LC45-summer, 1 1. 1.1, 7.1, 7., 7.5, 7.11, 7.1, 7.15, 7.1 1.1. TIR and FTIR a) B considering the electric field component in medium B in Figure 1. (b), eplain how ou can adjust the amount of transmitted light. Reflected light B = Low refractive inde transparent film ( n ) Reflected (a) n n n 1 1 TIR n1 i > c Transm itted Incident A light Glass prism FTIR i > c C A Incident light (b) (a) A light incident at t he long face of a glass prism suffers T IR; t he prism deflects the light. (b) T wo prisms separated b a thin low refractive inde film forming a beam-splitt er cube. T he incident beam is split into two beams b FTIR. 1999 S.O. Kasap, Optoelectronics (Prentice Hall) Figure 1.. from Optoelectronics, S. O. Kasap b) What is the critical angle at the hpotenuse face of a beam splitter cube made of glass with n₁ = 1.6 and having a thin film of liquid with n₂ = 1.. Can ou use 45 prism with normal incidence? c) plain how a light beam can propagate along a laer of material between two different media as shown in Figure 1.4 (a). plain what the requirements are for the indices n₁, n₂, n₃. Will there be an losses at the reflections? d) Consider the prism coupler arrangement in Figure 1.4 (b). plain how this arrangement works for coupling an eternal light beam from a laser into a thin laer on the surface of a glass substrate. Light is then propagated inside the thin laer along the surface of the substrate. What is the purpose of the adjustable coupling gap?
LC45-summer, 1 Air n n 1 Thin laer Laser light Thin laer Prism d = Adjustable coupling gap Glass substrate (a) n Glass substrate (b) (a) Light propagation along an opt ical guide. (b) Coupling of laser light int o a t hin laer - optical guide - using a prism. The light prop agates along t he t hin laer. 1999 S.O. Kasap, Optoelectronics (Prentice Hall) Figure 1.4. from Optoelectronics, S. O. Kasap Solution. a) Consider the prism A when the neighboring prism C in Figure 1. (b) in far awa. When the light beam in prism A is incident on the A/B interface, hpotenuse face, it suffers TIR as θ i > θ c. There is however an evanescent wave whose field decas eponentiall with distance in medium B. When we bring prism C close to A, the field in B will reach C and consequentl penetrates C. (The tangential field must be continuous from B to C). One cannot just use the field epression for the evanescent wave because this was derived for a light beam incident at an interface between two media onl; no third medium. The transmitted light intensit from A to C depends on the thickness of B. b) For the prism A in Figure 1. (b), n₁ = 1.6 and n₂ = 1. so that the critical angle for TIR at the hpotenuse face is c sin 1 n n 1 sin 1 1. 54. 1.6 in this case 45 prism cannot be emploed, Figure 1. (a). c) If the angle of incidence θ i at the n₁/n₂ laer is more than the critical angle θ c1 and if angle of incidence θ i at the n₁/n₃ laer is more than the critical angle θ₁₃ then the light ra will travel b TIR zigzagging between the boundaries as sketched in Figure 1. (a). For eample, suppose that n₁ = (thin laer), n₂ = 1 (air), n₃ = 1.6 (glass),
LC45-summer, 1 c1 sin 1 n n 1 sin 1 1 8.8 1 n 1 1.6 c1 sin sin 5.1 n1 So that θ i > 5.1 will satisf TIR. There is no loss in TIR as the magnitude of the amplitude of the reflected wa is the same as that of the incident wave. Note: There is an additional requirement that the waves entering the thin film interfere constructivel, otherwise the waves will interfere destructivel to cancel each other. Thus there will be an additional requirement, called the waveguide condition (Chapter ). d) The light ra entering the prism is deflected towards the base of the prism. There is a small gap between the prism and the thin laer. Although the light arriving at the prism base/gap interface is reflected, because of the close proimit of the thin laer, some light is coupled into the thin laer as it was discussed in part (a) due to frustrated TIR. This arrangement is a much more efficient wa to couple the light into the thin laer because the incident light is received b the large hpotenuse face compared with coupling the light directl into the thin laer. 7.1. Polarization. Suppose that we write the and components of a light wave generall as: cos t kz and cos t kz Show that at an instant and satisf the ellipse equation on the vs. coordinate sstem: cos sin Sketch schematics what this ellipse look like assuming =. When would this ellipse form an (a) ellipse with its major ais on the -ais, (b) a linearl light at 45 to the -ais, (c) right and left circularl light?
