Chapter 35: Q23P (page 1075)

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

Expert verified

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

Step by step solution

The wavelength of the radio waves is $位 = 400 m$ .

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

The initial phase difference of the two waves is $蠒_{0} = 90 掳$ .

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

$蠒 = \frac{2 蟺}{位} x$ ..... (1)

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

$\begin{array}{rcl} 蠒 & = & \frac{2 蟺}{400 m} 脳 100 m \\ = & \frac{蟺}{2} \\ = & 90 掳 \end{array}$

Thus, the net phase difference at the detector is

$\begin{array}{rcl} 蠒_{D} & = & 蠒_{0} + 蠒 \\ = & 90 掳 + 90 掳 \\ = & 180 掳 \end{array}$

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

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