Knowledge test

We consider a rectangular aperture of amplitude transmittance t:  

Lx and Ly are the aperture dimensions along x1 and y1 .

If the aperture is illuminated under normal incidence by a monochromatic plane wave of unit amplitude (Ui(x1,y1)) =1,the field distribution at the aperture (Ut(x1,y1)) is equal to the transmittance function:

Using the expression (III-3) characterizing the field distribution in the Fraunhofer diffraction figure, we can write:

where the integral can be identified as the FT at the point u=x0/λz ; v=y0/λz. Using table 2 (section “2-dimensional Fourier analysis”, paragraph “Couples of transforms relating to several separable functions in Cartesian coordinates”), we easily obtain:

As a result, the expression of the field at the observation plane becomes:

and the diffracted light intensity can be written:

The image aspect in the observation plane (in false colors and with 2Lx=Ly ) is shown on the following figure:

A 3D plot of the same function is shown on figure KT3.

One can note the presence of small diffraction lobes which are not centered on the axes   x0 et y0, ; those lobes are invisible in the two previous figure because of their small intensity. To visualize those lobes, we represented the intensity logarithm on figure KT4:

We note that the sinc function (defined in table 1 in the section “Two-Dimensional Fourier Analysis”) takes zero values for integer arguments. The first zeros obtained on both sides of the origin are given by:

The central lobe width along the x0 axis in the observation plane is therefore:

We consider a circular aperture of radius R (see figure KT5).

We define   r1 and r0 as the radial coordinates at the aperture plane and at the observation plane (respectively). Then:

If the aperture is illuminated under normal incidence with a monochromatic plane wave of unit amplitude (Ui(r1)=1), the field distribution at the aperture (Ut (r1)) is equal to the transmittance function:

By using expression (III-3) which characterizes the field distribution in the Fraunhofer diffraction pattern, we can write:

where we recognize under the integral sign:

Here, we have:

therefore:

We recall a result demonstrated in the paragraph I.4.4 :

and by applying the scaling theorem, we easily obtain that:

We multiply the numerator and the denominator by and we note the aperture surface. The distribution of the field amplitude at the observation plane becomes:

The intensity can therefore be written:

This intensity distribution is called «Airy pattern». By normalizing the radial variable  : r=2Rr0/λz, we obtain a simplified expression of the diffraction pattern intensity:

This intensity is shown on figure KT6.

We note that the diameter of the central lobe is equal to:  

We represented on figure KT7 the diffraction pattern at the observation plane, in false colors.

We consider a system with a square entrance (or exit) pupil of side length a, and another system with a circular pupil of diameter a.

The expression (V-1) instantly gives the corresponding coherent transfer functions (see figure KT9):

We note that in each case the cutoff frequency is f0 =a/2λdi

Orders of magnitude:

• a=2cm, di=10cm, λ=10-4cm --> f0=100periods/mm.

• Eye cutoff frequency : Φpupille=4-6mm, di =15mm, λ=0,5.10-6m --> f0=300periods/mm.

Period size:

This result is in good agreement with the space between cones in the retina, which is 3µm.

We consider a square pupil of side length BB'=a. At the point of spatial frequency (u,v) (see figure KT10), the common surface is given by: S(u,v)=AB×BC. Since AB=a-B'A and, for geometrical considerations,

we can express the common area as:

( f0 is the cutoff frequency in the coherent regime.)

We note that the cutoff frequency in the incoherent regime is twice larger than the cutoff frequency in the coherent regime (but the contrast is not better). For a circular pupil, the calculation is more complex but we obtain the same result for the cutoff frequency fi .