Chapters 14 & 18: Matrix methods

1. Chapters 14 & 18: Matrix methods Chapters 14 & 18: Matrix methods

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

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

5. y y x x 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 • The state of polarization of light is completely determined by • the relative amplitudes (E0x, E0y) and • the relative phases (D = y- x) of these components. • The complex amplitude is written as a two-element matrix, or Jones vector for complex field components

7. Jones calculus (1941) Indiana Jones Ohio Jones

8. y x Jones vector for 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 • Normalized form:

9. y x Jones vector for 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 • Normalized form:

10. y x Jones vector for linearly polarized light at an arbitrary angle • The electric field oscillations make an angle a with respect to the x-axis • If a = 0°, horizontally polarized • If a = 90°, vertically polarized • Relative phase must equal 0: • x = y = 0 • Perpendicular component amplitudes: • E0x = A cos a • E0y = A sin a • Jones vector: 

11. Circular polarization • Suppose E0x = E0y = A, • and Ex leads Ey by 90o = /2 • At the instant Ex reaches its maximum displacement (+A), Ey is zero • A fourth of a period later, Ex is zero and Ey = +A • The vector traces a circular path, rotating counter-clockwise

12. Circular polarization • The Jones vector for this case – where Ex leads Ey 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 Replace /2 with -/2 to get

13. Elliptical polarization If E0x E0y, e.g. if E0x = A andE0y = B, the Jone vectors can be written as Direction of rotation? counterclockwise clockwise

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

16. matrices • 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

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

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

19. Jones matrix for a linear polarizer 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. Operating on horizontally polarized light, For a linear polarizer with TA vertical.

20. Jones matrix for a linear polarizer 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 q

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

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

23. Jones matrix for a phase retarder • 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

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

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

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. Multiplying Jones matrices 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.

29. x z x-pol y y-pol rotated x-pol y-pol Multiplying Jones matrices Crossed polarizers: so no light leaks through. Uncrossed polarizers: SoIout≈ e2Iin,x

30. Matrix methods for complex optical systems 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

31. The ray vector A light ray can be defined by two coordinates: optical ray position (height), y islope, a a y optical axis 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. System ray-transfer matrices Optical system ↔ 2 x 2 ray matrix

33. matrices • 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

35. Multiplying ray matrices O2 O1 O3 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 (Pedrotti3). 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