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the question is A plane wave of 400-nm light is incident on a 25-µm slit in a screen, as shown in the figure below. At what incident angle will the first null of the diffraction pattern be on a line perpendicular to the screen?http://img130.imageshack.us/img130/9766/inter9ax.jpg
I really do not where to start. Please help me out, i am so lost.
Hootenanny
Jun9-06, 04:07 PM
Well, do you know the condition / formula for single slit interfence?
it's the one with the sin theta right or something right? please point me to the right track
Hootenanny
Jun9-06, 04:22 PM
I think this can be solved using the Fraunhofer Single Slit Diffraction relationship;
a\sin\theta = m\lambda
Where a is the size of the slit, \lambda is the wavelength and m is the order of the minima.
ok ok i think i am remembering this now, and m would be 1 right for first null. when they say null they mean the minima fully destructive right
Hootenanny
Jun9-06, 04:37 PM
ok ok i think i am remembering this now, and m would be 1 right for first null. when they say null they mean the minima fully destructive right
I am assuming that's what null means a minima (I haven't heard that terminology for a while). Yes, the fist minima would be at m = 1. This is different to maximas where the first maxima is at n = 1.
ok thank you so much now i can find the angle easily
Hootenanny
Jun9-06, 04:44 PM
My pleasure :smile: Just for your information, this is applying the Fraunhofer equation 'backwards' if you like. More info is available at the link below;
http://hyperphysics.phy-astr.gsu.edu/hbase/phyopt/fraungeo.html#c1
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