# Incidence angle in a Fraunhofer single slit problem

I have to find the first minima for a single slit diffraction problem. I know the wave lenght and the slit size, there is an unknow angle between the incidence wave and the perpendicular line to the slit plane.

I konw that i have to use the Fraunhofer euqtion for a single slit, i already have the correct answer. But I must to explain why the incident angle does not affect the result.

The equation applies for a monochromatich wave, wich front is parallel to the slit, and the wave lenght<<slit size. right?

If the wave front is parallel to the plane is difference in the phase of the secundary sources, ok? ( I´m not sure of this)

If there is an angle this phase difference is another, but the Fraunhofer equation gives us the answer anyway?

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OlderDan
Homework Helper
I think you would want to look at two effects and follow the usual derivation for the equation for single slit diffraction.

http://hyperphysics.phy-astr.gsu.edu/hbase/phyopt/sinslit.html#c1

There will be a phase shift of the secondary sources across the width of the slit due to the later arrival of the incoming rays at these points, but there will also be a change in the path length from these secondary sources to the screen where the rays come together to form the diffraction pattern.

I think there is an additional effect that as you rotate the slit the total amount of light energy that makes it through the slit diminishes, but this might only affect the intensity of the pattern and have no effect on the distribution of energy within the pattern.

Thanks Dan

I already cheked hyperphisycs. com, but obviuslly I couldn´t get that. All I see about Franhofer talk about normal wave fronts

OlderDan