Chapter 10: Acousto-Optic Modulators ==================================== Source: Hunsperger, sixth edition (2009), Problems 10.1--10.10, printed pages 217--219. Optical wavelength inside the guide is :math:`\lambda_g=2\pi/\beta`, not the vacuum wavelength. Problem 10.1 — Why optical confinement reduces acoustic drive ------------------------------------------------------------- **Working.** The photoelastic relation gives :math:`\delta n\propto\sqrt{M_2P_a/A_a}`. To obtain a fixed phase or coupling strength over a fixed interaction length, a similar index perturbation is required. Reducing the acoustic cross section :math:`A_a` while preserving optical overlap reduces necessary total power. Guiding also keeps light overlapped with the acoustic perturbation over a long interaction distance. **Answer.** Concentrate acoustic energy where the optical mode resides, rather than driving an unnecessarily large bulk volume. **Check.** Compare equal interaction lengths, efficiencies, and overlap. Confinement does not eliminate acoustic attenuation or electrical transducer conversion loss. Problem 10.2 — Acoustic index modulation in quartz -------------------------------------------------- **Working.** Use the book's fused-quartz value :math:`M_2=1.51\times10^{-18}\,{\rm s^3/g} =1.51\times10^{-15}\,{\rm s^3/kg}`. The driven cross section is :math:`A=(3\times10^{-6})(100\times10^{-6})=3\times10^{-10}\,{\rm m^2}`. .. math:: \delta n=\sqrt{\frac{M_2P_a}{2A}} =\sqrt{\frac{1.51\times10^{-15}}{6\times10^{-10}}} =1.5864\times10^{-3}. **Answer.** About 0.00159 under the book's uniform-overlap convention. **Check.** The cgs-to-SI mass conversion is essential. The value depends on the selected material orientation and photoelastic coefficient. Problem 10.3 — Bragg angle and diffracted-beam width ---------------------------------------------------- **Working.** The supplied propagation constant gives :math:`\lambda_g=2\pi/(20.85\,\mu{\rm m}^{-1})=0.301352\,\mu{\rm m}`. With acoustic period 2.5 micrometres, .. math:: \theta_B=\sin^{-1}\frac{\lambda_g}{2\Lambda}=3.45533^\circ. This angle is from the acoustic wavefront, so the incident beam makes 86.5447 degrees with acoustic propagation; its diffracted counterpart makes 93.4553 degrees on the opposite side. For a uniform 4 mm optical aperture, the exact sinc-squared far-field full width at half maximum is :math:`0.885893\lambda_g/b=0.003824^\circ`. The book's approximate Eq. (10.15), :math:`\lambda_g/b`, gives 0.004317 degrees. **Answer.** Incidence 86.545 degrees from acoustic propagation; diffraction at the complementary Bragg direction, with the above full half-power width. **Check.** The 2 mm interaction length instead controls angular/spectral acceptance; it is not the optical-aperture width. Use the measured modal propagation constant, rather than substituting the bulk index 2.295. Problem 10.4 — Refraction before and after Bragg diffraction ------------------------------------------------------------ **Working.** Assume the butt faces are perpendicular to the nominal optical axis and acoustic propagation is transverse to it. The incident GaAs ray is tilted 3 degrees from that axis. Tangential momentum at the GaAs/LiNbO3 face gives :math:`n_{\rm LN}\sin\theta_B=3.59\sin3^\circ`. Thus .. math:: \Lambda=\frac{1.15}{2(3.59)\sin3^\circ}\,\mu{\rm m} \simeq3.06\,\mu{\rm m},\qquad \theta_{\rm air}=\sin^{-1}(3.59\sin3^\circ)\simeq10.83^\circ. **Answer.** About 3.06 micrometres acoustic wavelength; the diffracted beam exits about 10.83 degrees from the output-face normal, on the deflected side. **Check.** Treating the initial 3 degrees as an internal LiNbO3 angle ignores the stated GaAs input medium. A different face orientation would require a different Snell construction. Problem 10.5 — Transit-time bandwidth ------------------------------------- **Working.** Acoustic transit across 50 micrometres takes .. math:: \tau=\frac{50\times10^{-6}}{250} =2.0\times10^{-7}\,{\rm s}. The inverse transit time is only 5 MHz; detailed transfer functions change the numerical coefficient, not the two-order-of-magnitude mismatch. **Answer.