100$

 $X=90+\frac{150}{360}*100$

 $X=165 \textdegree$

Therefore Phase difference X is given as

 $X=165 \textdegree$