Fakultät für Physik und Geowissenschaften Physikalisches Grundpraktikum Experiment O11e Optical Polarisation Tasks 0. During preparation for the laboratory experiment, familiarize yourself with the function of a λ/4 plate. Calculate the light intensity for the following forms of polarised light passing through a linear analyser as a function of the analyser angle θ with the vertical direction. Consider a linear polarised light angle α between polarisation direction and vertical, b circularly polarised light, c linearly polarised light passing first through a λ/4 plate before entering the linear analyser, and d circularly polarised light passing first through a λ/4 plate before entering the analyser angle ϕ between fast axis of the λ/4 plate and the vertical. 1. Measure the light intensity as function of the analyser angle θ for linear polarised light that passed through a λ/4 plate under angles of a ϕ = 0, b ϕ = 30 and c ϕ = 45. Compare to the calculations from the first task by making nonlinear fits to the data. 2. Characterise the state of polarisation of three unknown black boxes in front of a light source using appropriate measurements. Literature Physikalisches Praktikum, 14. Auflage, Hrsg. W. Schenk, F. Kremer, Springer Spektrum, Optik, 4 in German Optics, 4th edition, Eugene Hecht, chapter 8, pp. 325. Instrumentation LED λ Peak = 590 nm, polariser, λ/4 foil, laboratory power supply, photodiode, digital multimeter, frames with unknown polarisers 1
Experimental Setup The experimental setup Fig. 1 consists of a collimated LED light source, b polariser/black box, c λ/4 plate, d analyser, e lens and f photodiode. Abbildung 1: Top: schematic setup of the polarisation experiment, bottom: image of the actual setup Jones Vectors and Jones Matrices An electromagnetic wave with wave vector k = kẑ propagates in z-direction. The electric field vector E is perpendicular to the propagation direction and can be represented by E = E x ˆx + E y ŷ, where E x and E y are complex numbers. In polar form with real valued positive amplitudes E 0x and E 0y as well as phase angles δ x and δ y one obtains: E = E0x exp iδ x E 0y exp iδ y, 1 where the common position- and time-dependent phase factor exp iωt k r was omitted. When the electric field vector is normalized to its absolute value E = E0x 2 + E 0y 2, one obtains a unit vector 2
in the direction of the electric field: J = Jx J y = 1 E E0x exp iδ x E 0y exp iδ y. 2 These unit vectors are called Jones vectors. Since the zero-point of the phase can be arbitrarily chosen, only the phase difference δ = δ y δ x of both electric field vector components is physically relevant. In the following δ x = 0 is chosen. Linear Polarisation In case of linear polarisation the phase angles δ x = δ y are the same and one might conveniently choose as basis vectors the horizontal vector J x and the vertical vector J y : 1 Jx =, 0 Jy =. 3 0 1 The Jones vector of a general polarisation direction making an angle α with the vertical y-axis is Jα = sin α J x + cos α sin α J y =. 4 cos α Circular Polarisation In case of circular polarisation the amplitudes E 0x = E 0y are the same and the phase angles differ by δ = ±π/2. As basis vectors the Jones vectors for rightδ = π/2 J + and leftδ = π/2 J circularpolarised light are chosen: J+ = 1 1, J = 1 1. 