91Ó°ÊÓ

The distance between the two antennas d = 600m

The frequency of the signal f= 3.0MHz

The speed of the signal is c (3x10⁸m/s)

The signals are out of phase, thus the phase difference between the signals is $\mathrm{Ï€}$

The wavelength of the signal is expressed as

$$\begin{gathered} \mathrm{λ}=\frac{c}{f}=\frac{3×{10}^{8}m/sec}{3×{10}^{6}se{c}^{-1}} \\ \mathrm{λ}=100m \end{gathered}$$

Thus, the wavelength of the signal is 100m.

The following diagram shows the given situation

The distance AB = 600m

The point C is (300m, 800m)

Thus, the vertical distance BC= 800m

From Pythagoras theorem, AC is

$$\begin{gathered} AC=\sqrt{{600}^2+{800}^2} \\ AC=1000m \end{gathered}$$

Thus, the distance of point C from the radio signal A is 1000m

The path difference of the signal at point C is

$$\begin{gathered} ∆x=AC-BC \\ ∆x=(1000-800)m \\ ∆x=200m \end{gathered}$$
The path difference is 200m

If the interference is destructive at point C

The condition satisfied by the path difference is given

role="math" localid="1649934716581" $∆x=\left(m+\frac{1}{2}\right)\mathrm{λ}$

here, m is an integer. Now, substitute the values in the above relation to determine m

$$\begin{gathered} 200=\left(m+\frac{1}{2}\right)100 \\ m=2-\frac{1}{2}=1.5(notpossible) \end{gathered}$$
Since the value of m is always an integer, thus at point C, there is no destructive interference.

If the interference is constructive at point C

The condition satisfied by the path difference is given

$$\begin{gathered} ∆x=\left(m\right)\mathrm{λ} \\ 200=m\left(100\right) \\ m=2 \end{gathered}$$
Here, the value of m is an integer, therefore, at point C the constructive interference of second order is formed