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How many dark fringes will be produced on either side of the central maximum if light ( = 568 nm) is incident on a single slit that is 5.27 10-6 m wide?

2007-03-14 12:42:49 · 4 answers · asked by marinatedpickles 2 in Science & Mathematics Physics

4 answers

I think that's a very hard question for anyone to answer. I thought about the question for a while, and I still couldn't answer it. I think another option would be to search similar questions on Yahoo or Google search. If you know anyone else in the subject, try calling them or e-mailing them with the question. I am sorry if I am not helping much, it's the best I could do. I hope it helps, even if it was a little.

2007-03-15 15:20:42 · answer #1 · answered by Anonymous · 0 0

It is an interesting but answerable question. There are several ways to get to the answer, depending on how rigorous you want to be. Let's first consider the least rigorous. The diffraction minima for a single slit occur under the condition:

a*sin(theta) = m*lambda, where a=slit width and m>0

solving for m, m = a/lambda * sin(theta)

since we assume theta can only go up to 90 degrees, the valid integers for m must meet the condition:

m <= a/lambda * sin(90)

In the present example, m < 5.27/0.568, or m < 9.27.

Hence there are 9 minima on each side.

If you want to be a bit more complete, there is a classic way to explain the formula for finding the minima. Essentially, at a given diffraction angle, each position across the slit will cancel with the next position in the slit which is lambda/2 out of phase with it. Starting with one edge and working across, eliminate each position and the position which cancels it out. You find that unless the phase between the end points is m*lambda, there will be positions remaining which are not cancelled out. (if you want a better or more detailed explanation with diagrams, do a search for single slit diffraction).

For a more rigorous approach, you can use the Fourier transform of the slit aperture to get the diffracted intensity:

constant * (sinc(pi*a*sin(theta)/lambda)^2
where sinc(x) = sin(x)/x (equal to 1 at x=0)

and calculate the maximum argument of the sin^2 function to see how many minima you get. If you prefer, plot it and count the minima. You run out of minima after 9 because the sin(theta) can't exceed 1.

2007-03-19 20:21:46 · answer #2 · answered by or_try_this 3 · 0 0

Laser is greater effective because of the fact it rather is easy waves are monochromatic, meaning all easy debris are in section with one yet another. jointly as the alternative is real with a white easy source. besides the actuality that it has colored clear out, easy diffraction relies upon on the interference of the easy and not the wavelength.

2016-10-02 03:28:05 · answer #3 · answered by ? 4 · 0 0

None, it takes two slits to produce a fringe pattern.

2007-03-16 08:53:05 · answer #4 · answered by catarthur 6 · 0 2

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