Maximizing and Minimizing Wavelength Contributions in Radiowave Reflection

In summary: I'll try to draw it. Here's what I think happens. Suppose the wavefront hits the edge of the slit at an angle θ. The path length from the point where the wavefront hits the ground to the point where it reflects is L1. The path length from the point where the wavefront hits the edge of the slit to the point where it reflects is L2. The phase difference between the two Ls is (L1-L2) cos θ.In summary, the wavelength that contributes the most to strengthens the direct ray is λ=5.865 GHz.
  • #1
Const@ntine
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Homework Statement


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The transmitter is on the left, and the receiver is on the right. The transmitter transmits radiowaves. The distance between them is d = 50.0 m & each one's height is h = 35.0 m. The receiver can receive the radiowaves directly from the transmitter, or through reflection by the ground. When the radiowaves hit the ground, there is a phase difference of 180 degrees.

Find which are the maximum wavelength that contribute:

a) Strengthening.
b) Weakening.

Homework Equations

The Attempt at a Solution



I'll be honest here, I've been staring at this for a while and I just don't know where to start. It's probably something simple, but for the life of me I don't know what to do. It's in my Optics Part of the book, Chapter with Thin-Films and contribution by two openings on a board and all that. I'd really appreciate a nude or "directions" on what to look for or something.

Any help is appreciated!
 
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  • #2
If a reflected wave arrives with the same phase, it strengthens the direct ray. Is that a useful hint for you ?
 
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  • #3
Darthkostis said:
Find which are the maximum wavelength that contribute:
I am not at all sure what this question is asking. Never mind, let's just analyse the set-up and see where it leads.
Suppose the wavelength used is λ. When the wave has traveled distance s, what is the phase difference between the point the wave has reached and the present state at the transmitter? In the diagram, what phase difference will there be between the wave that takes the direct path and the wave that reflect from the ground?
Darthkostis said:
I'd really appreciate a nude
You will need a different forum for that.
 
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  • #4
BvU said:
If a reflected wave arrives with the same phase, it strengthens the direct ray. Is that a useful hint for you ?

Sure, but I'm mostly lost on the "what do I even do here" part. I know what "happens" based on logic, but I'm not sure what I'm looking for, or how to find it.

haruspex said:
I am not at all sure what this question is asking.

Well, it's a direct translation, so I'm not sure myself. Best I can come up with is "find the wavelengths with the biggest numerical value that result in either a strengthening or cancelling confluence".

haruspex said:
Never mind, let's just analyse the set-up and see where it leads.
Suppose the wavelength used is λ. When the wave has traveled distance s, what is the phase difference between the point the wave has reached and the present state at the transmitter? In the diagram, what phase difference will there be between the wave that takes the direct path and the wave that reflect from the ground?

So I should treat this as a "double slit" problem? Like this:

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haruspex said:
You will need a different forum for that.

Darn it, I meant to write nudge (keyboard's a tad busted).
 
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  • #5
Darthkostis said:
So I should treat this as a "double slit" problem? Like this:

Yes. Not exactly the same but the same ideas (eg path lengths differ by multiple of the wave length).
 
  • #6
So, uh, anybody have any more hints on this? I'm blanking out. I tried treating it as a double slit, with one slit being the "head" of the transmitter and the other being the place where the wavelength reflects, but it didn't work.
 
  • #7
Darthkostis said:
So, uh, anybody have any more hints on this? I'm blanking out. I tried treating it as a double slit, with one slit being the "head" of the transmitter and the other being the place where the wavelength reflects, but it didn't work.
In both cases, double slit and the direct ray and the reflected one in this problem, the key word is phase difference between the rays which meet. It comes from the path difference and the phase change at reflection. If it is integer multiple of 2π, the rays add up to maximum intensity. If the phase difference is odd number times π, minimum intensity is obtained.
 
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  • #8
Darthkostis said:
So, uh, anybody have any more hints on this? I'm blanking out. I tried treating it as a double slit, with one slit being the "head" of the transmitter and the other being the place where the wavelength reflects, but it didn't work.
Did you try to answer my questions in post #3?
 
  • #9
haruspex said:
Did you try to answer my questions in post #3?
Yeah, I assumed it was a nudge to get me to look at it as a double-slit problem. Find the angle the two "paths" create and construct the phase-difference from that.
 
  • #10
Darthkostis said:
Yeah, I assumed it was a nudge to get me to look at it as a double-slit problem. Find the angle the two "paths" create and construct the phase-difference from that.
So post expressions for the two path lengths.
 
  • #11
haruspex said:
So post expressions for the two path lengths.
I'm a tad stuck on that. If I look for an angleat the starting point, aren't I omitting part of the path of the second radiowave?
 
  • #12
Darthkostis said:
I'm a tad stuck on that. If I look for an angleat the starting point, aren't I omitting part of the path of the second radiowave?
What angle?
It is a simple matter of geometry. In the diagram, how far is it along the straight line between the tops of the two towers?
How far is it along the other route? Then add half a wavelength for the second case for the reflection phase shift.
What is then difference between the two path lengths?
 

FAQ: Maximizing and Minimizing Wavelength Contributions in Radiowave Reflection

What are radiowaves?

Radiowaves are a type of electromagnetic radiation that have longer wavelengths and lower frequencies than visible light. They are commonly used in communication technologies such as radio and television broadcasting, as well as in radar and satellite communication.

How do radiowaves reflect?

Radiowaves can reflect off of certain materials, just like how light reflects off of a mirror. This is known as reflection. The smooth and flat surface of the reflecting material causes the radiowaves to bounce off at an angle equal to the angle at which they hit the surface. This allows for the transmission of radiowaves over long distances, as they can be bounced off of multiple surfaces before reaching their intended destination.

What is the relationship between the frequency of a radiowave and its ability to reflect?

The frequency of a radiowave is directly related to its ability to reflect. The higher the frequency, the more easily the radiowave can penetrate materials and the less it will reflect. On the other hand, lower frequency radiowaves, such as AM radio waves, have a better ability to reflect off of surfaces and can travel further distances.

How is the phenomenon of reflection utilized in everyday life?

Reflection is utilized in a variety of ways in everyday life. One of the most common uses is in communication technologies, where radiowaves are bounced off of surfaces to transmit signals over long distances. Reflection is also used in satellite communication, radar systems, and even in our own eyes, where light is reflected off of objects to allow us to see them.

Can reflection of radiowaves be manipulated for specific purposes?

Yes, reflection of radiowaves can be manipulated for specific purposes. This is done through the use of specialized materials, such as reflectors, which are designed to reflect radiowaves in a specific direction. This is commonly used in satellite communication to direct signals towards a specific location. Reflection can also be manipulated through the use of mirrors and other surfaces with different shapes and angles, allowing for control over the direction and intensity of reflected radiowaves.

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