1. May 11, 2013

### Saitama

1. The problem statement, all variables and given/known data

2. Relevant equations

3. 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?

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2. May 11, 2013

### rude man

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

3. May 11, 2013

### Saitama

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

4. May 11, 2013

### rude man

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

5. May 11, 2013

### Saitama

What is the value of n?

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

6. May 11, 2013

### 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?

7. May 11, 2013

### rude man

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!

8. May 11, 2013

### Saitama

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 $\lambda/(4\cos \alpha)$.

9. May 11, 2013

### rude man

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.

10. May 11, 2013

### Saitama

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?

11. May 11, 2013

### TSny

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

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• ###### Radio Waves 1.jpg
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12. May 13, 2013

### Saitama

Sorry for the late reply.

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?

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• ###### radio waves.png
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13. May 13, 2013

### TSny

Yes, that looks ok.

14. May 13, 2013

### Saitama

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.

15. May 13, 2013

### TSny

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

16. May 13, 2013

### Saitama

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

17. May 13, 2013

### 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.

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• ###### Radio Waves.jpg
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18. May 13, 2013

### Saitama

That makes it a lot easier, thanks!