4500T304 - General Questions(18 Name Slab...

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Unformatted text preview: General Questions (18) ' Name Slab Waveguide (20) (Please Print) Interference Filter (22) Circular Aperture (16) Transmittances (24) E C E 4500 Third Hour Examination April 27, 2004 Rules for exam: 1.. 2. 3 4. 5 0°99 10. 11. 12. 13. 14. 15. The time allowed is 60 minutes. The examination will start at 2:50pm. 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 six new single-sided 8'1/ " x 11" information sheets that youhave prepared. You may also use the six single-sided information sheets that you prepared for the first examination as well as the six single—sided information sheets that you prepared for the second examination. Some physical constants are given below. You may use one "book of math tables" or calculus textbook. . 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, infrared, 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 16021773349 x 10‘19 coul k = 138065812 x 10'23joule / K (905' General Optical Engineering Questions For each question below, circle the one best answer. 1. At a freespace wavelength of 600 nm, the refractive index of gold is approximately a) 1 b) 1.5 f) 0.5 — 3'3 g) all of the above 2. Elliptically polarized light of freespace wavelength A can be converted to linearly po- larized light by a a) quarter waveplate b) half waveplate c) full waveplate d) an interference filter e) a parabolic mirror f) none of the above 3. At a freespace wavelength A, the TED mode is most tightly bound in the slab waveg— uide (where us is the cover refractive index, nf is the film refractive index, and n3 is the substrate refractive index) described by a)nc=1,nf=1.5,n5=1 b) nc=2,nf=3, 113:2 0) nc = 1.5, nf = 2.5, ns = l5 d)nc=1,fif=2,ns='_—1 e) none of the above 4. A thin slab of thickness d and refractive index n with flat and parallel surfaces will transmit normally incident light of freespace wavelength A for m being an integer if a) 2nd 2 mA b) 2nd —._- (m — ax c) nd 2 mA d) and z (m — a A e) 4nd 3 mA f) 4nd 2 (m — %),\ g) none of- the above 5. For a dielectric substrate of refractive index n‘.J in air, for a single—layer coating to be antireflecting, it should have a refractive index of f) ns+1 g) (“a ‘l' 1)/2 h) none of the above 6. A reflection hologram is recorded when two mutually coherent beams are incident on (the same side) (opposite sides) of a high—resolution photographic recording plate. (can— not be determined from given information) (none of the above) Single TEo Mode Slab Waveguide A singlemode titanium-diffused lithium niobate slab waveguide (nf = 2.234) is to be constructed for telecommunications applications. It is surrounded by a lithium nio— bate substrate (ns = 2.214) and an air cover (nc = 1.000). Light of freespace wavelength 1-00pm is to be guided in. the waveguide. Calculate, showing all work, the range of thicknesses, h, for which only a sin- gle TEO mode will propagate. The TMO mode, the T191 mode, the TMl mode and all higher-order modes are to be cut oil. Express the minimum and maximum thicknesses in um accurately to four significant figures. um<h< pm Wavelength Selection with an Interference Filter A “one half wavelength” thick interference filter is passes blue light from an argon laser of freespace wavelength 488.0 nm when it is rotated so that the angle of incidence in air is or = 30°. This is shown in the diagram below. The interference filter is com— posed of parallel layers of refractive indices 1.5 and 1.6 as shown in the figure. In a sec— ond angular orientation, it is desired that the interference filter pass green light from an argon laser of freespace wavelength 514.5-nm. In a. third angular orientation, it is desired that the interference filter pass violet light from an argon laser of freeSpace wavelength 476.5 nm. Calculate, showing all work, whether or not it is possible for this given filter to pass 514.5 nm light and whether or not it is possible to pass 476.5 nm light. If either or both of these cases are possible, calculate, showing all work, the required angles of incidence, a, to accomplish them. Put your answers in the spaces provided. If angles of incidence are realizable to pass these wavelengths, express these angles accurately to four significant figures. Is it possible for this filter to pass 514.5 nm light? (yes) (no) (cannot be determined from information given) (Circle one) _ If it is possible, then a (514.5.nm) 2 ° Is it possible for this filter to pass 476.5 nm light? (yes) (no) (cannot be determined from information given) (Circle one) If it is possible, then a (476.5nm) = ‘ ° Eraunhofer Diffraction by a Circular Aperture The characteristic Eraunhofer diffraction pattern due to a circular aperture illu— minated at normal incidence is known as an Airy disk and the radiant intensity is given by JflstinQ/A) (11’ D sin 9/ M2 where J1 is the first—order ordinary Bessel function of the first kind, D is the diameter 1(9) = 410 of the aperture, and 6 is the angle measured from the center of the aperture to an arbi- trarg,r point in the diffraction pattern. A circular aperture is shown at left in the diagram below. The resulting far-field Airy disk pattern is shown at right in the diagram below. l A ___...... Lx ——- {ML l~————Lz——i For the particular case of a circular aperture of 200 nm diameter illuminated by a helium—neon laser of freespace wavelength 