1. Chapters 14 & 18: Matrix methods

2. http://www.youtube.com/watch?v=kXxzkSl0XHk Welcome to the Matrix

4. Let’s start with polarization… light is a 3-D vector field linear polarization circular polarization y x z

5. y x x y Plane waves with k along z direction oscillating (vibrating) electric field Any polarization state can be described as linear combination of these two: “complex amplitude” contains all polarization info

6. Vector representation for complex field components The state of polarization of light is completely determined by the relative amplitudes ( E 0x, E 0y ) and the relative phases (  =  y   x ) of these components. The complex amplitude is written as a two-element matrix, or Jones vector

7. Jones calculus (1941) Indiana JonesOhio Jones

8. 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 Normalized form: Jones vector for horizontally polarized light

9. 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 x y Normalized form: Jones vector for vertically polarized light

10.  Jones vector for linearly polarized light at an arbitrary angle The electric field oscillations make an angle  with respect to the x-axis If  = 0°, horizontally polarized If  = 90°, vertically polarized Relative phase must equal 0:  x =  y = 0 Perpendicular component amplitudes: E 0x = A cos  E 0y = A sin  Jones vector: x y

11. Circular polarization Suppose E 0x = E 0y = A, and E x leads E y by 90 o =  /2 At the instant E x reaches its maximum displacement ( +A ), E y is zero A fourth of a period later, E x is zero and E y = +A The vector traces a circular path, rotating counter- clockwise

12. Replace  /2 with -  /2 to get Circular polarization The Jones vector for this case – where E x leads E y is The normalized form is - This vector represents circularly polarized light, where E rotates counterclockwise, viewed head-on -This mode is called left-circularly polarized (LCP) light Corresponding vector for RCP light: LCP RCP

13. Direction of rotation? counterclockwise clockwise Elliptical polarization If E 0x  E 0y, e.g. if E 0x = A and E 0y = B, the Jone vectors can be written as

15. General case resultant vibration due to two perpendicular components if,a line else, an ellipse Lissajous figures

16. various optical elements modify polarization 2 x 2 matrices describe their effect on light matrix elements a, b, c, and d determine the modification to the polarization state of the light matrices

17. Optical elements: 1. Linear polarizer 2. Phase retarder 3. Rotator

18. Selectively removes all or most of the E-vibrations except in a given direction TA x y z Linear polarizer Optical elements: 1. Linear polarizer

19. Consider a linear polarizer with transmission axis along the vertical (y). Let a 2x2 matrix represent the polarizer operating on vertically polarized light. The transmitted light must also be vertically polarized. For a linear polarizer with TA vertical. Operating on horizontally polarized light, Jones matrix for a linear polarizer

20. For a linear polarizer with TA vertical For a linear polarizer with TA horizontal For a linear polarizer with TA at 45° For a linear polarizer with TA at 

21. Introduces a phase difference (Δ  ) between orthogonal components The fast axis (FA) and slow axis (SA) are shown when  /2 :quarter-wave plate  :half-wave plate FA x y z Retardation plate SA Optical elements: 2. Phase retarder

22.  /2  ¼ and ½ wave plates net phase difference: Δ  =  /2 — quarter-wave plate Δ  =  — half-wave plate

23. We wish to find a matrix which will transform the elements as follows: It is easy to show by inspection that, Here  x and  y represent the advance in phase of the components Jones matrix for a phase retarder

24. Jones matrix for a quarter wave plate Consider a quarter wave plate for which |Δ  | =  /2 For  y -  x =  /2 (Slow axis vertical) Let  x = -  /4 and  y =  /4 The matrix representing a quarter wave plate, with its slow axis vertical is, QWP, SA vertical QWP, SA horizontal

25. HWP, SA vertical HWP, SA horizontal Jones matrix for a half wave plate For |Δ  | = 

26. x y Rotator SA  Optical elements: 3. Rotator rotates the direction of linearly polarized light by a 

27. Jones matrix for a rotator An E -vector oscillating linearly at  is rotated by an angle  Thus, the light must be converted to one that oscillates linearly at (  +  )

28. To model the effects of more than one component on the polarization state, just multiply the input polarization Jones vector by all of the Jones matrices: Remember to use the correct order! A single Jones matrix (the product of the individual Jones matrices) can describe the combination of several components. Multiplying Jones matrices

29. Crossed polarizers: x y z x-pol y-pol so no light leaks through. Uncrossed polarizers: rotated x-pol y-pol So I out ≈  2 I in,x Multiplying Jones matrices

30. In dealing with a system of lenses, we simply chase the ray through the succession of lens. That is all there is to it. Richard Feynman Feynman Lectures in Physics Matrix methods for complex optical systems

31. The ray vector A light ray can be defined by two coordinates: position (height), y islope,  optical axis optical ray y  These parameters define a ray vector, which will change with distance and as the ray propagates through optics: To “chase the ray,” use ray transfer matrices that characterize translation refraction reflection …etc. Note to those using Hecht: vectors are formulated as

32. Optical system ↔ 2 x 2 ray matrix System ray-transfer matrices

33. 2 x 2 matrices describe the effect of many elements can also determine a composite ray-transfer matrix matrix elements A, B, C, and D determine useful properties matrices

35. O1O1 O3O3 O2O2 Multiplying ray matrices Notice that the order looks opposite to what it should be, but it makes sense when you think about it.

36. Exercises You are encouraged to solve all problems in the textbook (Pedrotti 3 ). The following may be covered in the werkcollege on 20 October 2010: Chapter 14: 2, 5, 13, 15, 17 Chapter 18: 3, 8, 11, 12