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bruceinwayne
2009-Apr-03, 04:34 PM
In the Radio Astronomy episode on Astronomy Cast, Pamela stated that a radio telescope dish can have holes up to 10% (I think) the length of the wave.

I have been grappling with fully understanding the Electromagentic spectrum for a year now, (ever since I got into Astronomy) and I feel like this episode really helped me, but I don't get the 10% hole thing.

It would seem that it would be the HEIGHT of the wave that causes some part of the wave to hit the dish and reflect the wave back to the detector. Is it simply that wave height is a function of wave length so longer waves are "taller" and can't fit through the holes in the reflecting dish? Or do I have it all wrong.

I will understand Electromagnetic waves before I die!!!!

Thanks,

Bruce

Argos
2009-Apr-03, 05:31 PM
When you´re dealing with reflection of EM waves it is the frequency that matters, and not the amplitude ["height" as you say, which, in a graph, represents the strength of the signal]. Think of an optical mirror: low frequency waves, like visible light, are reflected, while high frequency ones, like X-ray, pass through. For radio waves it goes along the same lines.

See this (http://www.setileague.org/askdr/dishmesh.htm), and this (http://www.setileague.org/askdr/accuracy.htm[/url), and this (http://www.setileague.org/hardware/blkdish.htm) [I´m quoting the latter]:

If you choose to obtain a surplus antenna, dish condition becomes an important factor. The main consideration here is surface accuracy. In order to perform up to expectations, a dish's surface cannot deviate from the parabolic by more than a tenth of a wavelenght. At 1420 MHz, that's about 2 cm of allowable surface error. If the surface of the dish is dimpled, dented, or distorted beyond 2 cm, avoid that dish! Look for something which approximates a smooth parabolic curve. If panels are missing or bent, performance is going to suffer.

Now, if it were the amplitude ["height"] of the signal that determined the reflection, a weak signal [one of "low height"] would hardly be reflected [it would pass through the holes, according to your misconception], and a strong signal [with a big "height"] would be highly reflected. It does not work that way.

bruceinwayne
2009-Apr-03, 06:35 PM
Argos, thanks for the response, but in all honesty I just still don't get it.

A colleague of mine, we are high school math teachers, and I spend our free time trying to understand astronomy and all that goes with it, especially EM.

To better understand your explanation I need to understand the answer to the question, why doesn't a mirror reflect X-rays? Does the wave/particle pass between the atoms that make up the mirror?

thanks for helping the new kid, (even though I turn 40 in 2 weeks)

Argos
2009-Apr-03, 06:45 PM
To better understand your explanation I need to understand the answer to the question, why doesn't a mirror reflect X-rays? Does the wave/particle pass between the atoms that make up the mirror?

Yes, exactly. The wave passes through because of its tiny length, which enables it to avoid being scattered by atoms in the way.

You can reflect X rays if the angle of incidence is very shallow, as in the grazing-incidence telescopes. This link (http://encarta.msn.com/encyclopedia_761579309/x-ray_astronomy.html) contains a simple explanation, although it is off-topic in relation to your original post.

Jeff Root
2009-Apr-03, 06:52 PM
As far as I have been able to learn, the amplitude of an electromagnetic
wave cannot be measured as a distance or height, like the height of a
water wave. The amplitude is a measure of the strength of the electric
and magnetic fields, not an extension in space. It isn't entirely clear to
me whether electromagnetic waves have any extension in space at all,
perpendicular to the direction of motion. Even in the direction of motion,
they are often treated as point particles, and when treated as waves,
they are considered infinitely long, to make Fourier analysis possible.

-- Jeff, in Minneapolis

bruceinwayne
2009-Apr-03, 11:03 PM
Thanks guys,

Sometimes I feel like I get it, EM Radiation that is, but then when I start talking it out I wonder if I have any idea what I am saying. Pretty cool stuff though. The Encarta X-ray site definitely helped.

undidly
2009-Apr-03, 11:32 PM
As far as I have been able to learn, the amplitude of an electromagnetic
wave cannot be measured as a distance or height, like the height of a
water wave. The amplitude is a measure of the strength of the electric
and magnetic fields, not an extension in space. It isn't entirely clear to
me whether electromagnetic waves have any extension in space at all,
perpendicular to the direction of motion. Even in the direction of motion,
they are often treated as point particles, and when treated as waves,
they are considered infinitely long, to make Fourier analysis possible.

-- Jeff, in Minneapolis

A photon (electromagnetic particle) is as wide as it is high as it is long.
But it has no sharp edges ,just fades away inversely as the square of distance from the center.

"considered infinitely long, to make Fourier analysis possible."
And wide,that is how it gets through both slits at the same time to make the interference pattern.Bigger the slit separation the dimmer the pattern.

As you say the amplitude is not a physical (distance) measure at least not in any of the dimensions we are aware of.

bruceinwayne
2009-Apr-04, 03:55 PM
If a photon is "as wide as it is high and long," then I think that clears up a lot of confusion on my part.

Awesome stuff this light.

korjik
2009-Apr-04, 05:42 PM
Why this is can be treated classically using wave mechanics. Any EM wave trying to pass through a hole in a surface will be diffracted. If the hole is large compared to the wave, the wave passes through. If the hole is about the same size as the wave (same size as the length of the wave) then there will be alot of diffraction. If the hole is small, the wave is totally diffracted. This prevents the wave from going through the hole, but since the wave has to go somewhere, it acts like the hole isnt there.