furthest point at which light interferes


by kahwawashay1
Tags: furthest, interferes, light, point
kahwawashay1
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#1
Mar28-12, 06:25 PM
P: 96


This is from a hw problem and i kno how to solve it with formulas but i dont kno whats actually going on. So in the diagram above, light is emitted from S1 and S2, and you want to find the furthest point P at which the light interferes destructively. S1 and S2 are separated by a few micrometers, and the answer was also on the scale of micrometers. But, I dont see why the light cant interfere beyond P. I mean, if you draw concentric circles representing the wavefront around each of the points S1 and S2, then it seems the wavefronts still intersect each other in various ways in all directions, including all along x axis...
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kahwawashay1
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#2
Mar28-12, 08:15 PM
P: 96
should i post this in the hw section? is this why no one is answering me?
nasu
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#3
Mar28-12, 08:31 PM
P: 1,904
I suppose you should show your solution based on formulas. The answer will emerge from there.

However, if you just want to understand why there is some maximum distance, think about what happens with the path difference when the point P moves towards +∞.
Increases, decreases?
In order to get completely destructive interference, the path difference must be at least half wavelength.

kahwawashay1
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#4
Mar28-12, 09:29 PM
P: 96

furthest point at which light interferes


Quote Quote by nasu View Post
I suppose you should show your solution based on formulas. The answer will emerge from there.

However, if you just want to understand why there is some maximum distance, think about what happens with the path difference when the point P moves towards +∞.
Increases, decreases?
In order to get completely destructive interference, the path difference must be at least half wavelength.
The path difference would then decrease.
My solution was:
dsinθ = (m+0.5)λ
x2+d2 = (x + dsinθ)2
Combining these two equations and solving for x gives
x = [itex]\frac{d^{2}-\lambda^{2}(m+0.5)^{2}}{2λ(m+0.5)}[/itex]
(where d is the distance between the points emitting the light, dsinθ is path difference, x is the furthest point)

So just looking at the formula I know the furthest point is when the path difference is 0.5 wavelengths, but I am having trouble visually seeing how the light waves behave after this point. Like I said, if you draw concentric spheres around each of the points in accordance with Hyugen's principle, it seems that the waves continue to interfere forever

Also, isn't this basically the same thing as the double slit experiment, with the two point sources acting as if they just can through a slit? In deriving the formula for this experiment it is assumed that the screen on which the interference takes place is infinitely far away....and in either case, in this particular problem, the separation d was a few micrometers and the furthest point was also a few micrometers, but when you actually do the experiment, you get bright/dark spots a meter away and more...
nasu
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#5
Mar28-12, 11:13 PM
P: 1,904
The question is about points along that given line.
Nobody said that there is no more interference or that a screen at a larger distance will not catch the interference pattern.

If you are familiar with double slit interference you have a good starting point.
The problem proposed is equivalent with a a double slit setup. Just put a screen parallel to the line S1S2 and passing through P. The requirement is to have a minimum at a distance of d/2 from the center. If you move the screen farther from P, the first minimum won't disappear from the screen but will be farther from the center so not on the S1P line anymore.


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