Relating Path Difference and Intensity in Radar Signals

[Saitama]

Homework Statement

Homework Equations

The Attempt at a Solution

The path difference between the two rays being received by antenna is $h/cos(\alpha)$ but how to relate the path difference with the intensity?

 

[rude man]

What is the phase difference requirement between two signals for totally constructive interference?

 

[Saitama]

Phase difference is equal to $2n\pi$ or path difference is $n\lambda$.

 

[rude man]

Phase difference is equal to $2n\pi$ or path difference is $n\lambda$.

Correct. So you need n*lambda path difference. Solve for h?

 

[Saitama]

What is the value of n?

I end up with $h=n\lambda \cos \alpha$.

 

[TSny]

n can be any positive integer. Are you sure that you got the correct expression for the path difference? Did you take into account the law of reflection? I wonder if you are meant to include a phase shift due to reflection?

 

[rude man]

What is the value of n?

I end up with $h=n\lambda \cos \alpha$.

Do you want to make the height higher than it has to be? You can make it $h=\lambda \cos \alpha$ or $h=2\lambda \cos \alpha$ or ...

I would make it as low as possible while satisfying your equation!

 

[Saitama]

n can be any positive integer. Are you sure that you got the correct expression for the path difference? Did you take into account the law of reflection? I wonder if you are meant to include a phase shift due to reflection?

I don't know about considering the phase shift. There is nothing mentioned in the question about it. I think I got the right expression for path difference considering no phase shift. The given answer is $\lambda/(4\cos \alpha)$.

 

[rude man]

n can be any positive integer. Are you sure that you got the correct expression for the path difference? Did you take into account the law of reflection? I wonder if you are meant to include a phase shift due to reflection?

Good point. I assumed the OP gave the correct path difference and I didn't consider any phase change due to reflection.

Judging from the given answer the path length was not what the OP came up with.

 

[Saitama]

Good point. I assumed the OP gave the correct path difference and I didn't consider any phase change due to reflection.

Judging from the given answer the path length was not what the OP came up with.

The light reflected from the surface of water which reaches the antenna has to travel an extra distance of $h/\cos \alpha$ (which can be easily calculated from simple trigonometry). This is the path difference. What have I done wrong?

 

[TSny]

The light reflected from the surface of water which reaches the antenna has to travel an extra distance of $h/\cos \alpha$ (which can be easily calculated from simple trigonometry). This is the path difference. What have I done wrong?

The two waves travel the same distance to the red dots shown in the figure.

 

[Saitama]

Sorry for the late reply.

The two waves travel the same distance to the red dots shown in the figure.

The length AC is $h/\cos \alpha$.

$\angle ACB=2\alpha-\pi/2$
$\sin \angle ACB=AB/AC=AB\cos \alpha/h \Rightarrow -\cos 2\alpha=AB\cos \alpha/h$
$\Rightarrow AB=-h\cos 2\alpha/\cos \alpha$

Am I doing this right?

 

[TSny]

Yes, that looks ok.

 

[Saitama]

Yes, that looks ok.

Path difference is $AC-AB=\frac{h}{\cos \alpha}+\frac{h\cos 2\alpha}{\cos \alpha}$
= \frac{h(1+\cos 2\alpha)}{\cos \alpha}=2h\cos \alpha

The path difference should be equal to $n\lambda$ and for min height, $n=1$
\Rightarrow h=\frac{\lambda}{2\cos \alpha}

But this is wrong.

 

[TSny]

That all looks good. You might be expected to include a phase shift of the wave that reflects off the water.

 

[Saitama]

That all looks good. You might be expected to include a phase shift of the wave that reflects off the water.

If I include the phase shift, I do get the right answer. Thank you TSny!

 

[TSny]

Good!

There's a trick that can make finding the path difference easier. In the figure, the tower is extended a distance h below the ground. You can use the law of reflection to show that the triangles abd and cbd are congruent. Then ba = bc and ec is the path difference.

 

[Saitama]

There's a trick that can make finding the path difference easier. In the figure, the tower is extended a distance h below the ground. You can use the law of reflection to show that the triangles abd and cbd are congruent. Then ba = bc and ec is the path difference.

That makes it a lot easier, thanks!