Diffraction: Distinguishing Fresnal & Fraunhofer Diffraction

[mntb]

how do you distinguish Fresnal and fraunhofer diffraction mathematically?

 

[jtbell]

Can you be more specific about what you're looking for? Are you asking about the difference in the assumptions involved, or the final equations, or what? Are you asking about locations of maxima and minima, or about the intensity of the diffraction pattern as a function of position on the viewing screen? Do you have single-slit diffraction in mind, or something else?

 

[Dr Transport]

A very good explanation of this is in Introduction to Fourier Optics by Joseph Goodman. Essentially the difference between Fraunhofer and Fresnel diffraction is that the Fraunhofer pattern is obtained by taking the Fourier transform of the wave across the aperature and calculating the intensity. The Fresnel pattern does not involve expanding the expression for the plane wave. Look on page 71 of Goodman.

If you want more, I'll be happy to further explain it.

 

[Claude Bile]

The equation describing Fraunhofer diffraction is an approximated version of the equation describing Fresnel diffraction (which is, itself, an approximation). Recall that Fraunhofer diffraction assumes a planar incident wave, hence;

- The obliquity function (the function that describes transmission through an aperture as a function of propagation angle) is approximated to equal 1.
- The phase change term is omitted.
- The distance terms are constant over the area of integration.

Claude.

 

[mntb]

so is this right?
R>a^2/λ for Frauhofer diffraction, and R<a^2/λ for Fresnal diffraction
is there a math equation to express their difference?

 

[Claude Bile]

The equations you specify do not describe Fraunhofer or Fresnel diffraction per se, but rather a quantitative criterion that specifies which regime the diffraction falls under. The important factor buried within this criterion is the maximum phase error.

Equality defines the point where there is a \lambda/8 maximum phase error. The reason this point is chosen as a delineation is because for phase errors less than this, the phase effects are negligible. For phase errors greater than this, phase errors become significant.

Claude.

 

[mntb]

yes, I'm looking for regime the diffraction falls under, I read that the λ has to be much larger for the Fresnal diffraction, so is R<a^2/λ for Fresnal diffraction right?

 

[Claude Bile]

One lies in the Fresnel regime if

F=\frac{a^2}{R\lambda} &gt; 1

http://scienceworld.wolfram.com/physics/FresnelNumber.html

So, yes, you are correct.

Claude.