Question

In Fig. 35-38, sourcesand emit long-range radio waves of wavelength$\mathbf{400}\mathbf{}\mathbf{m}$ , with the phase of the emission from ahead of that from source Bby $\mathbf{90}\mathbf{°}$ .The distance ${\mathit{r}}_{\mathbf{A}}$ from Ato detector Dis greater than the corresponding distance localid="1663043743889" ${\mathit{r}}_{\mathbf{B}}$by $\mathbf{100}\mathbf{}\mathbf{m}$ .What is the phase difference of the waves at D ?

Short answer

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

Step-by-step solution

Given data

The wavelength of the radio waves is $\lambda =400\text{m}$.

The path difference of the two waves to the detector is $x=100\text{m}$.

The initial phase difference of the two waves is ${\varphi }_0=90°$.

Relation between path difference and phase difference

The phase difference of two waves of a wavelength $\lambda$ having path difference x is

$\varphi =\frac{2\pi }{\lambda }x$ ..... (1)

Determining the phase difference at the detector

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

$\begin{array}{rcl}\varphi & =& \frac{2\pi }{\text{400}\text{m}}\times \text{100}\text{m}\\ & =& \frac{\pi }{2}\\ & =& 90°\end{array}$

Thus, the net phase difference at the detector is

$\begin{array}{rcl}{\varphi }_D& =& {\varphi }_0+\varphi \\ & =& 90°+90°\\ & =& 180°\end{array}$

Hence, the required phase difference is $180°$.