Microwave Emitting Star, and detector with diffraction

[TFM]

Homework Statement

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?

Homework 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})

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

Thanks in advance,

TFM

 

[turin]

... detector ... located 0.5m above ... surface ...
... a star, emitting ... 21cm wavelength, rises ... above the horizon, ... detector indicates successive maxima and minima in ... intensity.

My main problem is that I am unsure where the diffraction is coming from?

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?

 

[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

 

[turin]

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

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.

 

[TFM]

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

Would that not need two sources though?

TFM

 

[turin]

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

Would that not need two sources though?

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

 

[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

 

[TFM]

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

TFM

 

[TFM]

Any helpful ideas?

TFM

 

[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.

 

[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

 

[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.

 

[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

 

[Carid]

Draw one ray from your star to the detector.
Draw another arriving at the detector after being reflected at the lake surface.
Add in the information about the path length difference.
Do a bit of trigonometry.
Find angle.

 

[TFM]

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

TFM

 

[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?

 

[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

 

[turin]

... I don't believed I have studied much on flt mirrors.

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?

 

[TFM]

We're in more trouble than I thought.

I Know, I can't spell flat!

Do you have a freshman physics book with a chapter on mirrors and a chapter on wave phenomena?

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

 

[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?

 

[TFM]

Your help was most useful.

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

TFM

 

[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.

 

[TFM]

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

TFM

 

[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?

 

[TFM]

Well, it would be, using trigonometry,

50*tan(90 - \theta)

TFM

 

[Carid]

Yes, which is about 4m32cm or about 20.6 times the wavelength.

 

[TFM]

Out of interest, is the 1 metre an arbitrary number, or is there a specific reason this number was chosen?

 

[Carid]

Take a look at my new diagram and tell me how tall the dotted line should be.

 

[TFM]

I can't see the diagram yet,, but ill look as soon as it is on.

Also, is the phase change as it reflects due to the fact that its going from a low density medium to high density medium, and some of the wave is reflected and some transmitted, like so:

http://www.kettering.edu/~drussell/Demos/reflect/reflect.html

(near the bottom)

Makes sense to me.

TFM

 

[TFM]

TFM
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.

The wavelength is 21cm, making half a wavelength 10.5 cm, not 11.5cm. will this effect the 1m (I still can't see the diagram yet)

TFM

 

[Carid]

Well done, you spotted my "deliberate mistake"...gulp...So what is the proper value for angle theta?

 

[TFM]

It would be:

\theta = asin(10.3/100)

Which gives the angle to be: 5.9 degrees.


Does this look okay?

 

[Carid]

10.3 ? I'm not the only one making "deliberate mistakes" this morning...

 

[TFM]

Should be 6.02 degrees.

Silly Me!

TFM

 

[TFM]

Have we made an error - I think we have found the minima, but the question asks:

At what angle above the horizon is the star when the first maximum is received?

?

TFM

 

[Carid]

Why do you think we have found a minimum?

 

[TFM]

I'm not sure, I suddenly saw the question and panicked. I think I remembered putting destructive interference in a previous post.

I also wrote down how destructive interference is when the waves out half wavelength out of phase written done in my work, despite the fact that the question just above said maxima. Suddenly gt very worried and panicked. - I get worried far too easily

Sorry,

TFM

 

[Carid]

Panic is good.
It means "let's go back to the fundamentals and see if we have understood and be sure we haven't neglected something along the way".

 

[TFM]

So can I just check something, just to see. I wrote down

destructive interference is when the waves out half wavelength out of phase

But we made the calculations with half a wavelength added on. Would this not make the final wave a minima

Just Checking

Also, Panic is good, unless the homework is due in in less then 1 hr...

TFM

 

[Carid]

OK let's examine the fundamentals.

We will get our maximum if the path length between the two rays differs by a single wavelength. Now we found an extra half-wavelength in the lower ray path. We need another half-wavelength. Where's that coming from? It comes from the reflection which advances the phase of the lower ray by half a wavelength. Now we've got two half wavelengths adding up to one whole wavelength. We should get that first maximum.

I hope you get to see the diagram I sent earlier; very simply the 1 metre height is NOT arbitrary. It's twice the 50cm height and is a question of symmetry.

 

[TFM]

Yeah I worked the 1m from symmetry after thinking about. So the water reflection from the water makes the wave half out of phase. See, I didn't realize that, should have though, since the the link showed the reflection wave going from a peak, reflected into a trough.

That makes sense. Thanks,

Out of interest, not part of the question, but I am intrigued, how would you find, say a second or third maxima? Would you just add on a whole wavelength onto the half in the triangle, eg:

\theta_n = (n-1)\lambda + \frac{1}{2}{\lambda}

where n is the nth maxima?

TFM

 

[Carid]

As the star rises in the sky we'd arrive at a moment when the extra path length is now 3/2 of a wavelength (31.5cm) and the angle would be about 17.5degrees.

That equation you quoted looks like something you'd use in optics where the wavelength is extremely short and the angle very small so that trig functions are discarded.

 

[TFM]

I could have sworn I put the asin in. Should have been:

\theta_n = asin\left((n-1)\lambda + \frac{1}{2}{\lambda}\right)

Thanks for all your assistance.

TFM

 

[turin]

So, I see Carid has got you covered, TFM. Great discussions, too. I just thought you might benefit from an alternative perspective. Can you see how to identify this with a Young's double slit phenomenon? In your problem, you have a detector, a water surface, and a single light source. In Young's experiment, there is a screen and two slits. Can you identify the analogies? Can you identify the crucial difference?

 

[TFM]

I can definatley see the similarities. The Two slits would be where the wave is reflected, straight up, on Carid's Diagram, the isosceles triangle. The crucial difference is that in Young's, the two waves would be in phase, whereas this version has the two waves out of phase.

TFM