# Two radio antennas separated by d = 289 m, one just north of the other, simultaneously transmit...

## Question:

Two radio antennas separated by d = 289 m, one just north of the other, simultaneously transmit identical signals of the same wavelength. A radio in a car traveling due north receives the signals. (a) If the car is at the position of the third maximum when 1000 m to the east of the antennas and 400 m north of their midline, what is the wavelength of the signals? (b) How much farther must the car travel to encounter the next minimum in reception? (Hint: Determine the path difference between the two signals at the two locations of the car.)

## Interference:

Interference refers to the superposition of waves to form a resultant wave of greater or lower amplitude. If the resultant wave produced is of greater amplitude, then the waves that superimpose must have been in-phase, this process is called Constructive Interference. For the resultant wave produced of lower magnitude, the waves must have been out-of-phase and this process is Destructive Interference.

## Answer and Explanation:

Given:

{eq}d = 289 \ m {/eq} Distance between the two antennas

{eq}x = 1000 \ m {/eq} Horizontal distance of the center point of the two antennas and the car

Part A) We first need to determine the distance of the car from the two antennas using the Pythagorean theorem as,

Distance between the car and the top antenna

{eq}r_1 = \sqrt {(400 \ m - 144.5 \ m)^2 + (1000 \ m)^2} = 1032.12 \ m {/eq}

Distance between the car and the top antenna

{eq}r_2 = \sqrt {(400 \ m + 144.5 \ m)^2 + (1000 \ m)^2} = 1138.63 \ m {/eq}

Path difference = {eq}r_2 - r_1 = 1138.63 \ m 1032.12 \ m = 106.51 \ m {/eq} This is the position of the 3rd maximum.

To solve for the wavelength,

{eq}path \ difference = 3 \lambda {/eq}

{eq}\lambda = \frac {path \ difference}{3} = \frac {106.51 \ m}{3} = 35.503333333 \ m = \boxed {35.5 \ m} {/eq}

Part B) The next minimum refers to a destructive interference. To solve we need first to determine the angle in the equation given by,

{eq}d sin \theta = (m + \frac {1}{2}) \lambda {/eq}

{eq}289 sin \theta = (3 + \frac{1}{2}) (35.5) {/eq}

{eq}\theta = sin^{-1} \frac {3.5 (35.5)}{289} = 25.46^\circ {/eq}

Given the horizontal distance, {eq}x = 1000 \ m {/eq}

Hence,

{eq}tan 25.46^\circ = \frac {y}{1000 \ m} {/eq}

{eq}y = 1000 tan 25.46^\circ = 476.12 \ m {/eq}

Finally,

{eq}\Delta y = 476.12 \ m - 400 \ m = \boxed {76.1 \ m} {/eq}

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from CLEP Natural Sciences: Study Guide & Test Prep

Chapter 8 / Lesson 16