Incidence angle in a Fraunhofer single slit problem

In summary, the speaker is looking for the first minima in a single slit diffraction problem with known wavelength and slit size, and an unknown angle between the incident wave and the perpendicular line to the slit plane. They know that they must use the Fraunhofer equation for a single slit and have the correct answer, but need to explain why the incident angle does not affect the result. The equation applies for a monochromatic wave with a front parallel to the slit and a wavelength much smaller than the slit size. There will be a phase shift and change in path length of the secondary sources, but these effects only affect the intensity of the diffraction pattern and not the distribution of energy. The speaker has checked online resources but cannot find
  • #1
Resero
2
0
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?
 
Physics news on Phys.org
  • #2
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.
 
  • #3
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
 
  • #4
Resero said:
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.
 

1. What is the definition of incidence angle in a Fraunhofer single slit problem?

The incidence angle in a Fraunhofer single slit problem refers to the angle at which a plane wave of light hits the slit. It is measured from a reference line perpendicular to the surface of the slit.

2. How does the incidence angle affect the diffraction pattern in a Fraunhofer single slit problem?

The incidence angle determines the spacing and intensity of the bright and dark fringes in the diffraction pattern. As the angle increases, the spacing between fringes decreases and the intensity of the central maximum decreases.

3. Can the incidence angle be adjusted in a Fraunhofer single slit experiment?

Yes, the incidence angle can be adjusted by changing the position of the light source or the position of the slit. However, the slit should always be perpendicular to the incident light for accurate results.

4. How does the width of the slit affect the incidence angle in a Fraunhofer single slit problem?

The width of the slit has no direct effect on the incidence angle. However, a wider slit will result in a wider diffraction pattern and a smaller angle between fringes.

5. Is the incidence angle the same as the diffraction angle in a Fraunhofer single slit problem?

No, the incidence angle and the diffraction angle are not the same. The diffraction angle is the angle at which the diffracted light emerges from the slit, while the incidence angle is the angle at which the incident light hits the slit.

Similar threads

  • Introductory Physics Homework Help
Replies
34
Views
2K
  • Introductory Physics Homework Help
Replies
8
Views
1K
Replies
11
Views
2K
  • Introductory Physics Homework Help
Replies
11
Views
1K
  • Advanced Physics Homework Help
Replies
1
Views
900
  • Introductory Physics Homework Help
Replies
3
Views
1K
  • Introductory Physics Homework Help
Replies
1
Views
6K
  • Introductory Physics Homework Help
Replies
1
Views
721
Replies
4
Views
216
  • Introductory Physics Homework Help
Replies
14
Views
904
Back
Top