1. April 3, 5 Fresnel zones 10.3 Fresnel diffraction
Free propagation of a spherical wave
Fresnel diffraction: For any R1 and R2.
Fraunhofer diffraction is a special case of Fresnel diffraction.
The integration for Fresnel diffraction is usually complicated. Though we can do it by computers, Fresnel zones have been very effective in estimating the diffraction patterns. They provide a mean to deepen our knowledge in wave diffraction. S P a R1 R2 P
Wave front k
K(q) q
Normal of the wave front
Directionality of secondary emitters:
Inclination factor (obliquity):
(To be proved. )

2. The area of the slice ring is
Free propagation of a spherical monochromatic wave:
Primary spherical wave:
Question: What is the field at P which is r0 away from the sphere?
Contribution from the sources inside a slice ring dS: S x P r O r0 r r dS
The area of the slice ring is

3. Fresnel (half-period) zones:x
r0+l/2
r0+ll/2 r Z1 Zl dS P
Contribution from the l th zone to the field at P:
Note:
1) Kl is almost constant within one zone. 2)
Each zone has the same contribution except the sign and the inclination factor.
The contributions from adjacent zones tend to cancel each other.

4. Sum of the disturbance at P from all zones on the sphere:
Note:
1) Same if
2) If m is even,

5. Inclination factor Km(p) = 0  |Em| ≈ 0 
Exact solution (simple spherical waves):
Note: Huygens-Fresnel diffraction theory is an approximation of the more accurate Fresnel-Kirchhoff formula.

6. Divide the zone into N subzones.
The vibration curve
A graphic method for qualitatively analyzing diffraction problems with circular symmetry. Phasor representation of waves. S O O' r0 x
r0+l/2N
r0+l/2 r Z1 P
p/N Os
Zs1 E1
The first zone
For the first zone:
Divide the zone into N subzones.
Each subzone has a phase shift of p /N.
The phasor chain deviates slightly from a circle due to the inclination factor.
When N ∞, the phasor train composes a smooth spiral called a vibration curve.

7. Each zone swings ½ turn and has a phase change of p.
The vibration curve:
Each zone swings ½ turn and has a phase change of p.
The total disturbance at P is
The total disturbance at P is p/2 out of phase with the primary wave (a drawback of Fresnel formulation).
The contribution from O to any point A on the sphere is
Os'
Zs1
Zs2
Zs3 As Os S O O' r0 x
r0+l/2
r0+ll/2 r Z1 Zl P A

8. So you won’t forget it.

9. Joseph von Fraunhofer
(1787–1826) German optician.
Augustin-Jean Fresnel
(1788–1827) French physicist.

10. Read: Ch10: 3
Homework: Ch10: 69,70,71
Due: April 14

11. April 7 Circular apertures
I. Spherical waves
1) P on axis:
a) The aperture has m (integer, not very large) zones.
i) If m is even,
ii) If m is odd,
which is twice as the unobstructed field.
b) If m is not an integer, This can be seen from the vibration curve.
Os'
Zs1
Zs2
Zs3 As Os P S O O' r0 x r Zm

12. 3) The area of one of the first few zones:
2) P out-of axis:
As P moves outward, portions of the zones (defined by P, S and O) will be uncovered and covered, resulting in a series of relative maxima and minima. (The integration will be very complicated.) O x P S O' r0 r
3) The area of one of the first few zones:
Number of zones in an aperture:
Example:
Fraunhofer diffraction condition:

13. Fraunhofer diffraction
II. Plane waves P O r0
r=r0+ml/2 Rm
Example:
On-axis field:
Fraunhofer diffraction
Fresnel diffraction

14. Read: Ch10: 3
Homework: Ch10: 72,73,74,77,78
Due: April 14

15. April 10 Fresnel zone plate
Circular obstacles
Poisson’s spot: Bright spot always appears at the center of the shadow of a circular obstacle.
Poisson intended to use this unusual conclusion to deny Fresnel’s wave description of light, but this prediction was soon verified to be true. The spot is ironically called Poisson’s spot. May have been observed by ancient people.
The spot is everywhere along the axis except immediately behind the obstacle.
The irradiance is not very different from that of the unobstructed wave.
Os' As Os P S O O' r0 x r A

16. Example: Transparent only for odd (or even) zones.
Fresnel zone plate
Zone plate: A device that modifies light by using Fresnel zones. Modification can be either in amplitude or in phase.
Example: Transparent only for odd (or even) zones.
The first 10 odd (even) zones will result in an intensity of 400 times larger compared to the unobstructed light.
I. For spherical waves:
Radii of the zones: r0 rm S P Rm Am
Identical to thin-lens equation. S is imaged at P.

17. Third-order focal length: because
II. For plane waves: F1 Rm F3 f3 f1
Radii of the zones:
Primary focal length:
Third-order focal length:
because
Fabrication of zone plates:
Photographically reduce large drawings.
Newton’s rings serves as good pictures for this purpose.

