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Angular Width of Electromagnetic Waves emerging from two buildings

1. Homework Statement
Hello everyone,

I have one more problem left of my homework and I seemed to be stumped. I don't even know where to begin with this problem, which comes from the section of my book labeled Single-Slit Diffraction. Anyways, here is how the problem reads:

Problem 22.18

You need to use your cell phone, which broadcasts an 800 MHz signal, but you're behind two massive, radio-wave-absorbing buildings that have only a 15 m space between them.

What is the angular width, in degrees, of the electromagnetic wave after it emerges from between the buildings?

I calculated the wavelength at 800mhz = .381m
a= 15m (distance of the slit OR distance between two buildings.

2. Homework Equations



3. The Attempt at a Solution

Due to the slit being so large, I am not sure how to approach the problem. I assume that we cannot use a small-angle approximation due to how large the slit and wavelengths are.

I think the equation to use would be the following
delta r= a/2 *sin(theta) = wavelength/2
However, I am not sure how to rearrange to get theta.

I tried rearrangeing to get aSin(theta) = wavelength => theta = sin-1(wavelength/a) and got 1.43 degrees. Would this be correct or did I just hammer the problem?

If anyway can give me a hand (or somewhere to start) I would really appreciate it.

Thanks,
Randy
 

Answers and Replies

Tom Mattson
Staff Emeritus
Science Advisor
Gold Member
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You need to use your cell phone, which broadcasts an 800 MHz signal, but you're behind two massive, radio-wave-absorbing buildings that have only a 15 m space between them.

What is the angular width, in degrees, of the electromagnetic wave after it emerges from between the buildings?
I've never heard of the "angular width...of the electromagnetic wave" before. I'm assuming that they meay the angular width of the central maximum.

I calculated the wavelength at 800mhz = .381m
a= 15m (distance of the slit OR distance between two buildings.
Yes, a would be the slit width.

Due to the slit being so large, I am not sure how to approach the problem. I assume that we cannot use a small-angle approximation due to how large the slit and wavelengths are.
Approach it like any other single-slit diffraction problem.

I tried rearrangeing to get aSin(theta) = wavelength => theta = sin-1(wavelength/a) and got 1.43 degrees. Would this be correct or did I just hammer the problem?
I agree that you need to find [itex]\theta=\sin^{-1}(\lambda/a)[/itex]. But you need to think about what angle that represents. That is not the angular width of the central maximum. So let me ask you, what angular width is it?
 
I do not know what angular width that represents. my book does not go into anymore details about this type of problem. Instead of just using a = 15m would I use a/2 = 7.5m? I do not know what else to do for this problem. I am kind of stumped.
 

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