What Is the Incident Angle for the First Null in a Diffraction Pattern?

[sssddd]

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]

Well, do you know the condition / formula for single slit interfence?

 

[sssddd]

it's the one with the sin theta right or something right? please point me to the right track

 

[Hootenanny]

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.

 

[sssddd]

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]

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.

 

[sssddd]

ok thank you so much now i can find the angle easily

 

[Hootenanny]

My pleasure 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