Central fringe of blue light

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Homework Statement


Hi

It is I again! Can someone please tell me where I went wrong? My answer is all over the place!



A narrow beam of light is directed normally at a single slit. The diffracted light forms a pattern on a white screen. A blue filter is placed in the path of the beam before it reaches the slit. The distance across five fringes including the central fringe is 18mm. Calculate the width of the central fringe




Homework Equations



Using W= (2 x wavelength x D)/a





The Attempt at a Solution


Solution:

Wavelength = 475nm (450-495 range)

D = 18mm

a= 5 (?)





I got 3.32nm!



Anybody???
 

Answers and Replies

  • #2
collinsmark
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A narrow beam of light is directed normally at a single slit. The diffracted light forms a pattern on a white screen. A blue filter is placed in the path of the beam before it reaches the slit. The distance across five fringes including the central fringe is 18mm. Calculate the width of the central fringe
[...]
Using W= (2 x wavelength x D)/a
That's the equation for the width of the central fringe, yes. But I don't think it's all that necessary for this problem.
Wavelength = 475nm (450-495 range)
That's a nice shade of blue. :smile:
D = 18mm
I think that D is the distance from the slit to the screen. That distance isn't given in the problem statement, but it turns out that you don't really need it for this problem.

[Edit: the 18 mm value is the distance spanning the central fringe plus 4 other fringes. You do need that for this problem. My original point is that the variable D is the distance from the slit to the screen, which is not necessary to know for this problem.]
a= 5 (?)
I think that a is the width of the slit. It's not given in the problem statement either, but it turns out that it's not necessary to solve this problem.
I got 3.32nm!
Something is not right! :eek:

Keep in mind the problem statement is asking for the width of the central fringe, not the width of the slit!

The problem gives you with width of the central fringe plus four other fringes.

Your job is to calculate the width of the central fringe, without the four other fringes.

The solution is pretty easy if you know basic characteristics of what the diffraction pattern looks like. (Hint: the trick is to realize that the central fringe is special compared to the other fringes. :wink:)
 
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