632.8 nm, calculate the diameter L:c of the first dark ring in the Airy disk pattern when the observation plane (shown at the right in the figure) is L2 = 10 meters. Express your answer in mm accurately to four signifi- cant figures. Put your answer in the space provided. L3 2 mm Far-Field Diffraction by Rectangular and Triangular 'Iransmittance Slits A rectangular transmittance slit and a triangular transmittance slit have ampli- tude transmittances as shown below. transmittance The rectangular slit has a width a and a peak amplitude transmittance of 0.5. It is illuminated in air with a collimated laser beam of freespace wavelength A. In the far-field diffraction pattern, a bright central peak surrounded on both sides by nulls and a series of dimmer peaks is observed. The bright central peak is taken to be at 6 = 0 where 9 is the angle measured from the center of the slit to the far field. For this diHrac- tion pattern, write down (it is not necessary to derive) the expression for the far-field radiant intensity I Express the radiant intensity as a function of a, 9, A, and I0 only where I0 is the value of I (0) at 6 = 0. Develop the exact expression for the angle in ra- dians from the central peak to the first null (61%”). Express the angle as a function of a and A only. Put your final answers in the spaces provided. Rectangular slit radiant intensity I (6) = For rectangular slit, far-field angle from central peak to first null = radians The triangular slit has a width (1 and a peak amplitude transmittance of 0.5. It is illuminated in air with the same collimated laser beam of freespace wavelength A. In the far—field diffraction pattern, a bright central peak surrounded on both sides by nulls and a series of dimmer peaks is observed. The bright central peak is taken to be at 6 = 0 where 6 is the angle measured from the center of the slit to the far field. For this diffrac- tion pattern, write down (it is not necessary to derive) the expression for the far—field radiant intensity I Express the radiant intensity as a function of a, 6, A, and In only where I0 is the value of 1(6) at 6 = 0. Develop the exact expression for the angle in ra- dians from the central peak to the first null (9p_n). Express the angle as a function of a and A only. Put your final answers in the spaces provided. Triangular slit radiant intensity I (0) = For triangular slit, far-field angle from central peak to first null = radians General Optical Engineering Questions For each question below, circle the one best answer. 1. At a freespace wavelength of 600nm, the refractive index of gold is approximately f) 0.5 — 3'3 2. Elliptically polarized light of freespace wavelength A can be converted to linearly po- larized light by a a) quarter waveplate 3. At a freespace wavelength A, the TED mode is most tightly bound in the slab waveg- uide (where nC is the cover refractive index, nf is the film refractive index, and 11,3 is the substrate refractive index) described by d) nc=1,nf=2,n3=1 4. A thin slab of thickness d and refractive index n with flat and parallel surfaces will transmit normally incident light of freespace wavelength A if a) 2nd = m A 5. For a dielectric substrate of refractive index n, in air, a single-layer antireflection coating should have a refractive index of d) iii/2 6. A reflection hologram is recorded when two mutually coherent beams are incident on (opposite sides) of a high-resolution photographic recording plate. Single TED Mode Slab Waveguide nc = 1.000 nf = 2.234 125 = 2.214 A=1.00,um '2_ 2 em = “g "’3 = 43.8601 nf_n5 “4 2_ 2 am — —fi "3 n; = 1092.45 nc nf—ns VCOTEO = ten—w—em = 81.4130 = 1.42093md v.30”,o = tan-V—“em = 88.2670 = 1.54055md A WOTE A tan—lanE heoTEo = W = W = 0.75823 ,um _ AVCOTMD _ Atan'dwaTM _ hcoTMo — W — W — 0.82205pm and so for true Single-mode (TED) behavior, the thickness should be 0.75823pm < h < 0.82205pm Wavelength Selection with an Interference Filter Angle of incidence a1 = 30° passes /\1 = 488.0 nm Parallel glass plates do not affect the ray direction in air. Thus from Snell’s law 1.0 sin (11 = 1.5 sin [31 and so fil = 1947122". The angular tuning is described by 2 n (1 cos [5’ = m A For one half wavelength filters at normal incidence 2nd = A” and so /\ ‘ A cos [31 = A—1 and thus An = wslfll n 2 517.6021 nm at normal incidence For tuning to A2 = 514.5 nm A COSfig = A—2 and so [32 = 6.276080 1.03ina2 = 1.5sinfig and so 052 = 943794". For tuning to A3 = 476.5 nm % and so {33 = 22.987330 11. II 005 153 1.03ina3 = 1.53infig and so 053 2 3585892". Fraunhofer Diffraction by a Circular Aperture The Rayleigh criterion A9p_n = 1.22% gives the angular distance from the intensity peak to the first null. The angular distance null-to—null is M“, = 2.44%- = 0.007720md = 0.44233°. The null-to-null linear distance is L,._n = L3, = Lz A6n_n = (10m)-(0.007720rad) = 77.2mm Far-Field Diffraction by Rectangular and Triangular Transmittance Slits Rectangular slit sin2 (E32)2 F2(jkma:) = (12 Since kg; = ZT‘TSinfl and I0 = a2 SinZ («11:11:10 ) ( 1man ) 2 I(6) = ID The first null occurs at fiasinfl _ A — fl 9 = sin—1e) Triangular slit 2 mafia) _ (1 Max) = I k a 4 ‘11-) Since kg: = 271752716 and I0 = (12/4 Sin4(1mzsin6) nasinfl ) 4 2A 1(9) = I0 The first null occurs at wasin 9 _ 2 A _ 7T 9 2 sin—1(2 ...
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4500T304 - General Questions(18 Name Slab...

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