Incidence angle in a Fraunhofer single slit problem

[Resero]

I have to find the first minima for a single slit diffraction problem. I know the wave length 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, which 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?

 

[OlderDan]

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.

 

[Resero]

Thanks Dan

I think that is all I nedd, your aid is helpfull.
I already cheked hyperphisycs. com, but obviuslly I couldn´t get that. All I see about Franhofer talk about normal wave fronts

 

[OlderDan]

I think that is all I nedd, your aid is helpfull.
I already cheked hyperphisycs. com, but obviuslly I couldn´t get that. All I see about Franhofer talk about normal wave fronts

Wave fronts from a single source are normal to the direction of propegation. I think what you mean to say is everything you find talks about a wave encountering a barrier with a slit that is normal to the direction of propegation, and hence parallel to the front. If the wave front encounters a surface at an angle, then there is a time delay and phase difference across the front-surface intersection. This is exactly what happens when a front encounters a flat piece of glass (for example) at an angle. In that case, because the wave travels more slowly in the glass the front changes direction, but it is still normal to the direction of propegation in the glass.

In your problem, for the derivation of the diffraction equation the wave front that makes it through the slit is treated as a line of secondary sources, but if you look at the phase relationship in the forward direction (the original direction of propegation, there is no relative phase shift of the rays. All they have done is travel straight through the slit, even though the slit is tilted at an angle to the front. If you think of points within the slit as secondary sources, then each source is phase delayed by an amount that corresponds to the extra path length the incoming front has to travel before reaching the point. In the forward direction, this delay exactly makes up for the reduced path length from the secondary source to the screen.

What you need to do is to analyze the effect of the phase delayed sources in directions at small angles to the forward direction to see what effect (if any) the phase delay and position change of the secondary sources has on the diffraction minima and maxima on a distant screen that is still normal to the original direction of propegation.