Image dissipation as distance increases If light propagates outward from its source according to the inverse square law, why don't images dissipate quickly as the distance from the source increases?
 A: It sounds like you are interested in the intensity of light as a function of distance. 
In a nutshell: $ I \propto 1/r^2 $ leads to $ \frac{dI}{dr} \propto -1/r^3 $
As you point out, the intensity of light follows an inverse square law. That means that the derivative of the intensity behaves like $-1/r^3$. So if you are far away from something, then the change in intensity as you go a little further away will tiny. On the other hand, if you are close, then moving a little further away can make a big difference.
A: The inverse square law has nothing to do with the phenomenon you are describing. The inverse square law works in geometric optics where images do not "dissipate", so the reason is different. What you are describing is called diffraction. It is caused by the wave properties of light.
The Huygens-Fresnel principle states that every point on a wavefront is a source of a new wave. In the odd number of dimensions, the new waves cancel in the reverse direction, so light only moves firward and, under certain conditions, "dissipate", as you call it. The images indeed cannot be resolved too far beyond the diffraction limit.
So what defines the diffraction limit? Take your example of a star. It shines equally in all directions, so its light already goes everywhere and is "dissipated" completely from the start. Does it prevent us from seing a star? No. A star does not create its image. For example, when the Sun shines on the Earth, there is no projected image of the Sun on the Earth. To create the image, you need a lens.
And this is where the fun begins. Once you use a lens to create an image, diffraction starts working. If the lens is very small, like a pinhole, the image would be blurred and more so with the distance between the pinhole and the screen increasing. Exactly like in your question, except the distance that counts is not between the source of light and the screen, but between the lens and the screen.
With a larger lens, diffraction is smaller and the image is sharper, because the wavefront is larger and new waves on it interfere maintaining the imge quality better. The amount of diffraction (for a constant size of the film or camera sensor) is defined by the lens aperture, which is a ratio of the diameter to the focal distance.
Let's go back to the star example, because it is an excellent one. When you look at a star, do you actually see it? Well, your intuition was correct. The answer is no. What you see is not a star, but the diffraction circle created by the lens (pupil) of your eye. The actual star image is much much much smaller. If you take a 10x or 100x binocular to "magnify" a star, you would find that it does not become larger like planets do. You need a magnification of a powerful telescope with the mirror diameter of several meters to see a disk of another star. One example, the disk of Betelgeuse has been resolved by the Hubble space telescope with the mirror of 2.4 m.
So in some way, your intuition was correct. An image of a star is completely blurred due to both, the distance that makes the star appear extremely small, and by the optics of the eye that creates a bright circle much larger than the image of the star would be. And if you increase the distance between the screen and projector, the image does become increasingly blurry provided the lens size remains the same and the screen size remains the same as well. For real photo lenses, you can clearly see the effects of diffraction at the apertures of 1/10 or less (in the standard 35mm cameras).
However, none of this has anything to do with the reverse square law that simply defines the intensity.
