4500T204 - General Questions (24) Name Spherical Surfaces...

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Unformatted text preview: General Questions (24) Name Spherical Surfaces (24) (Please Print) Reproduction Ratio (14) Thick Lens (14) Trans. & Reflected (12) Fresnel Rhomb (12) E C E 4500 Second Hour Examination March 31, 2004 Rules for exam: 1. LII-bu) OOH-ION 10. ll. 12. l3. 14. 15. The time allowed is 60 minutes. The test will start at 9:003m (rather than at 9:05am) and end at 10:00am (rather than at 9:55am). . 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 1/2" x 11" information sheets that you have prepared. You may also use the six single—sided information sheets that you prepared for the first 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 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! = 66260755 x 10'34joule—sec 299792458 X 108 meter I see 16021773349 x 10'” coul = 1.38065812 x 10'23j0ule I K WOOD” General Optical Engineering Questions For each question below, circle the one best answer. 1. A microscope eyepiece is labeled “R 10X.” The focal length of this eyepiece is there- fore a) 5 mm b) 10mm c) 15 mm d) 20 mm e) 25 mm f) none of the above 2. A close-up attachment lens with an optical refractive power of 1.5 Diopters is added to a 50 mm focal length lens. The refractive power of the combination is therefore a) 3.0 Diopters b) 7.5 Diopters c) 18.5 Diopters d) 20.0 Diopters e) 21.5 Diopters f) 30.0 Diopters g) 48.5 Diopters h) 65.0 Diopters i) 75.0 Diopters j) none of the above 3. In an optical system used for white-light imaging, spherical aberration can be reduced by a) changing to monochromatic light b) decreasing the aperture stop diameter c) increasing the aperture stop diameter d) decreasing the field of view e) increasing the field of view f) none of the above 4. As discussed in class, an accurate mechanical positioning system can straightfor- wardly be constructed using the optical aberration a) spherical aberration b) coma c) astigmatism d) curvature of field e) distortion f) longitudinal chromatic aberration g) lateral chromatic aberration h) none of the above 5. Randomly linearly polarized light propagating in air (refractive index : 1.000) is in- cident upon a plane boundary with glass of refractive index 1.500. The reflected light is linearly polarized. The angle of incidence of the randomly linearly polarized light is therefore a) 0° b) 33.6900 c) 41.8100 (1) 45.0000 e) 56.3100 f) 90° g) none of the above 6. Linearly polarized light is propagating in air (refractive index of 1.00). This light is incident normally upon plane boundary with glass of refractive index of 1.50. Immedi— ately upon reflection at the boundary, the electric field of the electromagnetic wave has a phase shift (relative to the phase of the incident wave) of a) 0 degrees b) 180 degrees c) 0 < phase shift < 180° (1) none of the above Imaging of Object Through Two Spherical Surfaces Light is propagating from left to right. The object is 150 mm to the left of the first surface as shown on the diagram below. An image is formed by a first spherical sur- face (shown in the top figure) between air (to the left) and glass (to the right). The first spherical surface is convex and has a radius of curvature of 50 mm. The center of curva- ture is located to the right of the surface. The index of refraction of the glass is 1.500. To treat the image formed by the first surface, the second surface is disregarded. Thus the image formed by the first surface is considered to be inside the glass. Now, however, a second spherical surface is introduced into this optical system. The second spherical surface (shown in the bottom figure) is a concave surface and has a radius of curvature of 50 mm. The center of curvature is located to the left of the surface. This second sur- face is located 250mm to the right of the first surface. The image from the first surface now becomes the object for the second surface. The index of refraction to the right of the second surface is 1.0 (air). To work this problem, treat the first and