LC45-summer, 1 4 Solution. Consider the LHS of equation cos cos cos cos cos cos cos cos cos cos cos t kz cos cos t kz t kz cost kz t kz cos t kz cost kzcost kz t kzcost kzcos sint kzsin t kzcost kzcos sint kzsincos t kz cos t kzcos sin t kzsin t kzsint kzcos sin t kzcos cost kzsint kzcos sin t kz cos t kzcos sin t kzsin t kz1 cos sin t kz sin cos t kzsin sin t kzsin [cos t kz sin t kz]sin sin cos The equation is proved. Consider the general epression cos sin This is a quadratic equation in (or ), a where b 1 a o ; c b 1 cos; c sin Thus, for a given ϕ and a given and, the graph vs. can be sketched where determines b and c and is given b
LC45-summer, 1 5 b b 4ac a If = then a) for ϕ = π/, = 1 and = we will obtain an ellipse with its major ais on the -ais
LC45-summer, 1 6 b) For ϕ =, = 1 and = 1 we obtain the line (linear) polarization at an angle π/4 to the -ais c) For ϕ = π/, = 1 and = 1we obtain a circle, if ϕ= π/ right circular polarization; shift ϕ = -π/ results in left circular polarization
LC45-summer, 1 7 7.. Linear and circular polarization Show that a linearl light wave can be represented b two circularl light waves with opposite rotations. Consider the simplest case of a wave linearl along the -ais. What is our conclusion? Solution. if the wave is along the -direction then circularl light has retarded electric field in -direction and ₀ = = = 1 for right circularl light: R cost kz, R cos t kz for left circularl light: L cost kz, L cos t kz Total and components of two polarizations are
LC45-summer, 1 8 cost kz cost kz cos t kz There is onl one component which proves the linear polarization along -ais In case of the linear polarization in -ais, the retarded component will be in - direction ( R cost kz, cos t kz ) + = 7.5. Jones Matrices When we represent the state of polarization of a light wave using a matri, called a Jones matri (or vector) then various operations on the polarization state correspond to multipling this matri with another matri that represents the optical operation. Consider a light wave travelling along z with field components and along and have a phase difference ϕ between them. If we use the eponential notation then ep jt kz and ep j t kz Jones matri is a column matri whose elements are and without the common ep[j(ωt-kz)] factor
LC45-summer, 1 9 ep j (1) ep j Usuall q. (1) is normalized b dividing b the total amplitude 1/. We can also factor out ep(jϕ ) to further simplif to obtain the Jones matri: where ϕ=ϕ ϕ. 1 J () ep j a) Table 7. shows Jones vectors for various polarizations. Identif the state of polarization for each matri. b) Passing a wave of given Jones vector J in through an optical device is represented b multipling J in b the transmission matri T of the device. If J out is the Jones vector for the output light through the device, then J out = T J in. Given 1 T () j Determine the polarization state of the output wave given the Jones vectors in Table 7., and the optical operation represented b T. Hint: Use input for determining T. 1 as 1 Table 7. Jones vectors Jones vector 1 1 1 cos 1 1 1 1 J in 1 sin j j Polarization????? Transmission j 1 e 1 1 1 matri T j e j 1 j e Optical operation????? a) Polarization state for each matri Solution.
LC45-summer, 1 1 Jones vectors Jones vector J in Polarization 1 Linear; horizontal 1 1 1 Linear; at 45 to -ais cos sin Linear; at θ to -ais 1 1 j Right circularl 1 1 j Left circularl b) Optical operation represented b T is Jones vectors Jones vector 1 1 1 J in 1 Polarization Linear, Linear; at horizontal 45 to -ais Transmission matri T Optical operation 1 Linear polarizer; horizontal transmission ais j e j e Isotropic phase changer phase retarder or cos sin Linear; at θ to -ais 1 j Quarterwave plate 1 1 j Right circularl 1 1 Half-wave plate The polarization state of the output waves from quarter-wave plate are 1 1 j Left circularl 1 j e Wave retarder; fast ais along Jones vectors Jones vector J in Polarization Transmission matri T Jones vector J out 1 Linear horizontal polarization, horizontal 1 j 1 Linear horizontal polarization, horizontal 1 1 1 Linear; at 45 to -ais 1 j 1 1 j Right circularl cos sin Linear; at θ to -ais 1 j cos j sin Right circularl 1 1 j Right circularl 1 j 1 1 1 Linear; at (π-π/4) to -ais 1 1 j Left circularl 1 j 1 1 1 Linear; at π/4 to -ais
LC45-summer, 1 11 The polarization state of the output waves are Jones vectors Jones vector J in Polarization Transmission matri T Optical operation Jones vector J out 1 Linear, horizontal 1 Linear polarizer; horizontal transmission ais 1 Linear, horizontal polarization 1 1 1 Linear; at 45 to -ais j e j e Isotropic phase changer or phase retarder j e 1 1 Linear, at 45 to -ais with leading cos sin Linear; at θ to -ais 1 j Quarterwave plate cos j sin Right elliptical polarization 1 1 j Right circularl 1 1 Half-wave plate 1 1 j Left circular 1 1 j Left circularl 1 j e Wave retarder; fast ais along 1 1 j e Linear, retarded - component, fast ais along 7.11. Glan-Foucault prism Figure 7.7 shows the cross section of a Glan-Foucault prism which is made of two right angle calcite prisms with a prism angle of 8.5. Both have their optic aes parallel to each other and to the block faces as in the figure. plain the operation of the prisms and show that the o-wave does indeed eperience total internal reflection.
