Diffraction

In general, diffraction occurs when waves pass through small openings, around obstacles, or past sharp edges, as shown in Figure below. When an opaque object is placed between a point source of light and a screen, no sharp boundary exists on the screen between a shadowed region and an illuminated region. The illuminated region above the shadow of the object contains alternating light and dark fringes. Such a display is called a diffraction pattern.

Figure 1 Diffraction Pattern

Light from small source passes by the edge of an opaque object. We might expect no light to appear on the screen below the position of the edge of the object. In reality, light bends around the top edge of the object and enters this region. Because of these effects, a diffraction pattern consisting of bright and dark fringes appears in the region above the edge of the object.

A.    Fresnel Diffraction

In optics, the Fresnel diffraction equation for near-field diffraction, is an approximation of Kirchhoff-Fresnel diffraction that can be applied to the propagation of waves in the near field. The near field can be specified by the Fresnel number, F of the optical arrangement, which is defined, for a wave incident on an aperture, as:

where

 is the characteristic size of the aperture

 is the distance of the observation point from the aperture

 is the wavelength of the wave.

When   the diffracted wave is considered to be in the near field, and the Fresnel Diffraction equation can be used to calculate its form.

Figure 2 Fresnel diffraction showing central Arago spot

The multiple Fresnel diffraction at nearly placed periodical ridges (ridged mirror) causes the specular reflection; this effect can be used for atomic mirrors.

1.    Rectangular Aperture

Fresnel diffraction configuration with a rectangular aperture is shown in figure below

Figure 3 Fresnel Configuring with a rectangle slits

Let’s review the approximation formula of Kirchoff-Fresnel.

For Fresnel diffraction, Frounhofer approximation (r’  r0’, r  r0) does not apply. In figure 2, the line PP’ is taken perpendicular to the plane aperture surface. It is clear that the approximation which is one order higher than Fraunhofer approximation we have

So that

(1.1)

         

On the other hand,          

So that,

                                                                  (1.2)

And therefore,

                                 (1.3)

For a general configuration, the factor of Q = cos q must be take into account. Equation (1.3) become,

                  (1.4)

By using Cartesian coordinate within the aperture we will have,

                                              (1.5)

with

                       (1.6)

Furthermore, we will use the approximation q = 0, equation (1.3) will become

                                              (1.7)

The solution of Fresnel integral for u and v in equation (1.7) is performed based on the following conversion

                                                  (1.8)

with

                                                              (1.9)

and

                              (1.10)

So that E(P) become

                  (1.11)

In this case,           

                  

This means that the field of wave in P is without diffraction effect.

¥

Figure 4 Cornu Spiral

 

       One way to calculate E(P) is by graph method using cornu spiral as shown in (6.10). Fresnel integral is expressed by D(u) = C (u) + iS (u).. At the starting point; C(0) = S(0) = 0 ,l and at spiral eye; C(¥) = S (-¥) = - ½. The distance between the two is  From the graph it can be seen that the integral of equation (50) can be determine from its components C(u₂) - C(u₁) and S(u₂) - S(u₁) as shown by straight line connecting points u₁ and  u_2.

       The following will be mentioned some examples of Fresnel integral by using cornu spiral

(a)             Unlimited aperture

Since X₁ and X₂ is unlimited u_1, v₁ = - ¥,  u₂ = v₂ = ¥, and C (± ¥) = 1/2 ,           S (± ¥) = ± ½.  So that equation (53) will results in

So the intensity of P is equal with the intensity of wave which did not diffracted I (P) = I₀ (P).

(b)            Single slit

The size of slit is unlimited to the direction of X₂, so that v₁ = - ¥,  v₂ =  + ¥. For this case

And its intensity

   

(c)             Straight edge

This is similar with single slit case, with an addition of u₁ = - ¥, for this case we will have

                  

2.                Circular aperture

Here, aperture zone can be divided into concentric rings as shown in figure 5. For a ring with a radius of R we have,

                  

Figure 5 The area of circular aperture in Fesnel zone

 

(R/r₀)²  and (R/r₀’)²  can be ignored so that,

                                          

So that the phase difference across 0’ and 0 (R = 0) is given by

      

It is clear that the phase difference depends only on R. Based on this all aperture plane can be divided into some rings limited by both circles so that the path of light across the circle is equal to l/2. This means that R for n-th circles fulfil

                    

or

                  

The ring limited by two circles with a radius of R_n and R_{n + 1} is called the n-th of Fresnel which has an area of

                  

It is clear that, the area of all Fresnel zones are the same.

B.     Double Slit Difffraction

Figure 6 Multiple slit interference pattern

When more than one slit is present, we must consider not only diffraction due to the individual slits but also the interference of the waves coming from different slit. You may noticed the curve dashed line in figure 6, which indicates a decrease in intensity of the interference maxima as  increases. This decrease is due to diffraction. To determine the effects of both interference and diffraction, we simply has equation

Although this formula looks complicated,it merely represents the diffraction pattern (the factor in brackets) acting as an “envelope” for a two-slit interference pattern (the cosine-squared factor) as shown in Figure 7.

Figure 7 The combined effect of diffraction and interference. This is the pattern produced when 650 nm waves pass through two 3,0  slits that are 18  apart. Notice how the diffraction pattern acts as an “envelope” and controls the intensity of the regularly spaced interference maxima.

For interference maxima as  , where d is the distance between the two slits. Firdt diffraction minimum occurs when , where  is the slidth width. Dividing equation  by equation  (with m = 10 allows us to determine which interference maximum coincides with the first diffraction maximum:

In figure 7, . Thus, the sixth interference maximum (if we count the central maximum as m = 0) is aligned with the first diffraction minimum and cannot be seen.

bibliography

Anonymous. 2013. Fresnel Diffraction. http://en.wikipedia.org. Accessed on Friday, December 20^th 2013

Resnick, Halliday. 2005. Fundamental of Physics 8th Edition. Jearl Walker: New York

Subaer, Abdul Haris. 2010. WAVES. Physics Departement in State University of Makassar: Makassar