second surfaces separately. Do not treat the combination of surfaces as a lens. First Surface Second Surface Calculate, showing all work, the location and magnification of the new image due to the first surface alone. Express your answers accurately to four significant figures. Specify whether the image is left or right of the first surface. Specify whether it is a real or virtual image. Put your final answers in the spaces provided. Using a distance be— tween tick marks of 50 mm, draw on the top diagram, the parallel "ray, the chief ray, the focal ray for the first surface and the image. Magnitude of image distance from first surface = mm The image is to the (left) (right) of the first surface. (Circle one.) Magnitude of linear magnification due to first surface alone 2 This first image is (real) (virtual). (Circle one.) Calculate, showing all work, the location and magnification of the final image due to the combination of surfaces. Express your answers accurately to four significant figures. Specify whether the final image is left or right of the second surface. Specify whether it is a real or virtual image. Put your final answers in the spaces provided. Us- ing a distance between tick marks of 50mm, draw on the bottom diagram, the parallel ray, the chief ray, the focal ray for the second surface and the final image. Magnitude of final image distance from second surface :: mm The final image is to the (left) (right) of the second surface. (Circle one.) Magnitude of linear magnification of final image _ = This final image is (real) (virtual). (Circle one.) Changing Reproduction Ratio An optical lens system is used to make integrated circuit photomasks. The‘lens is an Ultra—Micro-Nikkor lens that has a focal length of 49.5 mm and a fixed aperture of f / 1.8. It is designed tobe used with e—line light of freespace wavelength 546.1 nm. It is optimized for imaging at a reproduction ratio of 5 to 1. Using the above optical lens system, it is desired to image integrated circuit art- work at a reproduction ratio of 4 to 1 (rather than 5 to 1). Calculate, showing all work, the distance that the artwork needs to be moved from its position that produces a 5 to 1 reproduction ratio to its new position that produces a 4 to 1 reproduction ratio. Specifiy whether the artwork should be moved toward the lens or away from the lens. Express the distance in mm accurately to four significant figures. Write your final answers in the spaces provided. Distance artwork should be moved 2 mm Artwork should be moved (toward) (away) from lens (Circle one.) Fixed Reproduction Ratio Imaging with 3 Thick Lens An object is placed to the left of the lens shown in the diagram below. All di- mensions given are in mm. It is desired to form a real image of the object such that the reproduction ratio is “6 to 1.” Calculate, showing all work, the distance of the object from the left—most glass surface and the distance of the image from the right—most glass surface. Express your answers in mm accurately to within i0.1 mm. Put your answers in the spaces provided. Distance of object from left—most glass 2 mm Distance of image from right—most glass : mm Transmitted and Reflected Amplitude and Power Linearly polarized light from a helium—neon laser of freespace wavelength 632.8 nm is propagating in air. This light is normally incident upon a planar glass boundary. The glass has an index of refraction of 1.6. For the transmitted and reflected waves, calculate, showing all work, the quanti— ties listed below. For these calculations, use 1.000 for the refractive index of air. Express fractions accurately to within 0.00001. Express phase shift angles in degrees accurately to within 0.0001°. Write your final answers in the spaces provided. Transmitted Wave Fraction of amplitude transmitted (magnitude) 2 Phase shift upon transmission : _________— ° Fraction of power transmitted (magnitude) : Reflected Wave Fraction of amplitude reflected (magnitude) : Phase shift upon reflection : .