LC45-summer, 1 1 Solution. Calcite is opticall anisotropic material, birefringent. This is negative uniaial crstal since two of their principle indices the same (n₁ = n₂) and n₃ < n₁. The light has two orthogonall components in uniaial crstal, (o) ordinar and (e) etraordinar waves. The refractive indices (from Table 7.1) are n o = 1.658 and n e = 1.486. The critical angles for TIR in e- and o-waves are c c o wave arcsin 1/ no e wave arcsin 1/ n 4. e 7.9 If the angle of incidence is θ at calcite/air interface then from Figure 7.7 9 +(9 -θ)+8.5 = 18 θ = 8.5 > θ c (o-wave) θ = 8.5 < θ c (e-wave) Thus, the o-wave suffers TIR while the e-wave does not. Hence the beam that emerges is the e-wave, with a field e along optic ais.
LC45-summer, 1 1 Absorbe r o -ra θ Calcite 8.5 e -ra Optic ais Air-ga p The Glan-Foucault prism provides linearl light 1999 S.O. Kasap, Optoelectronics (Prentice Hall) 7.1. Farada ffect. Application of a magnetic field along the direction of propagation of a linearl light wave through a medium results in the rotation of the plane of polarization. The amount of rotation θ is given b BL where B is the magnetic field (flu densit), L is the length of the medium, and ϑ is the so-called Verdet constant. It depends on the material and the wavelength. In contrast to optical activit, sense of rotation of the plane of polarization is independent of the direction of light propagation. Given that glass and ZnS have Verdet constants of about and minutes of arc Gauss ¹meter ¹ at 589 nm respectivel, calculate the necessar magnetic field for a rotation of 1 over a length 1 mm. What is the rotation per unit magnetic field for a medium of length 1 m? (Note: 6 minutes of arc = 1 and 1⁴ Gauss = 1Tesla). Rotation per unit magnetic field Solution. For glass: For ZnS: B L ' Gauss B L ' Gauss 6' meter 11 m 1 1 6' meter 11 m 1 1 Gauss or. Tesla 7 Gauss or.7 Tesla
LC45-summer, 1 14 7.15 Transverse Pockels cell with LiNbO₃ Suppose that instead of the configuration in Figure 7., the field is applied along the z-ais of the crstal, the light propagates along the -ais. The -ais is the polarization of the ordinar wave and z-ais that of the etraordinar wave. Light propagates through as o- and e-waves. Given that a = V/d, where d is the crstal length along z, the indices are n n, o, e n o n e 1 1 n r o 1 n r e a a Show that the phase difference between the o- and e-waves emerging from the crstal is, L o L 1 V n n n r n r d e e o e o 1 where L is the crstal length along the -ais. plain the first and second terms. How would ou use two such Pockels cells to cancel the first terms in the total phase shift for the two cells? If the light beam entering the crstal is linearl in the z-direction, show that n el L ne r V d Consider a nearl monochromatic light beam of the free-space wavelength λ = 5 nm and polarization along z-ais. Calculate the voltage V π needed to change the output phase ϕ b π given a LiNbO₃ crstal with d/l =.1 (see Table 7.). Solution. Consider the phase change between the two electric field components
LC45-summer, 1 15 L o L 1 V n n n r n r d e e o e o 1 The first term is the natural birefringence of the crstal and occurs all the time, even without applied electric field. The second term is the Pockels effect, applied field inducing a charge in the refractive indices. L L z a d Light a d ϕ z light Two transverse Pockels cell phase modulators together cancel the natural birefringence in each crstal. If the light is linearl with its field along z, we onl need to consider the etraordinar ra (ϕ₀ = ). e 1 ne L L nor The first term does not depend on the voltage. The voltage V that changes the output phase b π is 1 L n o r V d V d Or V d L n 9 51.1 1.187.81 or1 15.5 V 7.1. Optical Kerr effect Consider a material in which the polarization does not have the second order term: P or P / 1 1