5 2 i 2 +i The handedness is defined with respect to the viewing direction. If the observer looks in the direction of the incoming light, then the electric field vector rotates clockwise for right-circular and anti-clockwise for left-circular light. A comparison of equations 4 and 5 shows that the linear basis vectors can be expressed by the circular ones and vice versa. Optical Elements The effect of optical elements on the polarisation state can be represented by the multiplication with matrices. To this end the action of each optical element is modelled by a so-called Jones matrix. If the light beam passes various optical elements, the corresponding Jones vector is successively multiplied by Jones matrices. A linear polariser that only transmits light with a polarisation direction under an angle θ with the vertical is given by sin 2 θ sin θ cos θ M θ = sin θ cos θ cos 2. 6 θ If a linearly polarised light beam with vertical polarisation direction, represented by the vertical Jones vector J y, passes the polariser, the transmitted Jones vector is given by M θ sin 2 θ sin θ cos θ Jy = sin θ cos θ cos 2 θ 0 1 = sin θ cos θ cos 2 θ. 7 The transmitted light intensity I is proportional to the square of the absolute value of the transmitted Jones vector: I M θ Jy 2 = sin 2 θ cos 2 θ + cos 4 θ = cos 2 θ, 8 3
i.e. the law of Malus is correctly obtained from this formalism. A λ/n plate with main axes along the horizontal x- and the vertical y-axis is represented by expi δx 0 M =. 9 0 expi δ y In case of a λ/4 plate with x-axis as fast axis one has δ y δ x = π/2. With the choice δ x = π/4 and δ y = +π/4 one obtains Mλ/4 x 1 0 = exp iπ/4. 10 0 +i In case of a λ/2 plate with x-axis as fast axis one might choose δ x = π/2 and δ y = +π/2, such that Mλ/2 x 1 0 i 0 = exp iπ/2 =. 11 0 1 0 +i When the fast axis of the λ/n plate makes an angle ϕ with the vertical y-axis, the corresponding Jones matrix is obtained by transforming equations 10 or 11 with a rotation matrix. The angle between the x -axis of the rotated coordinate system and the x-axis of the original system is called ϕ ; then the rotation matrix is Dϕ cos ϕ sin ϕ = sin ϕ cos ϕ. 12 Since in the present experiment the angles are usually referred to the vertical direction, it is more convenient to introduce the angle ϕ between vertical y-axis and rotated x -axis with ϕ = π/2 ϕ. The rotation matrix is then sin ϕ cos ϕ Dϕ = 13 cos ϕ sin ϕ and the transformed Jones matrix of a λ/4 plate is: M ϕ λ/4 = DϕMx λ/4 DT ϕ, 14 where D T ϕ denotes the transposed matrix of the orthogonal matrix Dϕ. One obtains M ϕ sin 2 ϕ + i cos λ/4 = exp iπ/4 2 ϕ 1 i sin ϕ cos ϕ 1 i sin ϕ cos ϕ cos 2 ϕ + i sin 2 ϕ = 1 1 + i cos2ϕ i sin2ϕ, 15 2 i sin2ϕ 1 i cos2ϕ and especially M 0 λ/4 J y = exp iπ/4 J y 16 M π/4 λ/4 J y = i J 17 M π/4 λ/4 J y = exp+iπ/4 J y 18 M π/4 λ/4 J y = +i J +, 19 i.e. linear polarised light passing a λ/4 plate parallel to the fast ϕ = 0 or slow ϕ = π/2 axis remains in a linear polarisation state, whereas it becomes circularly polarised, when the fast axis of the λ/4 plate makes angles ±π/4 with the direction of the incoming linear polarization. On the other hand, in case of right-circularly-polarised light passing a λ/4 plate with the fast axis making an angle ϕ with the vertical: M ϕ λ/4j + = 1 [ cos2ϕ 1 + i 2 1 sin2ϕ ] 1 sin2ϕ cos2ϕ 4 = exp i π 2 + iβ sin β cos β 20
with cot β = cos2ϕ/ 1 sin2ϕ or β = ϕ π/4. Circularly-polarised light is always turned into linearly polarised light after passing a λ/4 plate. The polarisation direction of the linearly polarised light is determined by the direction of the fast axis of the λ/4 plate and is rotated against that direction anti-clockwise by 45. General Polarisation State and Stokes Parameter Elliptical Polarisation In this section we start again with Eq. 1, the convention δ x = 0 and the phase difference δ = δ y : Ex E E = = 0x. 21 E y E 0y exp iδ For general values of the amplitudes E 0x and E 0y as well as the phase difference δ the tip of the electric field vector lies on an ellipse, formalized by Ex E 0x 2 + Ey E 0y 2 2 E x E 0x E y E 0y cos δ = sin 2 δ. 22 This ellipse is rotated with respect to the horizontal E x -axis by the angle ψ with 0 ψ < π, see Fig. 2. The amplitude ratio might be characterized by an angle ɛ via the equation E y ' E y E x ' -E 0y E 0y E 0y ' E 0x ' E x -E 0x E 0x Abbildung 2: Elliptical polarisation state Then the ellipse rotation angle is given by tan ɛ = E 0y E 0x 0 ɛ π/2. 