4500T203 - General Questions Two Lenses Thick Lens Imaging...

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Unformatted text preview: General Questions Two Lenses Thick Lens Imaging ( 15) (21) Name (20) (Please Print) Microscope Objective (12) Hemispherical Lens Metal Reflection (18) (14) E C E 4500 Second Examination April 2, 2003 Rules for exam: 1. SAPS” 9°99 10. 11. 12. 13. 14. 15. The time allowed is 60 minutes. The test will start at 9:00am (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/ " 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, 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 chalk board. Check the chalk 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 2.99792458 x 108 meter / sec 1.6021773349 >< 10‘19coul = 138065812 >< 10'23joule / K h c e k General Optical Engineering Questions 1. A light ray is incident at an oblique (non-normal) angle on a boundary between di- electrics. The incident light ray is propagating in a relatively low refractive index dielec- tric. On the other side of the boundary is a relatively high refractive index dielectric. The ray refracted into the relatively high refractive index dielectric is (Circle one.) a) deviated toward the normal to the boundary b) undeviated relative to the normal boundary c) deviated away from the normal to the boundary d) none of the above 2. The linear magnification (or transverse magnification or lateral magnification) of an optical system is m, where m = s’/s and s is the object distance and s’ is the im— age distance. The longitudinal magnification, on the other hand, is As’/As. Write an expression for the longitudinal magnification as a function of m. The longitudinal mag— nification As’/As = 3. Of the aberrations coma and astigmatism, which is more important to be corrected in an astronomical telescope? (Circle one.) a) coma b) astigmatism c) none of the above 4. Linearly polarized light is propagating in glass of refractive index of 1.50. This light is incident normally upon plane boundary with air (refractive index of 1.00). 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 (Circle one.) a) 0 degrees b) 180 degrees c) 0 < phase shift < 180° d) none of the above 5. A TE polarized light wave is propagating in a dielectric of refractive index of m. This light is incident upon plane boundary with a second dielectric of refractive index 112 (m > 712). The angle of incidence is greater than the critical angle. Immediately 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 (Circle one.) a) 0 degrees b) 180 degrees c) 0 < phase shift < 180° d) none of the above 6. Circularly polarized light of freespace wavelength A can be converted to linearly po— larized light by which of the following waveplates designed for that wavelength. (Circle one.) a) zero waveplate b) quarter waveplate 0) half waveplate (1) full waveplate e) none of the above 7. For A being the freespace wavelength, in a typical glass of refractive index n in the visible portion of the electromagnetic spectrum (Circle one.) a) dn/dA < 0 b) dn/dA = 0 c) dn/dA > 0 d) none of the above Imaging with Two Lenses Two thin lenses in air are used to image an object. Lens no. 1 is a converging lens with a focal length of f1 2 +100 mm. Lens no. 2 is also a converging lens. Lens no. 2 has a focal length of f2 = +50 mm. The object is 150 mm to the left of lens no. 1 as shown in the figure. The distance between tick marks in the figure is 50 mm. The dis- tance between these two thin lenses is 200 mm. Lens Lens No. 1 No. 2 Lens No. 1 Lens No. 2 (continued on next page) Calculate, showing all work, the location and magnification of the image due to lens no. 1 only. Express your answers accurately to four significant figures. Put your final answers in the spaces provided. Determine the locations of the parallel ray, the chief ray, and the focal ray for the object and lens no. 1 only. Draw these three rays on the middle figure. Show the image due to lens no. 1 only. Calculate, showing all work, the location and magnification of the final image due to lens no. 1 and lens no. 2 acting together. Use the image from lens no. 1 as the object for lens no. 2. Express your answers accurately to four significant figures. Put your final answers in the spaces provided. Determine the locations of the parallel ray, the chief ray, and the focal ray