4500T106 - Acronyms - (6) Name Quantities (3) (Please...

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Unformatted text preview: Acronyms - (6) Name Quantities (3) (Please Print) - FI‘ IR (9) Gap and Freq. (l4) Blackbody ' (9) LCD Display (12) E C E 4500 Collimated Beam , (12) Plane Mirrors (12) . First Hour Examination Convex Refractive (18) I February 13, 2006 Rules for exam: 1. 2. 3 4. 5 0040 10. '. ll. 12. l3. 14. 15; The time allowed is 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. ' i. All work must be shown for full credit. Put your final answers in the locations specified. . You may use the six single-sided 8 1/ " x 11" information sheets that you have prepared. . 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 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. I ‘ Any changes to the examination will be written on the board. Check theboard 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! h = 6.6260755 X10'34joule—sec c = 2.99792458X108meter/sec e = 1.6021773349x10“9coul k = 1.38065812x10'23joule/K ‘ nu: 1 81mm LW Dru I Optical Engineering Acronyms . State in words the meaning of the aeronyms listed below. Put your answers in the ’spaces provided. ' AR: DBR = CCD Optical Engineering Quantities Write definitions of the quantities listed below. Give typical unitsfor the quan- '_tity. Put your answers in the spaces provided. Quantitym ' _ Definition I Typical Units luminous intensity beam divergence Fourier Transform Infrared Spectrometer _ The Fourier Transform Infrared (FTIR) spectrometer is based on a Michelson in— terferometer as shov'vn below. , . Unknown Source ~ ' Fixed Mirror An unknown source is to be analyzed by the FTIR spectrometer. It is desired to measure the spectrum of the source As the moving mirror scans, the detector records the intensity, I, as a function of the optical path length difference, opd, between the two arms of the interferometer? This recording, I (opd), is called an interferogram. The spec— trum of the Source may be represented by S which is a function of 2 7r/A where A is the ' freespace wavelength. E'om the measured interferogram, I (opd), state expliciter how to calculate the spectrum 3(2 7r/A). Put your answer in the space provided. S(21r//\) : Energy‘Gap Requency A gallium arsenide semiconductor pn-junction diode laser emits light centered at a freespace wavelength of 850 nm. The spectral lineshape is homogeneously broadened! ' The spectral'fuJI-width of the emission is 0.1 nm. The refractive index of the gallium arsenide over this range of wavelengths is 3.6. Rom the above information, calculate, showing all work, the energy gap of the semiconductor material in electron-volts (6V).- Calculate, showing all work, the fre ‘quency full-width of the emission in Gigahertz (GH z). Express your answers accurately to four significant figures and put your answers in the spaCes provided. Energy gap = ' A eV Frequency full—width of emission = GHz ‘ Blackbody Radiation Numerous light soucres behave as 'bla‘ckbO'dy radiators. Examples include the sun, incadescent lights, and human beings (emitting in the infrared). Due to the impor- tance of blackbody radiation, it is quantitatively characterized in radiometry. Answer the questions below regarding blackbOdy radiation. 1 Give the name of the quantity that typically used to describe the spectral dis— 7 tribution in blackbody radiation. This name was used in the class notes, Give the sym— bol used for this quantity. This symbol was used in the class notes. 'Give the typical units of this quantity. These were given the in the class notes. Name of quantity ¥ Symbol for this quantity =_ Typical units for this quantity = Color Liquid Crystal Display A Liquid Crystal Display (LCD) produces “white” by combining, at each pixel, ' equal amounts of luminous power in each of the three primary colors. The wavelengths corresponding to these primary colors are as follows: -/\R = 700 nm (red) AG i 550nm (green) AB = 440 nm (blue) For “white” in this display, calculate, shoWing all work, the fraction of the total radiant power that is in eachcolor. Express your answers accurately to three significant figures. Put your answers in the spaces provided. Fraction of total radiant power in red color 2 Fraction of total radiant power in green color : Fraction of total radiant power in blue color 2 Collirnate'd Laser Beam A' highly secure optical communications link is to be established between a low- orbit satellite and a ground station on earth. A block diagram of the optical system is shown below. Earth _ .Satelliie v 7 ' L3 L1 . . ‘ L2 VFrequency _ Doubler Beam—Expanding . Telescope l-‘—' Lit—’l The satellite is at an altitude of- L4 = 160 km above the surface 'of the earth. The transmitter uses a mode-locked neodymium YAG laser of fundamental wavelength Vof 1.0641 pm. The fimdamental Wavelength of laser is then frequency doubled for data communications. The beam exiting the frequency doubler has a diameter of L1 = 2.00 min. Then the laser beam is expanded and “collimated” using a beam—expanding telescope Finally, the beam is directed down to the ground station. A circular area with a diameter of only L3 = 0.80 m is illuminated on the earth’s surface by this laser beam. Ignoring atmospheric turbulence effects. calculate, showing all work. the-minimum