# Homework Help: Microwave Emitting Star, and detector with diffraction

1. Dec 12, 2008

### TFM

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

A microwave detector is located 0.5m above the surface of a large lake far from the shore. As a star, emitting monochromatic microwave radiation of 21cm wavelength, rises slowly above the horizon, the detector indicates successive maxima and minima in the signal intensity. At what angle above the horizon is the star when the first maximum is received?

2. Relevant equations

Not sure of relevance:

Youngs:

$$I(s) = I_0 cos^2 \frac{kDs}{2}$$

Infinite Grating:

$$I(s) = I_0 \sum^\{infty}_{n = -\infty}\sigma (s-\frac{n\lambda}{d})$$

3. The attempt at a solution

I am slightly unsure about this question. My main problem is that I am unsure where the diffraction is coming from? Any help would be most appreciated.

Included is a diagram of my interpretation of the situation

TFM

#### Attached Files:

• ###### Star and Microwave detector.jpg
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2. Dec 12, 2008

### turin

I can't see your image file yet, but I'll bet that there is no diffraction, but that there is reflection. What physics principle allows for the variation in intensity?

3. Dec 12, 2008

### TFM

I wouldn't worry to much about the diagram, its one I did myself, and not my best.

I didn't think it would be diffraction, but is the only thing I could think of that gives maxima and minima.

I am probably missing out on something simple, but I can't think exactly what else would give maxima and minima (regularly any way - I do Astrophyics as well, so Dust clouds spring to mind otherwise, but that's not relevant with this question)

TFM

4. Dec 12, 2008

### turin

Diffraction does not give maxima and minima, but it usually leads to another phenomenon that does. Anyway, like I said, it probably isn't diffraction. You need to read your book to find out what physical phenomenon I'm talking about. Consider two waves, and what happens when they try to occupy the same space.

5. Dec 12, 2008

### TFM

Ah Interference - I feel so silly for not thinking of it sooner.

Would that not need two sources though?

TFM

6. Dec 12, 2008

### turin

EXCELLENT question! Hint: method of images. Oh, yeah, don't neglect phase changing behavior at material interface.

7. Dec 12, 2008

### TFM

I don't think I have studied Method of Images?

This seems like an optics question - trouble is, I didn't do Optics (1st Year Astrophysics didn't do that course, only Theoretical, and non-Astro) - This question is from a 'Skills' Course.

TFM

8. Dec 14, 2008

### TFM

Looked up Method of Images, and it seems to be about electric charges???

TFM

9. Dec 15, 2008

### TFM

TFM

10. Dec 15, 2008

### Carid

It's so romantic to see the moon reflected in the surface of a lake. Pity it doesn't emit much in the way of microwaves.

11. Dec 15, 2008

### TFM

It is indeed.

But wouldn't the reflected waves be going upwards, and, although they would interfere with the incoming waves, they wouldn't reach the detector interfered - the incoming waves are only interfered whilst the other waves are there?

TFM

12. Dec 15, 2008

### Carid

In a harbour you'll see a wave bouncing off the quay and interfering with any other wave that happens to be about. The microwaves beaming down from the star behave in much the same way.
We'll get our first maximum when the path lengths differ by a single half wavelength; only half because of the phase inversion at the water surface.
I reckon the star will be 6 degrees above the horizon.

13. Dec 15, 2008

### TFM

Okay. So what sort of equations might be useful? This is supposed to be a big question, so I am assuming there is quite a few bits of work involved in getting out the right answer?

TFM

14. Dec 15, 2008

### Carid

Draw one ray from your star to the detector.
Draw another arriving at the detector after being reflected at the lake surface.
Do a bit of trigonometry.
Find angle.

15. Dec 15, 2008

### TFM

Okay so (when it is accepted), is this the right sort of diagram?

TFM

#### Attached Files:

• ###### Star and Microwave detector 2.jpg
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16. Dec 15, 2008

### turin

Sorry for leaving you high and dry, but I see that Carid adopted my post. Also, sorry about that "method of images hint"; I had thought you might have been able to find it more broadly applicable than static, charged monopoles. I'll try to give it a different way. Have you studied flat mirrors, yet?

17. Dec 16, 2008

### TFM

That's okay

I didn't do optuics, so no, I don't believed I have studied much on flt mirrors. I nknow snells law, and that the anlge of incidenc = angle of reflection, although that isn't shown well on the diagram.

TFM

18. Dec 16, 2008

### turin

We're in more trouble than I thought. Do you have a freshman physics book with a chapter on mirrors and a chapter on wave phenomena?

19. Dec 17, 2008

### TFM

I Know, I can't spell flat!

We have a result. I have a textbook, University Physics, and it has FOUR chapters dedicated to optics

Including one on interference and one chapter section on reflection (and refraction) on a plane surface, and on a sphereical surface.

Having a look...

Law of reflection:

$$\theta_a = \theta_b$$

where these are the angles that the light hits and is reflected (assuming non perpendicular) - this is what a stated in a previous post.

Also, destructive interference:

$$r_2 - r_1 = (m + \frac{1}{2})\lambda , (m = \pm1, \pm2, \pm3...)$$

Are these useful? I think they maybe...

TFM

20. Dec 18, 2008

### Carid

TFM,

I'm fairly new in these parts and it seems I have transgressed the unwritten law in barging into a thread where someone else was already trying to help. Is that the way it works?

By the way, did you find anything incorrect or misleading in the contributions I made?

21. Dec 18, 2008

### TFM

Doe the equations above look good? (I am starting to get worried, since this is due in tomorrow at Midday! )

TFM

22. Dec 18, 2008

### Carid

TFM
Another error on my part. i should have addressed the previous message to turin.
You diagram of the lake looked about right. The rays arrive parallel from a distant star (all stars are distant even the Sun!). Imagine a point somewhere on the lake where the upper ray is 1 metre above the water and the lower ray is reflected in the water, undergoing a phase change. The upper and lower ray paths from this point to the final object (50 cm high) are the same. Now the surface phase change effectively gives us a half wave length difference. So for a first maximum the path from the star to the water must be half a wavelength longer than the path from the star to the 1 metre high point. So we construct a right angled triangle. The hypotenuse is the one metre height. The base is half a wavelength long (11.5cm). So the angle theta at the top of the triangle, which is equal to the angle of the star above the horizon is given by sin theta = 11.5cm/100cm; this give us a value of 6.6 degrees for the angle.

#### Attached Files:

• ###### lake.jpg
File size:
11 KB
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23. Dec 18, 2008

### TFM

Is that all you really need to do? It seems so simple, yet this is suppose to be the harder question!

TFM

24. Dec 19, 2008

### Carid

The hard thing is to get a feel for the physics. Plugging numbers into equations is what you do when you understand what they mean!

Hey, supplementary question; how far is that 1 metre line from the 50cm post?

25. Dec 19, 2008

### TFM

Well, it would be, using trigonometry,

$$50*tan(90 - \theta)$$

TFM