for lens no. 2 and the final image. Draw these three rays on the lower figure. Show the final im- age due to the combination of the lenses. Location of image due to lens no. 1 only is mm measured (left or right, circle one) from lens no. 1. Magnification of image due to lens no. 1 only is Location of final image due to lens no. 1 and lens no. 2 acting together is mm measured (left or right, circle one) from lens no. 2. Magnification of image due to lens no. 1 and lens no. 2 acting together is Thick Lens Imaging A Nikkor—H 85 mm f / 1.8 lens is to be used for making photomasks for integrated circuit manufacture. The “optical formula” for this lens is shown below. All dimensions given are in mm. The light used is to be of wavelength of 546.1 nm (e—line of mercury lamp). The photomask artwork is placed to the left of the lens shown in the diagram below. A real image is formed at a reproduction ratio of 10 to 1. Calculate, showing all work, the distance of the object from the left—most glass surface. Calculate, showing all work, 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 2 mm Microscope Objective A microscope is constructed by combining a microscope objective and an eyepiece in an appropriate way. A particular microscope objective examined in class has engraved on it the following: Apo 10 / 0.25 160 / 0.17 Explain completely the meaning of each engraved marking by finishing the sentence started below. Add further complete sentences in each case as necessary. “Apo ” means “10” means “0.25 ” means “160” means “0.17” means Hemispherical Solid Immersion Lens - Magnification A hemispherical solid immersion lens for reading compact disks is shown in the figure. The refractive index of the lens is m. The center of curvature of the lens lies on the vertical flat surface at the intersection of the dashed lines. The center of curvature is taken as the origin of an (113,14) coordinate system (+11: to the right, +y up). The radius of curvature is R. The refractive index of the surrounding medium is n2. An object on the compact disk is at the flat surface of the lens extending from the axis of the lens to its tip at height h. Two rays from the tip of this object are shown in the diagram. In the region of refractive index 712, the equation for the upper ray is sin/61 R = _ t —— —— 581 CO (/61 0103/1 + 82.71061 _ a1) , where a1 and ,61 are defined in the drawing. In the region of refractive index 112, the equation for the lower ray is $2 = ——cot/62 3/2 + R, where ,62 is defined in the drawing. Calculate, showing all work, an analytic expression for the linear magnification m as a function of n1, 712, R, and h only (eliminate all angles). Do not use the paraxial ray approximation. Put your final answer in the space provided below. 3 || Reflection from a Metal Linearly polarized light in air is reflected from the plane surface of a metal (a highly lossy material). The incident linear polarization is at a 45° angle with repect to the plane of incidence. Thus, there are equal amplitudes of TE and TM polarization in— cident upon the metal. Of all the possible angles of incidence, there is one particular angle that is designated as the “principal angle of incidence” and it is designated as 0?. Corresponding to the prinicpal angle of incidence there is the “principal azimuthal an- gle” and it is designated as «#1,. Give definitions of 0,, and 111?. Use mathematical expressions and complete sen- tences as appropriate. Definition of 01,: Definition of app: General Optical Engineering Questions 1. A light ray is incident at an oblique angle on a boundary between dielectrics. The incident light ray is propagating in a relatively low refractive index dielectric. On the other side of the boundary is a relatively high refractive index dielectric. The refracted ray in the relatively high refractive index dielectric is (Circle one.) a) deviated toward the normal to the boundary 2. The linear magnification (or transverse magnification or lateral magnification) of an optical system is m, Where m = sl/s. The longitudinal magnification, on the other hand, is As’/As. Write an expression for the longitudinal magnification as a function of m. The longitudinal magnification As’/As = — m2. 