nykkyo Posts: 132, Reputation: 1 Junior Member #1 Apr 25, 2013, 10:06 AM
Why does light converge
On the observer and bathe him in light ?because light is emitted in all diections from the aource?
 ebaines Posts: 12,129, Reputation: 1307 Expert #2 Apr 25, 2013, 10:20 AM
If you're asking how it is that an observer on earth can see light emitted from a distant source - such as stars, planets, nebulae etc - it's because most objects emit light in all directions, so any observer would be able to see them regardless of location. But some sources actually emit light in a highly-directional way, so your ability to see it depends on your position relative to the source. An example of this would be pulsars, which are neutron stars that emit a directional beam of electromagnetic radiation, much like a spot light rotating. That beam can only be observed on Earth when it is pointing in our direction, so to us it appears as a pulsating light source that flashes in time with the rotation of the star.
 nykkyo Posts: 132, Reputation: 1 Junior Member #3 Apr 25, 2013, 04:51 PM
I mean why do we see points of light when light is emitted in all directions. It implies light rays from long distances are separate. In that case our understanding how light is emitted is suspect.
 ebaines Posts: 12,129, Reputation: 1307 Expert #4 Apr 26, 2013, 06:04 AM
Consider how the particles of light (photons) travel from the source to your eye. Photons are emitted from all parts of the star's surface, and if you are relatively close to the star (like we are to the sun) it appears as a round disc. If you moved further away the apparent size of the disc gets smaller. Get far enough away and the apparent size of the disc becomes too small to be distinguished by your eyes, yet there are enough photons hitting your eye so that you can still see that there is astar, but you can''t see the disc. Seeing light coming fom a source of negligible apparent size makes the object appear as a pin point of light.

I hope this makes it clear - again think of star light as a series of photon particles eminating in all directions from the star, some of which travel on a path that intersects with your eyes with the result that you see the star.
 nykkyo Posts: 132, Reputation: 1 Junior Member #5 Aug 25, 2013, 11:56 PM
Originally Posted by ebaines
Consider how the particles of light (photons) travel from the source to your eye. Photons are emitted from all parts of the stars surfac, and if you are relatively close to the star (like we are to the sun) it appears as a round disc. If you moved further away the apparent size of the disc gets smaller. Get far enough away and the apparanet size of the disc becomes too small to be distinguished by your eyes, yet there are enough photons hitting your eye so that you can still see that there is astar, but you can''t see the disc. Seeing light coming fom a source of negligible apparent size makes the object appear as a pin point of light.

I hope this makes it clear - again think of star light as a series of photon particles eminating in all directions from the star, some of which travel on a path that intersects with your eyes with the result that you see the star.
I understand the inverse square law; but why are some photons funneled to our aperture?
 ebaines Posts: 12,129, Reputation: 1307 Expert #6 Aug 26, 2013, 05:52 AM
Originally Posted by nykkyo
I understand the inverse square law; but why are some photons funneled to our aperture?
If you understand the inverse square law then you understand that as you get further away from a lighrt-emitting object the "density" of photons that passes through any given surface area (such as the surface area of a telescope's mirror) gets smaller as well, but it never goes all the way to 0. So some photons are going to hit the telescope no matter how far away you are. It's not that these photons are "channeled" to the scope, but rather that some numberof photons inevitably hit by chance.

Let's run some numbers - imagine that a light source emits N photons per second. Consider a spherical shell 1 meter in radius cenered on the source - N photons must pass through the surface of that shell every second. So if you stand 1 meter away with a telescope that has surface area of 1 m^2 the proportion of those photons that will hit the telescope is the ratio of the telescope's area to the surface area of the 1 meter shell:

$
n = N \left( \frac {1m^2}{4 \pi 1^2} \right) = 0.0796 N
$

If you stand temn meters away the ratioj gets smaller: n= 0.000796 N

At 100 meters: n= 7.96E-6 N

At 100 meters n = 7.96E-8 N

and so on. Note that the ratio never gets to zero.
 nykkyo Posts: 132, Reputation: 1 Junior Member #7 Aug 26, 2013, 07:15 AM
Thec inverse suar law decrease4s intensity (photon density) but does not decrease coverage of aprteure.
 ebaines Posts: 12,129, Reputation: 1307 Expert #8 Aug 26, 2013, 07:24 AM
Nykko - I don't understand the point of your last post - are you answering your own question from post #5?
 nykkyo Posts: 132, Reputation: 1 Junior Member #9 Aug 26, 2013, 07:42 AM
Originally Posted by ebaines
nykko - I don't understand the point of your last post - are you answering your own question from post #5?
Phton density is the same in all directions; but the farther away from the source the lower number of phtons hit an area (like an aperture). The divegence of the rays is what cause the density to decrease. Since the rays are radially continuous the area is wholly illuminated. If the rays are radially quantized some rays may miss the aperture.
 Tuttyd Posts: 53, Reputation: 4 Junior Member #10 Sep 6, 2013, 11:37 PM
Originally Posted by nykkyo
I understand the inverse square law; but why are some photons funneled to our aperture?
I think I get what your are saying.

Photons don't need to be "funneled" in any way in order for us to see them. Photons are force carriers- charged particles interacting result in charged particles being exchanged. Photons carrier just enough energy to activate the light sensitive parts of the eyes.

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