4500T105 - Illuminance(10 Name Laws of Refraction(10(Please...

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Unformatted text preview: Illuminance (10) Name. Laws of Refraction (10) (Please Print) D—to—A Converter (20) FI‘IR Spectrometer (20) DIC Microscopy (20) Concave Mirror (20) E C E 4500 First Examination February 14, 2005 Rules for exam: ,ox MAL» 00>] 10. 11. 12. 13. 14. 15. . The time allowed is 60 minutes. The test will start at 9:00am and end at 10:00am. Answer all questions. The value of each question is given in parentheses by the question. There are a total of 100 points possible. . All work must be shown for full credit. . Put your final answers in the locations specified. . You may use the six single-sided 8 l/ " x 11" information sheets that you have prepared. Some physical constants are given below. You may use one "book of math tables." . You may use a "pocket calculator" that can be put in a normal-size pocket and requires no external electrical power. Graphing calculators and programmable calculators are acceptable. You may not use portable, hand-held, lap-top, or notebook computers or wireless or network connections. . You may not use any reference materials other than those listed above. Therefore, you may not use the class notes, any textbooks, homework problems, reprints of papers, journals, prayer books, etc. There is to be no sharing of anything. If excess information is given in a question, ignore the unneeded information. If too little information is given in a question, assume the information needed and clearly note this with your work. Any changes to the examination will be written on the board. Check the board periodically during the examination. Any acts of dishonesty will be referred to the Dean of Students without prior discussion. The official written Institute procedures on academic honesty (entitled "Maintaining Academic Honesty" and available from the Dean of Students Office) will be followed in all cases. Have a happy exam! = 6.6260755 x 10'34joule-sec = 299792458 x 108 meter/ sec = 1.6021773349 x 10'19cou1 k = 1.38065812 x 10'23joule / K 005‘ Table 1 Susan tumour! DATA WW Lam-(M ‘VMM MW CM Illuminance It is desired to measure the illuminance at the screen of a front-projection high— definition television (HDTV) system. Using one complete sentence, state the definition of “illuminance” for this situation. Give two possible units for illuminance. Indicate the V , appropriateinstrument for measuring illuminance. Put your answers in the spaces pro— vided. Definition of Illuminance: Two Possible Units of Illuminance: Illuminance Measuring Instrument: (circle the one best answer) (spectrometer) (interferometer) (radiometer) (photometer) (refractometer) (microscope) (telescope) (demultiplexer) Laws of Refraction As described in the course notes and in the textbook, there are two laws of re- fraction. The first law states that the refracted wave is phase matched to the incident wave. For a wave in a dielectric of index n1 incident uponan interface with dielectric of index 712, the angle of refraction 82 is given by the relationship n1 sin 01 = mg sin 02. For isotropic materials this is Snell’s law. Using one or more complete sentences, state the second law of refraction. Using additional complete sentences, define all quantities that you have used in your statement of the second law of refraction. Put your answers in the spaces provided. Statement of Second Law of Refraction: Definiti0n(s) of All Quantity(ies): Electra-Optic Digital-to-Analog Converter Segmented interdigitated electrodes are deposited on top of a lithium niobate slab waveguide as shown. The diffraction efficiency (fraction of the incident power that is diffracted) for each segment is given by 77 = sin2(AV), where V is the voltage applied to a. grating segment and A is a constant. With four grating segments, this device is to be used to convert an electronic 4-bit binary word into an optical analog value as represented by the total power of the optical wave being focused onto the detector. The voltage for each bit is “on” for a 1 bit and “off” for a zero bit. Any segment of the interdigitated electrodes can have a diffraction efficiency of 100% by applying a voltage V 2 VB to that