Two vertical, high-frequency radio antennas are 20 m apart. 2....

Question

Two vertical, high-frequency radio antennas are 20 m apart. 2.0 km away, in a plane parallel to the plane of the antennas, 'bright' spots of radio intensity are spaced 5.0 m apart, separated by spots with almost no radio intensity. What is the radio frequency?

Accepted Answer

Hello, fellow physicists today, we're gonna solve the following practice problem together. So first off, let's read the problem and highlight all the key pieces of information that we need to use. In order to solve this problem. A mobile communications company constructs two towers with each of the antennas being 40.0 m apart. They decide to place a metal plate that is located 2.5 kilometers from the antennas to be used as a screen. The plate and the antennas are parallel to each other. A surveyor walking along the location between the towers notices that constructive interference occurs at every 6.0 m on the metal plate that acts as a screen determine what the frequency emitted by the antennas will be. So that's our end goals. We're trying to figure out what the frequency will be that's created by the antennas. So that's our end goal. So we're trying to figure out what the frequency value is that the antennas create. Fantastic. We're also given some multiple choice answers. Let's read them off to see what our final answer might be. A is 3.13 megahertz B is 0.800 gigahertz C is 3.13 gigahertz and D is 0.800 megahertz. Awesome. So first off, let us assume that the signals from both towers will behave like waves that pass through two adjacent slits. Also let us note that since we observe constructive interference every 6.0 m, we can assume that the waves are in phase with each other. Now let us recall and use the equation for the points of constructive interference created from a double slit that appear at certain angles. And that equation states that D multiplied by sine of theta M is equal to M multiplied by lambda. So note that the spacing between the points of constructive interference is small compared to the distance from the two towers or slits where or the towers are the slits where the signals are measured. Therefore, we can recall and use small angle approximation. And that states that D multiplied by sine of theta subscript M is approximately equal to D multiplied by tangent of theta subscript M. But let's make a little side note, Mrs is important, but we need to note that tangent of theta subscript M is equal to Y subscript M divided by capital L. Thus, we could go on to write that D multiplied by sine of theta M is equal to D multiplied by tangent of theta subscript M is equal to D multiplied by Y subscript M divided by capital L which is equal to M multiplied by lambda. And note that the frequency and wavelength are related by the following equation that states that C which is the speed of light is equal to lambda multiplied by F. We can now recall and use the expression for the MTH and the M plus one in points of constructive interference. Thus, we can write that Y subscript M plus one minus Y subscript M is equal to M plus one multiplied by Lambda multiplied by L all divided by D minus M multiplied by Lambda multiplied by L all divided by D is equal to Lambda multiplied by L divided by he. But we need to make another important little side note here. But we need to note that Lambda is equal to C divided by F Awesome. So therefore, we can write that Y subscript M plus one minus Y subscript M is equal to C multiplied by L divided by F multiplied by D Awesome. So now we can rearrange this equation to isolate and sulfur F. And then we can plug in all of our known variables to sulfur, the numerical value of F. So let's do that. So when we rearrange this equation to solve for F, we will get that F is equal to C multiplied by L divided by Y subscript M plus one minus Y subscript M multiplied by D. OK. So when we plug in our known variables, let's do that, we know that C is the speed of light. And the numerical value for the speed of light is 3.0 multiplied by 10 to the power of eight meters per second. And then our L value is given to us as 2.5 kilometers, but we need to convert kilometers to meters. So we need to take 2.50 and we need to multiply it by 10 to the power of three. Awesome. And then we need to divide all of this by six point 0 m multiplied by 40.0 meters. Awesome. So let me plug that into a calculator. We will get that f the frequency is equal to 3.125 multiply by 10. The power of nine hurts. But we need to convert Hertz to Gila to gigahertz. So to do that, to convert to gigahertz, all we need to do is divide our Hertz value by 10 to the power of nine, which will give us 3.13 because we need to round to two decimal places. So it's gonna be 3.13 giga Hertz. And that is our final answer. Hooray. We did it and this is rounded to two decimal places. So therefore the frequency emitted by the antennas is 3.13 gigahertz. So looking at our multiple choice answers, the correct answer has to be the letter C 3.13 gigahertz. Thank you so much for watching. Hopefully that helped and I can't wait to see the next video. Bye.

Two vertical, high-frequency radio antennas are 20 m apart. 2.0 km away, in a plane parallel to the plane of the antennas, 'bright' spots of radio intensity are spaced 5.0 m apart, separated by spots with almost no radio intensity. What is the radio frequency?

Key Concepts

Interference Patterns

Wavelength and Frequency Relationship

Path Difference

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