output diameter of the laser beam L2 from the beam—expanding telescope to achieve the above specification. Express your answer in millimeters. Express your answer accurately to three significant figures. Put your final answer in the space provided. Output diameter of. laser beam frombeam-expanding telescope L2 : ' mm. ‘ Plane NIirrors at. a Right Angle Two mirrors are mounted-togetherso that there is always a right angle-between them as shown in the figure. The two mirrors are named Mirror 1 and Mirror 2 as shown in the drawing. The two mirrorscan be rotated by an angle a about the line where the mirrors intersect. This axis of rotation is normal to the page. A ray of light _ of freespace wavelength A is propagating horizontally from left to right and is incident upon Mirror 1. as shown in the drawing. Angle of incidence .is defined as the angle measured With respect to the normal to the surface. Calculate the angle of incidence upon Mirror 1. Express your answer as a function of a. For the ray reflected from Mirror 1 that is incident upon Mirrdr 2, calcu- late its angle of incidence upon Mirror 2. Express your answer as a function of a. For . the ray reflected from Mirror 2 (doubly reflected), calculate its angle of propagation, ,6, with respect to the horizontal. Put your final answers in the spaces provided. Angle of incidence upon Mirror 1 2 Angle of incidence upon Mirror 2 2 Angle of propagation of doubly reflected ray, ,8 : For the case of 0: 225°, calculate numerical values for these quantities. Put your final answers in the spaces provided. Angle of incidence upon Mirror 1 2 ° Angle of incidence upon Mirror 2 2 ° I Angle of propagation of doubly reflected ray, fl = o Convex Refractive Surface . _ The interface between two lossless isotropic dielectric regions is a spherical con- vex s'urface as shown in the diagram. The spherical surface has a radius of curvature of 50 mm. The center ’of curvature is located to the right of the surface. The index of re- fraction of the left-hand region is 1.0. The index of refraction of the right—hand region is 1.5. A vertical planar object is located 50mm to the left if the refractive surface. Light is propagatingvfro‘m left to right. ’ Calculate, showing all work, the location of the image and the linear magnifica- ‘ tion of the image. Express your answers accurately to four significant figures. Specify whether the image is left or right of the interface. Specify whether it is a real or virtual image. Put your final answers in the spaces provided. Using a. distance between tick marks of 50 mm, draw on the above diagram, the object, the parallel ray, the chief ray, the focal ray, and the image. Magnitude of image distance from interface 2 mm The image is to the (left) (right) of the interface, ,(Circle one.) Magnitude of linear magnification = The image is (real) (Circle one.) Optical Engineering Acronyms AR : anti-reflection DBR = distributed Bragg reflector CCD = charge-coupled device Optical Engineering Quantities Quantity Definition Typical Units luminous intensity luminous power output per unit solid angle lumens/steradian beam divergence far-field angular Width of beam radians or degrees Fourier Transform Infrared Spectrometer +°° 2 7r (opd) 3(27r/A) = / I(0pd) cos[ opd = —-00 d(0pd) - Energy Gap and He'quency Gallium arsenide laser /\=850nm .A)‘ = 0.10nm EG = hc/A = 1.239846Vum/0.850pm % 1.4586'eV f = c/A df = —(c//\2)d)\ 41.495 GHz 1A fl = (c/A2)AA> = 41.495 x 109 Hz Blackbody Radiation Name of quantity 2 Blackbody radiant emittance per unit wavelength , or spectral radiant exitance Symbol for this quantity 2 V WA Typical units for this quantity = mW/cm2 pm iPColor Liquid Crystal Display Luminous Efficiencies A2 = 550nm LE2 = 0.993 A3 = 440nm LE3 = 0.023 Luminous power in the i—th color PM = Pm- LEz- (680 lumens/ watt) where P31 is the radiant power. The total radiant power is PR = PLl/LEI + PL2/LE2 + PL3/LE3 The luminous powers are equal. PL1 =. PL2 = PL3 Thus, the fractions of the total radiant powers are PR1 1/LE1 — : = 0.8489 VPR l/LEl + + 1/LE3 —— = M = 0.003420 PR l/LEl + l/LEg + PE : A = 0,1475 PR + + l/LEg Collimated Laser Beam" Frequency—doubled laser ,\ = 1.0641 pm/2 = 0.53205pm Angle subtended at earth 0 = 0.80m/160km = 5.0 x 10—6 rad Diffraction-limited spreadihg of beam _4_L 9—;L2 Therefore the minimum diameter of the ouptut beam is L2 =_ :73; = 135.485mm Plane Mirrors at a Right Angle Angle of incidence upon Mirror 1 a Angle of incidence upon Mirror 2 Angle of propagation of doubly reflected ray, ,6 For a : 25°, Angle of incidence upon Mirror 1 = 25° Angle of incidence upon Mirror 2 = 65° Angle of propagation of doubly reflected ray, 3 7r/2—a 00 Convex Refractive Surface Object is left of the interface. Light is traveling from left to right. R = +50mm, m 1.00, M = 1.50, s = +50mm R h = L— : +100mm n2 — “1 R f2 = L = +150mm n2 " n1 n1 n2 _ “2—711 3 + s’ w R ' s' = ————T—lg——T = —150 mm solving for image distance ...
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This note was uploaded on 04/29/2008 for the course ECE 4500 taught by Professor Gaylord during the Spring '08 term at Georgia Institute of Technology.

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4500T106 - Acronyms - (6) Name Quantities (3) (Please...

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