set of electrodes. The difference in the pow- ers produced by the A—to—D converter for the binary numbers 0000 and 1111 should be as large as possible. Electrical-Digital Input 20 21 22 23 Optical-Analog Output Uniform Guided Wave Calculate, showing all work, the voltage for each grating segment required for each bit of the A—to-D converter. Express your answers as fractions of V0. voltage for least significant bit, V0001 2 V0. voltage for next significant bit, V0010 2 V0. voltage for next significant bit, V0100 = Vb. voltage for most significant bit, V1000 = Vb. Fourier Transform Infrared Spectrometer A Fourier Transform Infrared (FTIR) spectrometer is based on a- Michelson in- terferometer. An FTIR spectrometer is capable of determining the spectral content of a light source. This is accomplished by scanning one of the mirrors of the Michelson in- terferometer and recording the output interferogram as a function of the optical path difference between the two interfering beams. In an FTIR, the wavelength of a wave is represented by its wavenumber v where l A U: and /\ is the freespace wavelength. A Fourier transform relationship exists between the spectrum in Wavenumber space ('0) and the amplitude interference pattern (interfero- gram) in optical path difference space (6). Interferograms obtained for two sources are shown in the attached figure. The interferogram for the first source represented by the upper figure is S1(6) = A1 cos(27rv1 6) + A2 cos(27rv2 6). The interferogram for the second source represented by the lower figure is 82(5) '2 A3 605(271'01 6) + A4 008(27T’Ug 6). From these two source amplitude interferograms, answer the questions given below. For S 1 (6) (circle the one best answer) (lAll = lAzl) (A1 = A2) (lAll # [A2|) (none of these) For 5'1 (6), estimate, showing all work, the ratio 112/121. Put your answer in the space pro- vided. ‘ 02/111 = For 82(6) (circle the one best answer) (IA3l : |A4|) (A3 = A4) (lAal 5‘ |A4l) (“0116 Of these) For these two interferograms, draw the corresponding amplitude spectra in the space provided. AMPUTUDfi‘ AMPUTUDE O 1©N1 OFTEAL WWH DWFERENCE AMPLHUDE AMPUTUDE WAVENUMBER Differential Interference Contrast (Nomarski) Microscopy 2 A Differential Interference Contrast (DIC) or Nomarski microscopy configuration is shown in the diagram. Analyzer I:::I Beam Combiner Objective Channel Wavguide Substrate Condenser Beam Splitter Polarizer :3 Light of freespace wavelength A is used in this microscope. The modified Wollaston beam splitter produces two orthogonally polarized beams (solid line and dashed line) that have a lateral displacement (shear) that is less than the resolution of the objective. This DIC microscope is used to examine a channel waveguide as shown in the figure. The waveguide material is a polymer with an index of refraction of nu, at wavelength A. The width of the waveguide is Wu, and the thickness of the waveguide is Lu, as shown. The polymer is deposited on a substrate that has a refractive index of ms at wavelength A. The thickness of the substrate is L s as shown. The waveguide and substrate are sur— rounded by air. In the figure, the two orthogonally polarized beams are shown at both the left edge of the channel waveguide and at the right edge of the channel waveguide. When the above channel waveguide is vieWed through the DIC microsCope, the beam combiner is adjusted so that the phase retardation bias introduced produces the max— imum transmittance (or intensity) contrast between the left edge of the channel wave— guide and the right edge of the waveguide. Under these conditions, derive, showing all work, the transmittance diflerence, AT, between the left edge and the right edge of the waveguide. Express your answer only as a ftmction' of the parameters given above. Sim- plify your answer as much as possible. Put your final answer in the space provided. Concave Mirror 2 A concave spherical mirror is in air as shown in the attached figure. The magni- tude of the radius of curvature of the mirror is 100 mm. A vertical planar object exists 75 mm to the left of the mirror. A point on the object is located 50 mm above the axis of the optical system. Calculate, showing all work, the focal length (including sign) of the