Fraunhofer Diffraction: Circular aperture

1. Fraunhofer Diffraction: Circular aperture Wed. Nov. 27, 2002

2. Fraunhofer diffraction from a circular aperture  y  P r x  Lens plane

3. Fraunhofer diffraction from a circular aperture Path length is the same for all rays = ro Do x first – looking down Why? 

4. Fraunhofer diffraction from a circular aperture Do integration along y – looking from the side  P +R  y=0  ro  -R r = ro - ysin

5. Fraunhofer diffraction from a circular aperture Let Then

6. Fraunhofer diffraction from a circular aperture The integral where J1() is the first order Bessell function of the first kind.

7. Fraunhofer diffraction from a circular aperture • These Bessell functions can be represented as polynomials: • and in particular (for p = 1),

8. Fraunhofer diffraction from a circular aperture • Thus, • where  = kRsin and Io is the intensity when =0

9. Fraunhofer diffraction from a circular aperture • Now the zeros of J1() occur at, • = 0, 3.832, 7.016, 10.173, … • = 0, 1.22, 2.23, 3.24, … • =kR sin = (2/) sin • Thus zero at sin  = 1.22/D, 2.23 /D, 3.24 /D, …

10. Fraunhofer diffraction from a circular aperture The central Airy disc contains 85% of the light

11. Fraunhofer diffraction from a circular aperture D  sin = 1.22/D

12. Diffraction limited focussing • sin = 1.22/D • The width of the Airy disc W = 2fsin  2f  = 2f(1.22/D) = 2.4 f/D W = 2.4(f#) >  f# > 1 • Cannot focus any wave to spot with dimensions <  f D  

13. Fraunhofer diffraction and spatial resolution • Suppose two point sources or objects are far away (e.g. two stars) • Imaged with some optical system • Two Airy patterns • If S1, S2 are too close together the Airy patterns will overlap and become indistinguishable S1  S2

14. Fraunhofer diffraction and spatial resolution • Assume S1, S2 can just be resolved when maximum of one pattern just falls on minimum (first) of the other • Then the angular separation at lens, • e.g. telescope D = 10 cm  = 500 X 10-7 cm • e.g. eye D ~ 1mm min = 5 X 10-4 rad

15. Polarization

16. Matrix treatment of polarization • Consider a light ray with an instantaneous E-vector as shown y Ey x Ex

17. Matrix treatment of polarization • Combining the components • The terms in brackets represents the complex amplitude of the plane wave

18. Jones Vectors • The state of polarization of light is determined by • the relative amplitudes (Eox, Eoy) and, • the relative phases ( = y - x) of these components • The complex amplitude is written as a two-element matrix, the Jones vector

19. y x Jones vector: Horizontally polarized light The arrows indicate the sense of movement as the beam approaches you • The electric field oscillations are only along the x-axis • The Jones vector is then written, where we have set the phase x = 0, for convenience The normalized form is

20. y x Jones vector: Vertically polarized light • The electric field oscillations are only along the y-axis • The Jones vector is then written, • Where we have set the phase y = 0, for convenience The normalized form is

21. y x Jones vector: Linearly polarized light at an arbitrary angle • If the phases are such that  = m for m = 0, 1, 2, 3, … • Then we must have, and the Jones vector is simply a line inclined at an angle  = tan-1(Eoy/Eox) since we can write  The normalized form is

22. Jones vector and polarization • In general, the Jones vector for the arbitrary case is an ellipse y Eoy b  a x Eox