Question

One radio transmitter A operating at 60.0 MHz is 10.0 m from another similar transmitter B that is $180^\circ$ out of phase with transmitter A. How far must an observer move from transmitter A toward transmitter B along the line connecting A and B to reach the nearest point where the two beams are in phase?

• A. $3I_0/7$

Answer 1

1.25 m

Answer 2

The waves from the two transmitters are in phase at a point if the total phase difference is an integer multiple of $2\pi$. The total phase difference is the sum of the phase difference due to the path difference and the initial phase difference between the sources. The initial phase difference is $\pi$. The phase difference due to path difference $r_B - r_A$ is $k(r_B - r_A) = \frac{2\pi}{\lambda}(r_B - r_A)$. The condition for constructive interference is $\frac{2\pi}{\lambda}(r_B - r_A) + \pi = 2\pi m$, which simplifies to $r_B - r_A = (2m-1)\frac{\lambda}{2}$. Let the observer be at distance $x$ from A, so $r_A = x$ and $r_B = 10-x$. Substituting these and $\lambda=5.0$ m, we get $10-2x = (2m-1)2.5$, which gives $x = 5 - (2m-1)1.25$. We seek the smallest non-negative value of $x$ in the range $[0, 10]$. Testing integer values of $m$, we find that $m=2$ gives $x = 5 - (3)1.25 = 5 - 3.75 = 1.25$ m, which is the smallest positive value in the range.