23 tan2ψ = tan2ɛ cos δ. 24 In general the angle ψ does not vanish, i.e. the x, y coordinate system is not the principal axis system of the ellipse. A principal axis transformation into the coordinate system x, y yields a major half-axis E 0x and a minor half-axis E 0y E 0x, see Fig. 2 with E 2 0x + E 2 0y = E 2 0x + E 2 0y. 25 5
The eccentricity of the ellipse is characterized by the angle χ with tan χ = E 0y E 0x, π/4 χ π/4. 26 The sign of the half-axis ratio yields the handedness of the elliptical polarisation with right-handed light: 0 χ π/4, sin δ > 0 and left-handed light: π/4 χ 0, sin δ < 0. A phase difference δ = mπ with m = 0, ±1, ±2,... corresponds to a linear polarisation; the condition E 0x = E 0y with a phase difference δ = ±π/2 corresponds to circular polarisation. Stokes Parameter The Stokes parameter s 0, s 1, s 2, s 3 of monochromatic light are given by s 0 = E 2 0x + E 2 0y 27 s 1 = E 2 0x E 2 0y 28 s 2 = 2E 0x E 0y cos δ 29 s 3 = 2E 0x E 0y sin δ. 30 These characterize the polarisation state of the light completely. The parameter s 0 is proportional to the measured intensity. Using the Jones matrices for linear polarisers 6 and the electric field vector 21 of the elliptical polarisation state, the parameters s 1 and s 2 can be calculated as s 1 = 2 M π/2 E 2 s 0 31 s 2 = 2 M π/4 E 2 s 0, 32 i.e. s 1 indicates whether the light is more horizontally or vertically linear polarised, whereas s 2 specifies the degree of polarisation with respect to a linear polariser with a direction under 45. s 3 yields the degree of right circular polarization which can be calculated using the Jones matrix M + M + = M with M + = 1 1 +i 33 2 i 1 such that The four parameters are not independent, but s 3 = 2 M + E 2 s 0. 34 s 2 0 = s 2 1 + s 2 2 + s 2 3. 35 The Stokes parameters s 1, s 2 and s 3 can be written in terms of the angles ψ and χ introduced above: s 1 = s 0 cos2χ cos2ψ 36 s 2 = s 0 cos2χ sin2ψ 37 s 3 = s 0 sin2χ 38 with 0 2ψ 2π and π/2 2χ π/2. This allows for a geometric definition. In the three-dimensional parameter space s 1, s 2, s 3 all polarisation states lie on a sphere with radius s 0 Poincaré sphere. The position of a polarisation state on this sphere is given in spherical coordinates with polar angle 2χ and azimuthal angle 2ψ just by equations 36 38. Using Jones vectors and Jones matrices the values of the Stokes parameters can be easily calculated for certain polarisation states. In case of a light beam with linear polarisation that has a polarisation direction under an angle α to the vertical, one obtains with the Jones vector 4 the Stokes parameters here written as a four-component vector s 0 s 1 s 2 s 3 = s 0 1 cos2α sin2α 0. 39 6
The vector s 1, s 2, s 3 = s 0 cos2α, sin2α, 0 lies in the equatorial plane of the Poincaré sphere and rotates for increasing angles α clockwise around the s 3 -Achse, starting from 1, 0, 0, i.e. this vector points along the negative s 1 -direction 1, 0, 0, when the light beam is vertically linear polarised α = 0, in the positive s 2 -direction for a polarisation direction under +45 0, 1, 0, in the positive s 1 - direction for horizontal polarization α = π/2 and in the negative s 2 -Richtung for a polarisation direction along 135. This is consistent with equations 36 38, 24 and 26. In case of a linear polarisation one has δ = 0 and a vanishing eccentricity