Final Exam Page 1/11 Spring 2017 Name SOLUTIONS Closed book; closed notes. Time limit: 120 minutes. An equation sheet is attached and can be removed. A spare raytrace sheet is also attached. Use the back sides i required. Assume thin lenses in air i not speciied. As usual, only the magnitude o a magniication or magniying power may be given. I a method o solution is speciied in the problem, that method must be used. Raytraces must be done on the raytrace orm. Be sure to indicate the initial conditions or your rays. You must show your work and/or method o solution in order to receive credit or partial credit or your answer. Provide your answers in a neat and orderly ashion. No credit i it can t be read/ollowed. Only a basic scientiic calculator may be used. This calculator must not have programming or graphing capabilities. An acceptable example is the TI-30 calculator. Each student is responsible or obtaining their own calculator. Note: On some quantities, only the magnitude o the quantity is provided. The proper sign convention must be applied. 1) (10 points) A 100 mm ocal length thin lens achromatic doublet is constructed using two glasses with glass codes 648338 and 569560. Determine the required ocal lengths o the crown and lint elements. Crown: 569560 Flint: 648338 n d 1.569 56.0 n 1.648 33.8 CROWN d CROWN For an Achromatc Doublet: 1/ 0.01/ mm CROWN CROWN 56. 0 FLINT FLINT 33. 8 56. 0-33. 8 560. - 33. 8 CROWN FLINT CROWN FLINT CROWN FLINT 2. 52 CROWN 0. 0252 / mm 1. 52 CROWN 0. 0152 / mm 39. 7mm 65. 8mm CROWN Note: CROWN FLINT FLINT CROWN = _39.7 mm FLINT = -65.8 mm
Final Exam Page 2/11 Spring 2017 2) (25 points) A 6X Keplerian telescope is comprised o two thin lenses separated by 140 mm. The objective lens is 48 mm in diameter. The eye lens is 14 mm in diameter. This telescope is to be used with a human eye, and the eye is placed at the exit pupil o the telescope. The eye has a 5 mm diameter entrance pupil. For distant objects, what is the unvignetted object ield o view (in degrees) o this system including the eye? A blank raytrace sheet is on the next page. First design the Keplerian Telescope: MP 6 6 t 7 140mm 20mm 120mm D 120mm 48mm D 20mm 14mm DEYE 5mm t 140mm EYE ER Because D 6D, the objective lens is the stop o the telescope. The eye relie is ound by imaging the objective through the eye lens: 1 1 1 t 140mm ER ER 23.33mm D XP D 48mm 8mm MP 6 The exit pupil o the telescope (8 mm) is larger than the entrance pupil o the eye (5 mm). When the eye is included in the system, the eye serves as the system stop.
Final Exam Page 3/11 Spring 2017 These results can be conirmed by tracing a potential chie ray and a potential marginal ray. The potential chie ray starts at the center o the objective lens and will cross the axis at the XP o the telescope and determines the ER. The potential marginal ray is launched at the edge o the objective lens (y = 24 mm) with a ero slope. As expected the potential marginal ray has a height o -4 mm at the eye. The system marginal ray is ound by scaling the potential marginal ray to the radius o the eye pupil (2.5 mm). The scale actor is 2.5mm Scale Factor 0.625 4mm The system EP is at the objective lens with a diameter o 30 mm. Continues
Final Exam Page 4/11 Spring 2017 Since vignetting cannot occur at a stop or pupil, the ield o view is limited by the eye lens. The condition or no vignetting must be satisied at the eye lens using a scaled chie ray: a y y y C y y 2.5mm y 14.0mm a D 2 7.0mm a 7.0mm 2.5mm C14.0 C Scale Factor 0.3214 The scaled chie ray has a height o 4.5 mm at the eye lens. The chie ray slope in object space is uo 0.03214 This slope can be converted into the unvignetted FOV o the system: 1 HFOV tan u O 1.84deg FOV 1.84deg 3.68deg It is also interesting to note that the telescope MP relates the chie ray angle in object space to the chie ray angle in image space: u MP u 0 in this case 0. 1928 6 0. 03214 Unvignetted FOV = +/- 1.84 degrees