______—__—_— ° Fraction of power reflected (magnitude) 2 Fresnel Rhomb A Fresnel rhomb is shown below. Such a rhomb may be made of glass of refrac- tive index m. The rhomb is surrounded by refractive index 712. It is observed that incident linearly polarized light (with polarization in a particular di- rection) is converted to circularly polarized light after two internal reflections. Thus it functions as a “quarter wave retarder.” The angle of incidence at each reflection is 61. Write the single—equation relationship between n1, ng, and 01 that causes the conversion of linear polarization to circular polarization in this device. Do not inlcude any other variables. Relationship between n1, ng, and 01: General Optical Engineering Questions For each question below, circle the one best answer. 1. A microscope eyepiece is labeled “R 10X.” The focal length of this eyepiece is there- fore e) 25 mm 2. A close-up attachment lens with an optical refractive power of 1.5 Diopte’r‘s is added to a 50 mm focal length lens. The refractive power of the combination is therefore e) 21.5 Diopters ‘3. In an optical system used for white-light imaging, spherical aberration can be reduced by b) decreasing the aperture stop diameter 4. As discussed in class, an accurate mechanical positioning system can straightfor- wardly be constructed using the optical aberration c) astigmatism 5. Randomly linearly polarized light propagating in air (refractive index = 1.000) is in— cident upon a plane boundary with glass of refractive index 1.500. The reflected light is linearly polarized. The angle of incidence of the randomly linearly polarized light is therefore e) 56.3100 6. Linearly polarized light is propagating in air (refractive index of 1.00). This light is incident normally upon plane boundary with glass of refractive index of 1.50. Immedi- ately upon reflection at the boundary, the electric field of the electromagnetic wave has a phase shift (relative to the phase of the incident wave) of b) 180 degrees Imaging of Object Through Two Spherical Surfaces First surface R = +50mm, m = 1.00, 722 = 1.5, s = 150mm R f1 = L = +100mm n2 —- n1 R f2 = L = +150mm n2 _' n1 n1 n2 __ nz—nl s + s’ _ R 3': n _n"2_ n = +450mm R ' s l m = — n18 = —2 (inverted real image) 1128 Second‘surface (250 mm to right of the first surface) R = —50mm, n1 = 1.50, 712 = 1.00, s = —200mm R f1 = ——"l——— : +150mm n2 — n1 R f2 = —"—2— = +100mm n2 — n1 n1 n2 __ n2_n1 s + s’ a R ' sefi—E‘EZ—E = +57.1429mm R 3 n1 3’ . . . m : —— = +0.42857 (same orientatlon as object for second surface) 7123 Final image magnification m = (+ 0.42857) (— 2) — 0.85714 (inverted real image) First Surface Changing Reproduction Ratio f = 49.5mm m1 = 1/R1'= —1/5 t —0.20 _ m2 = 1/R2 = —1/4 = —0.25 object distance 8 = f(1--,%) = f(1-R) 31 = f(1—R1) = f(1+|R11) 32 = f(1—-R2) = f(1+|R21) = 247.5mm 297 mm Distance artwork should be moved 2 49.5 mm Artwork should be moved toward the lens. Fixed Reproduction Ratio Imaging with a Thick Lens Focal length, f = 51.6mm Real image, m < 0 Linear magnification, m = —-1/6 = —0.16667 Object distance 1 s = (1 — E)f 2 7f 2 7(51.6mm) = 361.2mm Distance of object from left-most glass 2 361.2 — (51.6 — 23.7) = 333.3 mm Image distance 3’ = ——ms 2 —(—1/6)361.2mm = 60.2mm Distance of image from right-most glass 2 60.2 — (51.6 — 38.1) = 46.7mm Transmitted and Reflected Amplitude and Power 01 2 0° 02 = 0° Treat as TE polarized light and use Hesnel’s equations. TE Polarized Transmitted Wave Fraction of amplitude transmitted = t = grit—2- : 0.769231 Phase shift upon transmission 2 0° Fraction of power transmitted = t2? = 0.946746 1 TE Polarized Reflected Wave Fraction of amplitude reflected = r 2: 11121 = —0.230769 "1 +712 Phase shift upon reflection 2 180° Fraction of power reflected = 7'2 = 0.053254 Fresnel Rhonib The phase shift between TM and TE polarizations upon total internal reflection is 2 ' 2 2 2 - 2 2 _ n1 n1 sm 61 — n2 _1 n1 sm 01 — n2 - = 2tan 1 ——-—-—-—-— -— 2tan —————— ¢TM ¢TE ( n3 cos 01 ) ( n1 cos 01 = 45° ...
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4500T204 - General Questions (24) Name Spherical Surfaces...

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