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Width of the first dark fringe (2 slit interference)

by XanziBar
Tags: dark, fringe, interference, slit, width
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Feb15-10, 03:20 PM
P: 46
1. The problem statement, all variables and given/known data

In a double-slit experiment, the two narrow slits are spaced 0.20 mm apart. A viewing screen is put 1.2 m behind the slits. A parallel light ray is shining on the slits.

(a) If the wavelength of the incoming light ray is 633 nm. What is the width of the central bright fringe of the interference pattern on the viewing screen?

(b) If the wavelength of the incoming light ray is 633 nm. What is the width of the first dark fringe of the interference pattern on the viewing screen?

2. Relevant equations

y[m]=m*lambda*L/d (bright)
y'[m]=(m+.5)*lambda*L/d (dark)
theta [m]=m*lambda/d (bright)
y'[m]=(m+.5)*lambda/d (dark)

(using the small angle approximation in all equations.

3. The attempt at a solution

Okay so for part a I was thinking the following: the central maximum might be defined by the 1st dark fringes on either side of the central bright so y'[0]=1.899 mm and since the pattern is symmetric it is defined by the first dark on the left and the right which means the width is 2*y'[0]=3.8 mm. Is that on the right track at all???

For part b I am completely lost, I almost want to define the width of the first dark fringe as being between the central maximum and the 1st bright. But that would count a bunch of light area as being part of the dark! Then I want to say that maybe after all the first dark has a width of zero. After all the light is only completely canceled at one tiny point in space. What am I thinking about wrong? Help?
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Feb15-10, 05:02 PM
P: 46
Feb16-10, 12:38 AM
P: 46
If you want, I can try to explain the problem better. Or show more of a solution?

Don't leave me hanging here guys.

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