In Fig. 35-38, sourcesand emit long-range radio waves of wavelength400m , with the phase of the emission from ahead of that from source Bby 90° .The distance rA from Ato detector Dis greater than the corresponding distance localid="1663043743889" rBby 100m .What is the phase difference of the waves at D ?

Short Answer

Expert verified

The two waves have a 180° phase difference at the detector.

Step by step solution

01

Given data

The wavelength of the radio waves is λ=400m.

The path difference of the two waves to the detector is x=100m.

The initial phase difference of the two waves is ϕ0=90°.

02

Relation between path difference and phase difference

The phase difference of two waves of a wavelength λ having path difference x is

ϕ=2πλx ..... (1)

03

Determining the phase difference at the detector

From equation (I) the phase difference introduced in the path to the detector is,

ϕ=2π400m×100m=π2=90°

Thus, the net phase difference at the detector is

ϕD=ϕ0+ϕ=90°+90°=180°

Hence, the required phase difference is 180°.

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