# A radio transmitting station operating at a frequency of 170 MHz has two identical antennas that...

## Question:

A radio transmitting station operating at a frequency of 170 MHz has two identical antennas that radiate in phase. Antenna B is 7.20 m to the right of antenna A. Consider point P between the antennas and along the line connecting them, a horizontal distance x to the right of antenna A. For what values of x will constructive interference occur at point P?

## Interference:

If two waves of the same frequency and with the constant phase difference are interacting in the space, the interference pattern is produced, which consists of the series of maxima and minima, corresponding to the points of constructive and destructive interference. The constructive interference occurs when the path length difference between the point of observation and the two sources equals the integer number of waves. If that difference is equal to the odd number of half-wavelengths, the interference is a destructive one.

## Answer and Explanation:

The condition for the constructive interference is that the difference in distance between the point and two antennas is the integer number of wavelengths. If the point P is distance *x* from the antenna A, we obtain:

{eq}d - 2x = m \lambda {/eq}

Here

- {eq}d = 7.2 \ m {/eq} is the distance between two antennas;

- {eq}m {/eq} is the order of maximum;

- {eq}\lambda = \dfrac {c}{f} {/eq} is the wavelength of radiation;

{eq}c = 3\times 10^8 \ m/s {/eq} is the speed of light;

- {eq}f = 1.7\times 10^8 \ Hz {/eq} is the frequency of radiation;

Solving for the distance *x*, we get:

{eq}x = \dfrac {d - m\dfrac {c}{f}}{2} {/eq}

The numerator must be positive, which determines the number of possible maxima between two antennas;

Calculating, we get:

{eq}m < \dfrac {df}{c} = \dfrac {7.2 \ m \cdot 1.7\times 10^8 \ Hz}{3 \times 10^8 \ m/s} = 4.08 {/eq}

Therefore. we will have 4 maxima and four corresponding distances from the antenna A:

{eq}\begin{cases} x_1 = \dfrac {df - c}{2f} = \dfrac {7.2 \ m \cdot 1.7\times 10^8 \ Hz - 3\times 10^8 \ m/s}{2\cdot 1.7\times 10^8 \ Hz} \approx \boxed {\color{green}{2.72 \ m}} \\ x_2 = \dfrac {df - 2c}{2f} = \dfrac {7.2 \ m \cdot 1.7\times 10^8 \ Hz - 2\cdot 3\times 10^8 \ m/s}{2\cdot 1.7\times 10^8 \ Hz} \approx \boxed {\color{orange}{1.84 \ m}} \\ x_3 = \dfrac {df - 3c}{2f} = \dfrac {7.2 \ m \cdot 1.7\times 10^8 \ Hz - 3\cdot 3\times 10^8 \ m/s}{2\cdot 1.7\times 10^8 \ Hz} \approx \boxed {\color{blue}{0.95 \ m}} \\ x_4 = \dfrac {df - 4c}{2f} = \dfrac {7.2 \ m \cdot 1.7\times 10^8 \ Hz - 4\cdot 3\times 10^8 \ m/s}{2\cdot 1.7\times 10^8 \ Hz} \approx \boxed {\color{red}{0.071 \ m}} \end{cases} {/eq}

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from CLEP Natural Sciences: Study Guide & Test Prep

Chapter 8 / Lesson 16