Propagation of waves - UMD Physics

Propagation of waves - UMD Physics

Polarization Jones vector & matrices Phys 375 1 Matrix treatment of polarization Consider a light ray with an instantaneous Evector as shown E z, t iE x z , t jE y z , t y Ey Ex x E x Eox e i kz t x E y Eoy e i kz t y

2 Matrix treatment of polarization Combining the components i kz t y i kz t x E i Eox e jEoy e i y i x i kz t E i Eox e jEoy e e ~ i kz t E Eo e

The terms in brackets represents the complex amplitude of the plane wave 3 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 twoelement matrix, the Jones vector ~ i x Eox Eox e Eox

~ i x E o ~ e i i y Eoy Eoy e Eoy e 4 Jones vector: Horizontally polarized light The electric field oscillations are only along the x-axis The Jones vector is then written, ~ Eox Eox e i x A 1 ~ Eo ~ A 0 E

oy 0 0 where we have set the phase x = 0, for convenience The arrows indicate the sense of movement as the beam approaches you y E x The normalized form is 1 0 5 Jones vector: Vertically polarized light

The electric field oscillations are only along the y-axis The Jones vector is then written, ~ Eox 0 0 0 ~ E o ~ A i y Eoy Eoy e A 1 Where we have set the phase y = 0, for convenience y E x The normalized form is

0 1 6 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, Ex m Eox 1 Ey Eoy y x and the Jones vector is simply a line inclined at an angle = tan-1(Eoy/Eox) since we can write

~ E ~ m cos ox Eo ~ A 1 sin E oy 7 Circular polarization y Suppose Eox = Eoy = 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 x t=0, Ey = 0, Ex = +A t=T/8, Ey = +Asin 45o, Ex = Acos45o t=T/4, Ey = +A, Ex = 0 8 Circular polarization The Jones vector for this case where Ex leads Ey is i 1 ~ Eox e x A Eo i A i y 2 Eoy e Ae i

The normalized form is, This vector represents circularly polarized light, where E rotates counterclockwise, viewed head-on This mode is called left-circularly polarized light What is the corresponding vector for right-circularly polarized light? 1 2 1 i Replace /2 with -/2 to get 1 2 1 i 9

Elliptically polarized light If Eox Eoy , e.g. if Eox=A and Eoy = B The Jones vector can be written A iB counterclockwise A iB clockwise Here A>B 10 Jones vector and polarization In general, the Jones vector for the arbitrary

case is an ellipse ( m; (m+1/2)) y Eoy Eox A ~ Eo i B cos i sin E e oy b a tan 2 2 Eox Eoy cos E ox2 E oy2 x Eox

11 Optical elements: Linear polarizer Selectively removes all or most of the Evibrations except in a given direction TA y x Linear polarizer 12 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. Thus, a b 0 0 c d 1 1 Operating on horizontally polarized light,

a b 1 0 c d 0 0 Thus, 0 0 M 0 1 Linear polarizer with TA vertical. 13 Jones matrix for a linear polarizer For a linear polarizer with a transmission axis at 2 cos M sin cos sin cos

2 sin 14 Optical elements: Phase retarder Introduces a phase difference () between orthogonal components The fast axis(FA) and slow axis (SA) are shown FA y x SA Retardation plate 15 Jones matrix of a phase retarder We wish to find a matrix which will transform the elements as follows: E e i x int o E e i x x ox

Eoy e ox i y int o Eoy e i y y It is easy to show by inspection that, Here x and y components e i x 0 M i y 0 e

represent the advance in phase of the 16 Jones matrix of a Quarter Wave Plate Consider a quarter wave plate for which || = / 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, e i 4 M 0 i 1 0

4 e 0 i 4 e 0 i 17 Jones matrices: Half-wave Plate For || = e i 2 M 0 e i 2 M 0 i 1 0 2

e 0 i e 2 i 1 0 2 e 0 i e 2 0 1 0 1 HWP, SA vertical HWP, SA horizontal 18

Optical elements: Quarter/Half wave plate When the net phase difference = /2 : Quarter-wave plate = : Half-wave plate /2 19 Optical elements: Rotator Rotates the direction of linearly polarized light by a particular angle y x SA Rotator 20 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 ( + ) a c b cos cos d sin sin cos One then finds M sin sin cos 21

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