Separation distance of Michelson Interferometer Suppose I would like to use Michelson Interferometer to observe fringes of equal thickness by creating an angle between the mirrors.  Why is it vital for the path difference between the mirrors to be small in order to observe the fringes?
 A: Imagine the laser beam used for the interferometer. Picture it shining onto a distance wall. The beam expands with distance, like a cone. You can then see how the center of the beam has the shortest distance to the target. Now picture them in the interferometer. When the beams are at the exact same length you will get a perfect match - complete constructive and destructive interference, that is one bright spot or one black. As the mirror differences become less and less equal more rings will appear, but they will be smaller. There will also be less contrast between them. Eventually, if the difference between beam lengths become too great and the classic pattern is lost. 
Note: The image that appears on the screen always has an opposite image that is projected back onto the laser. So the “light” never really goes away.
A: "Fringes of equal thickness" probably means equal spacing.  You only get fringes of equal spacing in a Michaelson interferometer when the beams in the interferometer are collimated.
The only reason for it might be necessary for the paths in the two arms of the interferometer to be nearly equal is if the light is not temporally coherent.  If the coherence length of the light is greater than the largest path difference between the two beams, and if the beams have the same intensity, then you should get high contrast fringes.  A typical laser pointer has quite a long coherence length, sometimes greater than a couple of meters.
