# Electric Field and Optics questions

1. Mar 24, 2004

### Spectre32

I got a few simpel questions dealing with and electric field and and "optics" question

OK for the electric field we have an object that is out in space. THe max strength of the electric field form the communications tranmissions are 6.8E-10. THe first part of the questiona tell us that the radio recieving dish is 900m^2. We are to find the average power incident on the dish. The second part asks us to find the max strength of the magnetic field for this. For the first part of the question I was going to use the equation I = Eo*Bo/2Uo, and for some reason i can't get it to work. For the second part i was going to use E= cB but again, no luck.

For the second problem there is a lake with a width of 26M. You are staind 3M away from the lake and your eyes are 1.6M above the ground. There a a tree on the other side of the lake, and from your point of view u can only see the top of the tree's reflection in the lake. They want me to find out how tall the tree is. What i did is draw this picture and then draw my rays of the image from the lake to the tree, and then to me. This game me two triangles.( of which one I knew two sides. After figureing out the 3rd side for one of them i went to work on the second one. Yet again i couldn't figure out the height of the tree. I know i am doing something stupid.

Thanks for the help.

2. Mar 24, 2004

### Janitor

Is the transmitter understood to be isotropic? If so, then the energy of the transmission wave would have to drop off as the inverse square of distance. Is the energy of an e.m. wave proportional to the square of intensity? How far is the receiving antenna from the emitting source?

3. Mar 24, 2004

### Spectre32

TI's 10billion KM away. It says nothing about Isotropic conditions.Is the energy of an e.m. wave proportional to the square of intensity? The problem dosn't say.

4. Mar 24, 2004

### Janitor

5. Mar 24, 2004

### Spectre32

yeah, lol i think thats way to complex for what i'm going for here. My teacher was showing me what was going on with it, but i tired t recreate it and it's not happining.

6. Mar 24, 2004

### Janitor

Okay, I read your question more carefully, and I see that distance from the transmitter is really not an issue.

Here is a simpler site:

http://hyperphysics.phy-astr.gsu.edu/hbase/waves/emwv.html

One of the equations on it is:

S=(E_m)^2/(2 c mu_0).

They don't define E_m (E subscript m) anywhere that I can see, but if it is the "max strength of the electric field", and if S is calculated in units of watts per square meter, then multiplying S by 900 square meters is going to give you some number of watts.

Has your teacher talked about root mean squares (rms)? I am not sure if that is an issue here. Maybe the 2 in the denominator takes into account the averaging of the sinusoidal wave.

7. Mar 24, 2004

### Janitor

As far as your second question goes, I think the critical idea is that the angle of incidence equals the angle of reflection. So your two triangles are similar. Is it fair to say that the right triangle nearest to you has a base of 29 meters and a height of 1.6 meters? And that the top of the tree is seen at the far edge of the pond? It seems to me that this would only allow one to say that the tree satisfies

h/d = 1.6/29

where h is its height and d is its distance beyond the edge of the pond that is farthest from you. You would then have to know how far away the tree is. Have you listed all of the data given for that problem?

Last edited: Mar 24, 2004
8. Mar 24, 2004

### Janitor

If "other side of the lake" means right on the far bank of the lake, can you say that

1.6/3 = h/26 ?

This would mean that you are seeing the top of the tree on the near bank.

That would make a pretty easy equation to solve!

The only problem with this interpretation is that you wouldn't "just see the top" of the tree, would you? I would say in this case you would be able to see the whole danged tree's reflection. If "just see the top" is interpreted as: "you can see the whole tree, but you can't see a humminbird hovering a foot over it," then this is probably the way you are supposed to solve this problem.

Last edited: Mar 24, 2004
9. Mar 24, 2004

### Spectre32

well i know the height of the tree is 13.9 M. My teacher game me the answer but i wasn't sure how he got it.

10. Mar 24, 2004

### Janitor

Okay then, the second way I looked at the tree problem does indeed give that answer. One triangle has you as its side (left side, lets say), and the ground from you to the near shore as the base. The other triangle has its right side as the tree, and the pond as its base.