18. Read: Ch10: 3
Homework: Ch10: 79,80,81
Due: April 21

19. April 17 Rectangular apertures
Fresnel integrals and the rectangular apertures
Fresnel diffraction with no circular symmetry. The zone idea does not work. P r z y S r0 A
The contribution to the field at P from sources in dS:
K(q) =1 if the aperture is small (<<r0, r0).
In the amplitude r = r0, r = r0.
In the phase
Half of the unobstructed field: Eu/2
Fresnel integrals

20. Ep and Ip can be evaluated using a look-up table.
Fresnel integrals:
C(x)
S(x)
Ep and Ip can be evaluated using a look-up table.
Off-axis P points can be estimated by equivalently shifting the aperture and changing the limits (u1, u2, v1, v2) in the integrals according to the new values of (y1, y2, z1, z2).
It also applies to special apertures, such as single slit, knife-edge, and narrow obstacles. What we need to do is to find the values of u1, u2, v1 and v2.

21. Notes on how to find u1, u2, v1 , and v2 :
Plane wave incidence:
Notes on how to find u1, u2, v1 , and v2 : P z y S r0 P'
Project the observing point P onto the aperture plane. Call the projected point P'.
Let P' be the origin of the coordinate system. Let the y and z axes be parallel to the two sides of the aperture.
(y1, y2, z1, z2) are the coordinates of the four limits of the aperture, when viewed at P'. Please note that they are not the size of the aperture. The size of the aperture is given by | y2-y1 |=a and | z2-z1 |=b.

22. Aperture 2 mm×2 mm, l=500 nm. (a2/l = 8 m)
Example: Fresnel diffraction of a plane wave incidence on a rectangular aperture
Aperture 2 mm×2 mm, l=500 nm. (a2/l = 8 m)
For any point P (X in m, Y in mm, Z in mm): P X Y

23. Read: Ch10: 3
Homework: Ch10: 82,83,85.
Optional homework:
Using Mathematica, draw the following three diffraction patterns (contour plots) for a plane wave incidence on a rectangular aperture.
Aperture = 2 mm×2 mm; l = 500 nm; Screen = 0.4 m, 4 m and 40 m away.
Note:
Describe the procedures of how you calculate the intensity distribution.
In Mathematica the Fresnel integral functions are FresnelC[] and FresnelS[]. You may need to study ListContourPlot[] or ListPlot3D[].
For each distance adjust the screen area you plot so that you can see the main features of the diffraction pattern.
Use logarithmic scale for the intensity distribution. Let each picture span the same orders of magnitude of intensity down from its maximum.
Discuss the evolution of the diffraction patterns for the above three distances.
Due: April 28

24. April 19, 21 Cornu spiral 10.3.7 Cornu spiral
Cornu spiral (clothoid): The planar curve generated by a parametric plot of Fresnel integrals C(w) and S(w). B1 B2 u1 u2
B12(u)
1) Arc length = parameter w:
3) A point on the spiral:
4) A vector (phasor, complex number) between two points:
This is a constant vector, not a function.

25. Diffraction from a rectangular aperture:B1 B2 u1 u2
B12(u)
Similarity between Cornu spiral and the vibration curve:
Each point on the curve corresponds to a line (or a circle) on the aperture.
Each line element on the curve corresponds to the field from a slice (or a ring) on the aperture.
The total field is the product of two vectors between the end points on Cornu spirals, or one vector on the vibrational curve.
Practice:
Off-axis point: Slide arc (string) of constant length along the spiral.
Expanding the aperture size: Extend endpoints of arc along the spiral.
Very large aperture:

26. 10.3.8 Fresnel diffraction by a slit
From rectangular aperture to slit:
Practice:
Off-axis point: Slide arc of constant length Dv along the spiral.
Relative extrema occur. Small Dv has broad central maximum.
Expanding the aperture size: Extend ends of arc along the spiral.
Relative extrema occur. z P y z1 z2 a Z
0.5
1.0

27. Fresnel diffraction by a slit

28. 10.3.9 Semi-infinite opaque screenz P z1
z2= ∞ Z
From slit to semi-infinite edge:
Shadow
Edge B+
(C(v1), S(v1))
Edge
Shadow

29. 10.3.10 Diffraction by a narrow obstaclez P z1 z2 Z u1 u2 B+ B-
There is always an illuminated region along the
central axis:
For off-axis points, slide the obstacle along the spiral.

30. Example: Thin slit and narrow obstacle. Special cases:
Babinet’s principle
Babinet’s principle: The fields from two complementary diffraction apertures satisfy: u1 u2 B+ B-
Field from each screen
Unobstructed field
Example: Thin slit and narrow obstacle.
Special cases:
This happens in Fraunhofer diffraction when P is beyond the Airy disk.

31. Reading: 10.4 Kirchhoff’s scalar diffraction theoryP dS
Green’s theorem
Helmholtz equation
Kirchhoff integral theorem
Applying to unobstructed spherical wave from a point source:
Fresnel-Kirchhoff
diffraction formula
Obliquity factor
Differential wave equation 
Huygens-Fresnel principle

32. So you won’t forget it.

33. Read: Ch10: 3-4
Homework: Ch10: 91,92.
Optional homework:
1. Draw a Cornu spiral using Mathematica.
2. The diffraction pattern from a single slit with plane wave incidence is:
a) Draw the 3D picture of |B12|2/2 as a function of , for the
range of
b) Discuss the evolution of the diffraction patterns when you change (v2-v1)/2.
c) For a given slit width a, where and how much is the highest diffraction intensity?
Due: April 28