LC45-summer, 1 16 The first term with the electric susceptibilit χ₁ corresponds to the relative permittivit ε r and hence to the refractive inde n o of the medium in the absence of the third order term, i.e. under low fields. The ² term represents the irradiance I of the beam. Thus, the refractive inde depends on the intensit of the light beam, a phenomenon called the optical Kerr effect: n n n I and n o 4n o And η=(μ₀/ε₀) 1/ = 1π = 77 Ω, is the impedance of the free space. a) Tpicall, for man glasses, χ₃ 1 ²¹ m²/w; for man doped glasses, χ₃ 1 ¹⁸ m²/w; for man organic substances, χ₃ 1 ¹⁷ m²/w; for semiconductors, χ₃ 1 ¹⁴ m²/w. Calculate n₂ and the intensit of light needed to change n b 1 ³ for each case. b) The phase ϕ at a point z is given b n n ni t z t z It is clear that the phase depends on the light intensit I and the change in the phase along z due to light intensit alone is n I t z As the light intensit modulates the phase, this is called self-phase modulation. Obviousl light is controlling light. When the light intensit is small n₂i n₀, obviousl the instantaneous frequenc t / Suppose we have an intense beam and the intensit I is time dependent I=I(t). Consider a pulse of light traveling along the z-direction and the light intensit vs. t shape is a Gaussian (this is approimatel so when a light pulse propagate in an optical fiber, for eample). Find the instantaneous frequenc ω. Is this still ω₀? How does the frequenc change with time, or across the light pulse? The change in the frequenc over the pulse is called chirping. Self-phase modulation therefore changes the frequenc spectrum of the light pulse during propagation. What is the significance of this result?
LC45-summer, 1 17 c) Consider a Gaussian beam in which intensit across the beam cross section falls with radial distance in a Gaussian fashion. Suppose that the beam is made to pass through a plate of nonlinear medium. plain how the beam can become self-focused? Can ou envisage a situation where diffraction effects tring to impose divergence are just balanced b self-focusing effects? Solution. a) The fractional change in the refractive inde is n n n n ni n I 4n Thus I n 4n If 1 or.1% n For glasses: n₀ 1.5, χ₃ = 1 ²¹ m²/w, I = 1. 1¹⁶ W/m² for doped glasses: n₀ 1.5, χ₃ = 1 ¹⁸ m²/w, I = 1. 1¹³ W/m² for organics: n₀ 1.5, χ₃ = 1 ¹⁷ m²/w, I = 1. 1¹² W/m² for semiconductors: n₀.5, χ₃ = 1 ¹⁴ m²/w, I = 5.5 1⁹ W/m² large intensities. b) the phase ϕ at a point z is given b n n n I t z t z So that the instantaneous frequenc is t n z I t t
LC45-summer, 1 18 When the light intensit is small n₂i n₀, ω = ϕ/ t=ω₀. We have an intense beam with time dependent intensit I(t). Consider a pulse of light traveling along the z-direction and the light intensit vs. t shape is a Gaussian. Then the instantaneous frequenc is, t n z I t t So that the frequenc changes with time, or across the light pulse. Since I/ t is rising at the leading edge and falling in the trailing edge, the two ends of the pulse contain different frequencies. The change in the frequenc over the pulse is called chirping. Self-phase modulation therefore changes the frequenc spectrum of the light pulse during propagation. The second term above results in the broadening of the frequenc spectrum and hence leads to more dispersion for a Gaussian pulse propagating in an optical fiber. c) If an intense Gaussian optical beam is transmitted through a plate of nonlinear medium, it will change the refractive inde in the medium with the maimum change of the refractive inde in the center of the beam. Therefore, the plate will act as a graded-inde medium and change the curvature of the wavefront. Under certain conditions the plate can act as a lens with a power-dependent focal length, producing a co-called self-focusing of the beam. Similarl, if an intense Gaussian beam propagates through a nonlinear medium, the medium can act as a gradedinde waveguide. In this case, under certain conditions, the self-focusing effect can compensate the divergence of the beam due to diffraction, and the beam will be confined to its self-created waveguide. Such self-guided beams are called spatial solutions.
LC45-summer, 1 19 Figure 1. propagation of the light through nonlinear medium. Ref. http://www.absoluteastronom.com/topics/gradient_inde_optics The intensit variation across the beam cross section leads to a similar refractive inde variation in the nonlinear medium. Thus, the medium resembles a graded inde guide or a GRIN rod and can focus the beam.