of the ellipse, i.e. from 26 follows χ = 0 and from 24 one obtains ψ = ɛ. On the other hand, 23 and 4 yield E 0y /E 0x = tan ɛ = cot α. This is equivalent to cos2ψ = cos2α and sin2ψ = sin2α in agreement with the general representation of the Stokes parameters in equations 36 38 and the Stokes parameters of the linearly polarised state 39. The circular polarisation states are located at the poles of the Poincaré sphere, the right-circular at the North and the left-circular at the South Pole. In these cases the eccentricities of the ellipses are equal to ±1, see equation 26, i.e. the angle 2χ is +90 right-circular or 90 left-circular; the angle ψ is undetermined in this case. Elliptically polarised states with preferential right- or left-circular character lie on the northern or southern hemisphere of the Poincaré sphere. A simple operational guide The Stokes parameters are determined by six intensity measurements. To this end the intensities of the various polarisation states are measured: I 0 : Intensity of the light component with vertical linear polarisation. I π/2 : Intensity of the light component with horizontal linear polarisation. I π/4 : Intensity of the light component with linear polarisation under 45. I π/4 : Intensity of the light component with linear polarisation under 45. I + : Intensity of the light component with right-circular polarisation. I : Intensity of the light component with left-circular polarisation. With the electric field vector E from Eq. 21 and the Jones matrices one can easily show that I 0 M 0E 2 = E0y 2, I π/2 M π/2e 2 = E0x 2, I ±π/4 M ±π/4e 2 = E0x 2 + E 0y 2 ± 2E 0xE 0y cos δ /2 and I r,l M r,le 2 = E0x 2 + E 0y 2 ± 2E 0xE 0y sin δ /2, such that s 0 s 1 s 2 s 3 I π/2 + I 0 I π/2 I 0 I π/4 I π/4 I + I. 40 If it is further assumed that the light is not depolarised, as has been implicitly assumed in the definitions 27 30, the number of intensity measurements might be reduced to four, i.e. the measurement of the intensity I = I π/2 + I 0 = I π/4 + I π/4 = I + + I of the unpolarized light as well as the measurement of the intensities I π/2, I π/4 and I +. After Eqs. 31, 32 and 34 one has: s 0 s 1 s 2 s 3 I 2I π/2 I 2I π/4 I 2I + I. 41 If one has to analyze the unknown polarisation state of a light beam, one might measure the four intensities defined above and calculate the Stokes parameters. On the other hand, the following simple guide might be of help. This uses a rotatable linear analyser and a λ/4 plate. If the intensity variation is monitored when the linear analyser is rotated, with the λ/4 plate inserted before the analyser or not, the following conclusions can be drawn: 7
1. No intensity variation with analyser alone Insert λ/4 plate in front of analyser with fast axis in arbitrary direction a No intensity variation for all analyser angles natural unpolarized light b Intensity minimum for one analyser position i. Zero intensity for one analyser position circularly polarised light ii. Non-zero intensity for all analyser angles mixture of circularly polarised and unpolarised light 2. Intensity variation with analyser alone a Zero intensity for one analyzer position linearly polarised light b Non-zero intensity for all analyzer angles Insert λ/4 plate in front of analyzer with fast axis parallel to analyzer direction at maximum or minimum intensity i. Zero intensity for one analyzer position elliptically polarised light ii. Non-zero intensity for all analyzer angles A. Minimum intensity at the same analyzer setting as without λ/4 plate mixture of linearly polarised and unpolarised light B. Minimum intensity at different analyzer setting as without λ/4 plate mixture of elliptically polarised and unpolarised light 8