A radio station operating at 88.5 MHz broadcasts from two identical antennas at the same elevati

A radio station operating at 88.5 MHz broadcasts from two...

A radio station operating at 88.5 $\mathrm{MHz}$ broadcasts from two identical antennas at the same elevation but separated by a 9.0 -m horizontal distance $d,$ Fig. $28 .$ A maximum signal is found along the midline, perpendicular to $d$ at its midpoint and extending horizontally in both directions. If the midline is taken as $0^{\circ},$ at what other angle(s) $\theta$ is a maximum signal detected? A minimum signal? Assume all measurements are made much farther than 9.0 $\mathrm{from}$ the antenna towers.

A radio station operating at 88.5 $\mathrm{MHz}$ broadcasts from two
identical antennas at the same elevation but separated by a
9.0 -m horizontal distance $d,$ Fig. $28 .$ A maximum signal is
found along the midline, perpendicular to $d$ at its midpoint
and extending horizontally in both directions. If the midline
is taken as $0^{\circ},$ at what other angle(s) $\theta$ is a maximum signal
detected? A minimum signal? Assume all measurements are
made much farther than 9.0 $\mathrm{from}$ the antenna towers.

Key Concepts

Far-Field Approximation

The far-field approximation assumes that the observer is located at a distance much greater than the separation between the sources, allowing the waves to be approximated as parallel (plane waves). This simplification makes it possible to analyze the interference pattern using trigonometric relations between the angle of observation and the path difference.

Wave Interference

This concept refers to the interaction of waves from two or more coherent sources, where they superpose and combine to produce regions of enhanced (constructive) or diminished (destructive) amplitude. It is fundamental in understanding how two antennas broadcasting the same signal can create an interference pattern with alternating maxima and minima at various angles.

Path Difference

Path difference is the difference in distances traveled by two waves from their sources to a given point. It determines the phase difference between the waves, and thus whether they interfere constructively (if the path difference is an integer multiple of the wavelength) or destructively (if the path difference is a half?integer multiple of the wavelength).

Constructive and Destructive Interference

Constructive interference occurs when the path difference between waves is an integer multiple of the wavelength, leading to signals that add together and produce maximum intensity. Destructive interference happens when the path difference is a half-integer multiple of the wavelength, resulting in signals that cancel each other out and produce a minimum intensity. These conditions are pivotal in predicting the angles at which maximum and minimum signals are detected.

Transcript

00:01 All right, because the measurements are made far from the antenna, we can assume this is the same as a double slit experiment.

00:12 Okay, so this is essentially a double slit, but the two antenna acting as the two slits.

00:17 So the first step is obviously to find the wavelength of the broadcasting signal, which is just speed, speed of light over the frequency.

00:27 So the speed of light is 3 times 10 of the 8 meters per second, and frequency is 88 .5 megahertz, so it's 88 .5 times 10 to the 6 hertz.

00:37 Giving us wavelength of 3 .39 meters.

00:41 Note that this is much higher than the visual wavelengths in the nanometers.

00:47 So this is in the radio.

00:49 Okay.

00:50 So we have two cases, constructive and destructive interference.

00:55 The maxima will be, will occur in the constructive interference.

01:01 And so in the case of constructive interference, and i use blue ink here, we have d -sign theta, and theta is what we have to find, is equal to m -lamda, m being the order.

01:15 Therefore, theta is just the arc sign of m -lamda over d.

01:21 And so we try out different values of m to see what, what theta's we get right that's the idea here so for m equals zero data not or theta zero we have arc sine of zero and so that's zero degrees theta 1 you have arc sign of 1 times 3 .39 over 9 and so this gives us 22 degrees for theta 2 arc sine of 2 times 3 .39 over 9.

02:02 This gives us 49 degrees.

02:05 Now when we try m equals 3, you have arc sine of 3 times 3 .39 over 9.

02:15 Note that the numerator is greater than the denominator, which cannot happen.

02:21 Sign theta cannot take a value of greater than 1.

02:24 So this is impossible or imaginary.

02:30 This will give you a matter if you enter it on the calculator...

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