Fresnel diffraction

Fresnel Diffraction.nb

Optics 505 - James C. Wyant

1

13 Fresnel Diffraction
In this section we will look at the Fresnel diffraction for both circular apertures and rectangular apertures. To help our physical understanding we will begin our discussion by describing Fresnel zones.

13.1 Fresnel Zones
In the study of Fresnel diffraction it is convenient to divide the aperture intoregions called Fresnel zones. Figure 1 shows a point source, S, illuminating an aperture a distance z1 away. The observation point, P, is a distance to the right of the aperture. Let the line SP be normal to the plane containing the aperture. Then we can write

S z1

r1

Q

ρ

r2 z2 P
Fig. 1. Spherical wave illuminating aperture. SQP = r1 + r2
= = ‚!!!!!!!!!!!!!!!!! ‚!!!!!!!!!!!!!!!!! 2 2 22 1 +r + 2 +r

z

z1 + z2 +

þþþþ r

1 2

2

J þþþþ þ + þþ

1 z1

z 1 þþþþ þ N + þþ z2

Ÿ

The aperture can be divided into regions bounded by concentric circles = constant defined such that r1 r2 differ by ƒ 2 in going from one boundary to the next. These regions are called Fresnel zones or half-period zones. If z1 and z2 are sufficiently large compared to the size of theaperture the higher order terms of the expansion can be neglected to yield the following result.
r + l

n

l 2 þþ þþ þþþþ = þþþþ rn J þþþþ þ + þþþþ þ N

2

1 2

1 z1

1 z2

Fresnel Diffraction.nb

Optics 505 - James C. Wyant

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Solving for
rn =

r

n,

the radius of the nth Fresnel zone, yields
= ‚!!!!!!! ! l

‚!!!!!!!!!!! l

L=

þþþþþþþþþþþþþþþþþþ þ þ 1 1 þþþþþ +þþþþþ z1 z2

n L or r1 1

L,

r2 =

‚!!!!!!!!!!! l

2

L,

Ÿ,

where

(1)

Figure 2 shows a drawing of Fresnel zones where every other zone is made dark. Note that in the center the zones are widely spaced and the spacing decreases as the radius increases. As shown above, the radius of the zones increase as the square root of integers. 3

2

1

0

-1

-2

-3 -3 -2-1 0 1 2 3

Fig. 2. Fresnel zones. If
r

n

and

r

n 1 are
+

inner and outer radii of the nth zone then the area of the nth zone is given by
+

Area of nth Fresnel zone = p rn 1 2 - p rn 2 = p Hn + 1L l L - p HnL l L 2 = p l L = p r1 , independent of n.

(2)

That is, the area of all zones are equal. If the higher order terms in the expansion for SRQ are maintained the areaof the zones would slightly increase with increasing . Generally, it is assumed that z1 and z2 are sufficiently large compared to that the higher order terms can be neglected and the area of all zones are equal.
r r

Fresnel Diffraction.nb

Optics 505 - James C. Wyant

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13.2 Circular Aperture
On-Axis Irradiance
In discussing the Fresnel diffraction of a circular aperture it isconvenient to first look at the on-axis irradiance. Consider a circular aperture of radius R, and let S and P lie on the normal through the center of the circular aperture as shown in Figure 3.

S z1

r1 Q ρ R z2 P
Fig. 3. Circular aperture of radius R.

r2

If R is small compared with z1 and z2 the usual approximations can be made in the Kirchhoff integral that the obliquity factor is equal to1 and the 1 ƒ r1 r2 in the denominator can be replaced with 1 ƒ z1 z2 . The cylindrical symmetry about the axis SP suggests using polar coordinates in the integration. Then A z z
+ r

R
È
É

uo @PD = Since r1 2
r Çr =

þþþþþþþþþþþþþþþþ þþþþþ Å É l 1 2

k Hr1 +r2 L

2 p r Çr
2

(3)

0
=

=

z1 2

2

and r2 2

z2 2

+ r

,

r1 Ç r1

=

r2 Ç r2
+ þþþþ þ N r Çr þþ

ÇH

1 r1 + r2 L = J þþþþþþþ r1

1 r2

Therefore,
r Ç r = þþþþþþþþþþþþþþþ Ç H þ þ 1 + 2

r1 r2 r r

r1 + r2 L  

þþþþþþþþ þþþ þ Ç H þ þþþþ 1+ 2

z1 z2 z z

r1 + r2 L

Fresnel Diffraction.nb

Optics 505 - James C. Wyant

4

uo @P D =
where l@0D

p þþþþþþþþþþþþþþþþ þþþþþþþþþþþ þ þþþþ Å É l H 1 + 2L

2 A z z

l@ R D l@0D
È
É

k Hr1 +r2 L

ÇH

r1 + r2...