mirror. Cal— culate, showing all work, the location of the image along the optical axis to the left or to the right of the mirror. CaICulate, showing all work, the location of the image point above or below the optical axis. Express your answers in mm. For the image distance, indicate whether it is left or right of the mirror. For the image point, indicate whether it is above or below the optical axis. Put your answers in the spaces provided. In the attached figure, the spacing between tick marks is 50 mm. On the drawing, put a dot at the center of curvature and at the focal point. On the figure, draw a ray trace diagram that incorporates 1) the parallel ray, 2) the chief ray, and 3) the focal ray. Label the parallel ray, the chief ray, and the focal ray. 1 Focal length 2 mm (Sign) Object Image Distance Distance Distance along axis left or right along axis left or right from axis above or below 75 left Illuminance The “illuminance” is the luminous power (or luminous flux) per unit area impinging on the screen. Possible units of illuminance are lumens/m2, footcandles, phat, or lust. A photometer is an illuminance measuring instrument. Laws of Refraction The second law of refraction states that the refracted wavevector lies in the plane of incidence. ' The refacted wavevector is a vector of magnitude of 2 7r 112 /A Whose direction is the direction of phase propagation of the refracted wave. The quantity A is the freespace wavelength. ' The plane of incidence is the plane defined by the incident wavevector and the normal to the interface where the incident wavevector strikes theinterface. Electro-Optic Digital-to-Analog Converter Bragg regime 1} = sin2 (AV) 01' V = Zsin‘lfl = 271' and so V =%(sin-’1\/fi)Vb Therefore, V0001 = g— [Mflfifl V0 = 02301 V2, for n = 1/8. V0010 = %Vb = 0.3333Vo for n = 1/4. V0100 = %Vb = 0.500014; for n = 1/2. V1000 = V0 for 77 = 1- Fourier Transform Infrared Spectrometer The first interferog‘ram is 5'1 (6) = A1 cos(2 7r «)1 6) + A2 003(2 7r 02 6). From the first interferogram, it is observed that when the two waves are out of phase (at 5/ 121) they cancel completely. Thus From the interferogram, it is observed that the waves are in phase at 6 = 0. As the path difference increases, the waves are 180° out of phase at 6 2 5/121 (5A1 wavelengths) and then are back in phase at 6 = 10/ 01 (10M wavelengths): There are a total of about 11 amplitude peaks for 0 < 6 < 10/ 111. If A1 alone were present there would be 10 ampli- tude peaks. Therefore the seocnd wavelength A2 must be such that 11 A2 = 10 /\1. Thus A1 112 11 A—z ‘Ul E The second interferogram is 32(6) 2 A3 cos(2 7r v1 6) + A4 003(2 7r v3 6). From the second interferogram, it is observed that when the two waves are out of phase (at about 6 2 22/121) they do not cancel completely. Thus lAal 7é lA4l ‘ For these two interferograms, the amplitude spectra are shown. AMPUTUDE AMPUTUDE 0 . 1©N1 OPTEAL WWH DWFERENCE AMPUTUDE AMPUTUDE W‘Q V1V3 WAVENUMBER Differential Interference Contrast (Nomarski) Microscopy 2 freespace wavelength = A channel waveguide refractive index channel waveguide thickness =Lw left edge of channel waveguide (An)1 right edge of channel waveguide (An)2 T = II": —- sinz—g— where I‘ = I‘m-as + 277r-AnL For maximum contrast Fbias : 77/2 andso 7r 27r I‘ = — —A L 2 + A " 1r 27r I‘ = — — L 1 2 + )‘(A’nh 7r 271' I‘ = — — A L 2 2 + A( “)2 T1 — sin2[ 2 . I“; +2—’r T2 = szn2[ “8 3 ATle—Tzzsin This can be simplified using cos 20 1 1 «2 9____ 3m 2 2 ”111 nw—l = l—nw 7r 7T 4 A ’7? 7T 4 /\ (nu, —— 1) Lw] — sin2 [ sin2 [— + — (nu, —- 1) Lw] sinz [— + — (l — nw) Lw] 3+” 4 /\ Rood; + k(nw—1)Lw] - 0045 + k(1‘"w)L‘"]) AT 2 2 where k = 27r/A. This can be further simplified using cod; + 0) = -—sin0 1 . . AT = 5 (sm [k (1 — nw) Lw] —— 3m [k (nu, — 1) Lw]) Using sin(—0) = - sin0 AT 2 sin [k (1 — nw) Lw] If the arguments are sufficiently small, this could be further simplified using the small argument "approximation sin 0 = 0. Concave Mirror 2 Mirror parameters R 2 —100mm, 3 = 75mm f = e; = +50mm l+i__3 s s’ _ R. 1 3R I: = -- = 150 1ft f ' s _%_l 2s+R + mm (e omirror) s/ m = — g = —2.000 (inverted real image) Distance from axis I 2 m0 = (—2) (50mm) 2 —100mm (below axis) ...
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