** This geometry cannot reproduce a 100 MHz modulation envelope across the full optical interaction aperture. **Check.** Acoustic carrier frequency and modulation bandwidth are different. A fast carrier alone does not overcome aperture transit delay. Problem 10.6 — Interdigital pitch and optical deflection -------------------------------------------------------- **Working.** :math:`\Lambda=v_a/f_a=2800/10^8=28\,\mu{\rm m}`. In a conventional single-electrode alternating-polarity IDT, adjacent finger centres are separated by :math:`\Lambda/2=14\,\mu{\rm m}`; equal-polarity fingers repeat every 28 micrometres. For a 50 percent metal fill, finger width and edge gap are 7 micrometres. .. math:: \Delta\theta=2\sin^{-1}\frac{0.6328}{2(2.2)(28)} \simeq0.5886^\circ. **Answer.** 14 micrometre adjacent centre pitch and about 0.589 degrees internal optical deflection. **Check.** “Finger spacing” can mean edge gap or centre pitch; the dimensions above distinguish them explicitly. Problem 10.7 — One-degree Bragg operation ----------------------------------------- **Working.** The Bragg angle is half the optical deflection: .. math:: \Lambda=\frac{0.6328}{2(1.6)\sin1^\circ} =11.3308\,\mu{\rm m},\qquad f_a=\frac{2500}{11.3308\times10^{-6}}=220.637\,{\rm MHz}. **Answer.** An alternating IDT has adjacent centre pitch 5.6654 micrometres and same-polarity repeat 11.3308 micrometres. .. figure:: /_static/knowledge_base/worked_exercises/integrated_optics_6e/acousto_optic_geometry.svg :alt: Bragg input and diffracted wavevectors and alternating IDT fingers with half-wavelength adjacent pitch. :width: 100% The grating vector supplies transverse momentum; the two optical directions are separated by twice the Bragg angle. **Check.** If 1 degree meant total deflection instead of Bragg angle, the acoustic period would be approximately twice as large. Problem 10.8 — Analyzer at 500 megahertz ---------------------------------------- **Working.** With :math:`n=2`, :math:`\lambda_g=0.3164\,\mu{\rm m}`. At 500 MHz, acoustic period is 6 micrometres. The deflection is :math:`\Delta\theta\simeq\lambda_g/\Lambda=0.0527333` rad. Use the question's envelope-width definition of resolution: .. math:: N=\frac{2\Lambda b}{\lambda_g\ell} =\frac{2(6)(1000)}{(0.3164)(5000)}=7.5853. **Answer.** About 3.021 degrees deflection and 7.59 envelope-width resolution elements in the book's approximation. **Check.** This is not :math:`f_ab/v_a`; that expression incorrectly uses carrier frequency in place of the accepted acoustic bandwidth. The full 50--500 MHz operating range is not necessarily flat-efficiency bandwidth for one fixed Bragg alignment. Problem 10.9 — Analyzer at 100 megahertz ---------------------------------------- **Working.** Acoustic wavelength increases to :math:`\Lambda=3000/(100\times10^6)=30\,\mu{\rm m}`. Keep optical beam width and interaction length unchanged: .. math:: \Delta\theta\simeq0.3164/30=0.0105467\,{\rm rad} =0.604279^\circ,\qquad N=\frac{2(30)(1000)}{(0.3164)(5000)}=37.9267. **Answer.** About 0.6043 degrees deflection and 37.93 envelope-width resolution elements. **Check.** Relative to 500 MHz, deflection is divided by five, whereas the approximate Bragg acceptance envelope grows by five. Problem 10.10 — Measured modulation depth and inconsistent units ---------------------------------------------------------------- **Working.** The prompt labels the initial quantity as intensity (200 mW/cm2) but the final quantity as power (175 mW). These cannot be divided without an illuminated area. If the intended final intensity is 175 mW/cm2, the removed zeroth-order fraction is .. math:: \eta_{\rm diff}=\frac{200-175}{200}=0.125. **Answer.** Under that explicit same-intensity-unit interpretation, 12.5 percent of the original beam is modulated out of the zeroth order. If “modulation depth” means :math:`(I_{\max}-I_{\min})/(I_{\max}+I_{\min})`, it is :math:`25/375=6.667` percent. **Check.** With the mixed units taken literally, no unique numerical depth follows. The quoted acoustic power does not supply the missing illuminated area.