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

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

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

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

5 Fraunhofer diffraction from a circular aperture Let Then

6 Fraunhofer diffraction from a circular aperture The integral where J 1 (  ) 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 I o is the intensity when  =0

9 Fraunhofer diffraction from a circular aperture Now the zeros of J 1 (  ) occur at,  = 0, 3.832, 7.016, , …  = 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 < D f  

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 S 1, S 2 are too close together the Airy patterns will overlap and become indistinguishable S1S1 S2S2 

14 Fraunhofer diffraction and spatial resolution Assume S 1, S 2 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 cm e.g. eye D ~ 1mm  min = 5 X rad

15 Polarization

16 Matrix treatment of polarization Consider a light ray with an instantaneous E- vector as shown x y ExEx EyEy

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 (E ox, E oy ) and,  the relative phases (  =  y -  x ) of these components The complex amplitude is written as a two- element matrix, the Jones vector

19 Jones vector: Horizontally polarized light 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 x y The arrows indicate the sense of movement as the beam approaches you The normalized form is

20 x y 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 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 (E oy /E ox ) since we can write x y  The normalized form is

22 Jones vector and polarization In general, the Jones vector for the arbitrary case is an ellipse a b E ox E oy x y 