3. Of the aberrations coma and astigmatism, which is more important to be corrected in an astronomical telescope? (Circle one.) a) coma 4. Linearly polarized light is propagating in glass of refractive index of 1.50. This light is incident normally upon plane boundary with air (refractive index of 1.00). Immedi— ately upon reflection, the electric field of the electromagnetic wave has a phase shift (rel— ative to the phase of the incident wave) of (Circle one.) . a) 0 degrees 5. A TE polarized light wave is propagating in a dielectric of refractive index of m. This light is incident upon plane boundary with a second dielectric of refractive index n2 (m > n2). The angle of incidence is greater than the critical angle. Immediately upon reflection, the electric field of the electromagnetic wave has a phase shift (relative to the phase of the incident wave) of (Circle one.) c) 0 < phase shift < 180° 6. Circularly polarized light of freespace wavelength A can be converted to linearly po— larized light by which of the following waveplates designed for that wavelength. (Circle one. ) b) quarter waveplate 7. For A being the freespace wavelength, in a typical glass of refractive index n in the visible portion of the electromagnetic spectrum (Circle one.) a) dn/dA < 0 Imaging with Two Lenses f1=+100mm f2=+50mm Lens No. 1 51=+150mm 1+3. _ i 31 81' f1 51': flsl =‘300mm 81—f1 m =__8_1_’ = _fl9 = _2 1 31 150 Lens No.2 32=—100mm 1 1 _ 1 32 82' f2 52’ = f2” = 33.333mm 82-f2 _ 32' _ 33.333 _ m2— 82 _ _100 — +0.33333 Magnification of Lens No. 1 and Lens No. 2 acting together m = m1 m2 = —0.66667 Lens Lens No. 1 No. 2 Thick Lens Imaging Real image, m < 0 Linear magnification, m = —0.10 = —1/10 Focal length, f = 85.0mm Object distance 1 s = (1 — Rdf = 11f = 11 (85.0mm) = 935mm Distance of object from left-most glass 2 935 — (85 — 54.2) = 904.2 mm Image distance 3’ 2 —m3 = -—-(—1/10) 935 mm = 93.5 mm Distance of image from right-most glass 2 93.5 — (85 — 43.2) = 51.7mm Microscope Objective “Apo” means the objective is corrected for chromatic aberration at blue (486.1 nm), yel- low (587.6 nm), and red (656.3 nm) wavelengths. “10” means the linear magnification is 10X (m = 10) when the objective is used as in- tended. “0.25” means the numerical aperture is 0.25 (NA = 0.25). In air the objective accepts a .cone of light with apex angle 2 sin‘1(0.25) = 28.96”. “160” means the tube length is 160 mm (L = so’ — f0). This is{ the extrafocal image distance when the objective is used as intended. “0.17” means that the objective was designed to be used with a cover glass of thickness 0.17 mm; The use of a cover glass is needed in biological and medical applications. Hemispherical Solid Immersion Lens - Magnification Upper ray :1: — cot(fl a ) + Sin fil R 1 1 1 m sin(fi1 — a1) ’ . h - . . szn a1 = E for angle of 1nc1dence a1 a — sin"1 1"— 1 _ R n1 sin (11 = n2 sin fll Snell’s law _1 n ‘ ' _1 n h = — s n = s n fil sm (“2 i a1) 2 (“2 R) Lower ray $2 = _COtfi2 2-12 + R) h . . tan a2 = E for angle of 1nc1dence a2 a — tan"1 2 2 ‘ R n1 sin (12 = n2 sinfiz Snell's law n1 n1 h __ - —1 - _ - —1 '62 — S'ln S’L‘n a2) —- S’I/n The magnified image occurs at the intersection of the two ray lines. $1=$2=$m and y2:y1:ym sin fll sin(,61 — m) R -cot(fi1 — GIN/m + -cot.32 31m + R- and so RU — fl] 9'" 2 com“; — (:0th T a1) RU * 3%] m = Mhlcotfiz —- coma; 501)] where a1, [91, and 32 are all gii'en as functions of n1, 123, R, and h aboVe. Reflection from a Metal Definition of 01,: The principal angle of incidence, 0p, is the angle of incidence for which ¢TM—¢TE = 90°, where ¢T M and ¢TE are the phase shifts upon reflection of the TE and TM polarized components of the incident beam. Of all the angles of incidence, this angle produces the most nearly circularly polarized reflected light. The reflected elliptical polarization reaches its greatest width at this angle of incidence. Definition of 1,01,: The principal azimuthal angle is 11),, = tan—1|ITTTTTA§IL where ’I‘T M and rTE are the am- plitude reflection coefficients when the angle of incidence is 01,. The principal azimuthal angle is thus a measure of the ellipticity, III—Mi, of the elliptical polarization. Considering ITTEI . all possible angles of incidence, the principal azimuthal angle is the minimum value of the azimuthal angle. ...
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4500T203 - General Questions Two Lenses Thick Lens Imaging...

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