1.22

1. 1.22 This specifies the location of the first minimum and is also Rayleigh’s criterion for resolution limited by diffraction from a circular aperture.

2. Well resolved Barely resolved Rayleigh’s criteria for resolution • For two objects to be resolved, the peaks of their diffraction patters cannot overlap “too” much. • Rayleigh’s Criteria: when the central maximum of one just overlaps the first minimum of the other, then the two are barely resolved. • For a circular aperture the required angular separation is: (D=Diameter of aperture)

3. Far from the slit Close to the slit z Incident plane wave Fresnel Diffraction: Example Fresnel Diffraction from a single slit: Slit

4. Fresnel zone plates • For a small enough aperture diameter, waves passing through the aperture and arriving at a source point all have the same phase • Waves passing through the aperture at larger distances from the center experience a phase delay • The phase delay from each Fresnel zones (annulus of a specific radius) increases by π • By selecting even or odd zones, light waves will add constructively, but only at a single point. • This behaves as lens! • r0=distance between zone plate and field point

5. Focal points of a Zone • A 500 nm plane wave is incident on a 5 mm diameter circular slit which acts a Fresnel Lens. Determine locations on the optical axis where the light diffracted by the slit reaches a maximum value. Zone Radius = Rn When rn Rn r0 so Focal points at odd n There are more zones in a fixed radius aperture as you move closer (r0 decreasing), so the intensity will go through maxima and minima.

6. B Fresnel Diffraction • When we are not within the Fraunhofer limit, the quadratic terms that we neglected begin to contribute to the integral • We can write this as • Now • To find the intensity we need only calculate C and S where

7. r r’ R S P r0 r’0 Fresnel Cosine and Sine Integrals • Recall that • So

8. C(w) S(w) • Where and • So now define • So that • The complex integral can be written as • These are the well known Fresnel Sine and Cosine Integrals

9. The Cornu Spiral w=1 • These integrals can be calculated numerically and have been tabulated in Pedrotti • The Cornu spiral is a graphical calculation tool that can be a big help in solving Fresnel diffraction problems • The parameter, w, is the length of the arc measured from the origin • W is related to a position in the aperture • The eyes of the spiral are at (0.5,0.5) and (-0.5,-0.5) • These correspond to aperture positions of +/- ∞ • The Field can be found by reading the values off the chart and substituting them in to…. w = 0 ∞ w = -1 -∞

10. Diffraction through an infinite aperture • This is a limiting case of no diffraction • The edge of the aperture is located at • xmin= ymin = umin= vmin = - infinity • xmax= ymax = umax= vmax = + infinity • From the chart, • C(∞)=S(∞)=0.5 • C(-∞)=S(-∞)= - 0.5 • Since there is no diffraction, Ep =E0 ! • So, for Fresnel diffraction by any aperture:

11. Length = The infinite aperture umax (1+i) A similar curve exists for v i umin  1

12. A few examples • Diffraction by a Straight Edge • Umin = -∞, Umax = +∞ • Vmax =+∞ • While the edge of the razor is at vmin=0, we can find the intensity for any value of v by shifting the coordinate system:

13. V=1 V=2 V=0 V=-2 V=-1.25 Diffraction by a Straight Edge

14. Another Fresnel Example Fresnel diffraction is observed behind a a long horizontal slit of width 3.7 mm thick, which is placed 2 m from the light source and 3 m from the screen. If light of wavelength 630 nm is used, compute, using the Cornu Spiral, the irradiance of the diffraction pattern on the axis at the screen. Express the answer as some number times the unobstructed irradiance there. 3 m 2 m 3.7 mm

15. Solution • r’0= 2 m, r0= 3 m • L=1.2 m Horizontal slit

16. The Cornu Spiral Length2=((0.4963+0.4963)2 + (0.6058+0.6058)2=2.453 Length2=((0.5+0.5)2 + (0.5+0.0.5)2=2 • S(3.0)=0.4963 • C(3.0)=0.6058 • S(-3.0)=-0.4963 • C(-3.0)= -0.6058 • Length2=2.453 • S(inf)=0.5 • C(inf)=0.5 • S(-inf)=-0.5 • C(-inf)= -0.5 • Length2=2