- #1
Paavo T
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Homework Statement
There are at least two different situations. If you have N-2 middle minimums, the amount of slits is N. Please, see the bottom left http://scienceworld.wolfram.com/physics/DiffractionGrating.html" . The case with no such thing is the question about. How can I determine the amount of slits with the other cases? Please, see the other pictures in the link.
Homework Equations
[tex]a sin(\theta) = n \lambda[/tex]
where a is the width of the separtion of two side-by-side slits.
Intensityhttp://mathbin.net/47038"
[tex]I = I_{0} \left[ \frac{sin \left( \pi b sin \left( \frac{\theta}{\lambda} \right) \right) }{\pi b sin \left( \frac{\theta}{\lambda}} \right) \right]^{2} \left[\frac{sin\left(N \pi a sin\left(\frac{\theta}{\lambda}\right)\right)}{sin \left( \pi a sin\left( \frac{\theta}{\lambda}}\right) \right) \right]^{2}[/tex]
The Attempt at a Solution
If you have 2 slits, then you have 5 main maximums. My proposal that the amount of maximums is 2N+1, where N is the amount of slits was rejected, by my teacher. You can see that it is not the case from the Wolfram link above. Then I tried to delve into the wave function but I am really lost with it. Cannot see how it helps in finding the the number of slits.
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