Final Exam Page 5/11 Spring 2017 3) (15 points) An object-space telecentric system has a ocal length o 100 mm. The stop diameter is 10 mm (i.e. stop radius equals 5 mm). An object is located 250 mm to the let o the lens. The object sie is +/- 10 mm. Sketch the system and determine the required lens diameter or the system to be unvignetted. The method o solution is not speciied. For Object-Space Telecentricity, the stop must be located at the rear ocal point o the lens. Using a ray trace (next page), irst trace a potential marginal ray, scale this ray to the stop radius o 5 mm, and trace a chie ray that is parallel to the axis in object space. = 100 mm Stop y 250 mm F y = 100 mm 66.67 mm The image is located 66.67 mm to the right o the stop (166.67 mm to the right o the lens). Marginal ray scale actor: a y 5mm 5 1mm The image sie is ±6.667 mm. The chie ray goes through the center o the stop. Use the vignetting requirement to determine the lens radius: a y y y 12.5mm y 10.0mm a 22.5mm D 45.0mm Continues
Final Exam Page 6/11 Spring 2017 The problem can also be done geometrically. First determine the image distance: 1 1 1 250mm 166. 67mm The image is 66.67 mm to the right o the stop. Plot the image-side marginal ray: = 100 mm y Stop F y y 5mm y y 166. 67mm 66. 67mm = 100 mm 66.67 mm y 12. 50mm The chie ray height at the lens equals the object sie by telecentricity: y 10mm Vignetting: a y y a 22.5mm D 45.0mm Lens Diameter = 45.0 mm
Final Exam Page 7/11 Spring 2017 4) (10 points) A diameter stop is located in an optical system comprised o two thin lenses in air as shown. An object is located 100 mm to the let o the irst element. Determine the image location relative to the second element and the image magniication. NOTE: Use Gaussian Reduction and Gaussian Imaging or this problem. Cascaded imaging may not be used (you may not image through one lens and then use this image as an object or the other lens). Raytrace analysis is not permitted. Stop D = 1 = -40 mm 2 = 50 mm s P P d d s 1 / 40mm 0. 025 / mm 1 / 50mm 0. 020 / mm 1 2 System Power: t t 40mm 1 2 1 2 0. 015 / mm 66. 67mm 2 1 d t 53. 33mm d t 66. 67mm Imaging: s 100mm d 153. 33mm 1 1 1 117. 9mm rom P s d s 184. 6 mm to the right o L 2 Magniication: 117. 9mm m 0769. 153. 33mm Image Location: 184.6_ mm to the _R_ o the second element. Magniication = _-0.769_
Final Exam Page 8/11 Spring 2017 5) (15 points) For the same system consisting o two thin lenses and a diameter stop, determine the locations and diameters o the Entrance Pupil and the Exit Pupil. NOTE: Use Gaussian Reduction and Gaussian Imaging or this problem. Raytrace analysis is not permitted. Exit Pupil Solution: Stop D = 1 = -40 mm 2 = 50 mm XP 20mm 1 1 1 XP 33. 33mm XP The XP is to the let o L 2. 2 XP 33. 33mm m XP 167. 20. 0mm XP Sie: D m D D 20. 0mm XP XP D 33. 33mm XP XP: _33.33_ mm to the L_ o the second element. D XP = _33.33_ mm Continues or EP Solution
Final Exam Page 9/11 Spring 2017 Entrance Pupil Solution: Stop D = 1 = -40 mm 2 = 50 mm EP Since the Stop is a real object, the light goes rom right to let and n n 1 20mm n n 1 EP 1 1 1 20mm 40mm EP 13. 33mm EP The EP is to the right o L 1. 1 EP / n 13. 33mm/ 1 m EP 067. / n 20. 0mm/ 1 EP Sie: D m D D 20. 0mm EP EP D 13. 33mm EP EP: _13.33_ mm to the R_ o the irst element. D EP = _13.33_ mm
Final Exam Page 10/11 Spring 2017 6) (25 points) A ball lens is a glass sphere with a radius R that is used as a thick lens. The lens has index on reraction n = 1.500 and the lens is used in air. The ball lens is to be used as a 100X objective in a inite tube length microscope. The optical tube length OTL o the microscope is 200 mm. The OTL is the distance rom the rear ocal point o the objective (in this case, the rear ocal point o the ball lens) to the intermediate image produced by the objective. This intermediate image is presented to the eyepiece o the microscope. Determine the required radius o curvature o the ball lens R. For this objective, determine the object-side working distance (the distance rom the object to the ront vertex o the ball lens). First, consider the properties o a microscope objective: P P F F OTL = 200mm Imaging Requirements: 100X OTL 200mm m 100 100 1 1 1 100 1 101 1 101 200mm 100 200mm Alternate Newtonian Solution: m OTL 100 OTL 200. mm 100 200. mm 202mm 202. mm 100 Continues
Final Exam Page 11/11 Spring 2017 Now design the Ball Lens with a ocal length o 2.00 mm: n. 1 1 5 1 1 1 R R 2R n. 1 1 1 5 1 2 1 R R 2R t 2R t 12 1 2 n 1 1 1 2R 2 2R 2R 4R 1. 5 1 1 2 R 3R 3R 1 3R 2 WD n = 1.5 P P d R 2 R 200. mm 3 R 1. 333mm By inspection, both Principal Planes o the Ball Lens must be at the Center o Curvature o the Ball (coincident with the Nodal Points o the sphere). 2 t 1/ 2R 2R d d R or d n 2/ 3R 1. 5 R Working Distance: WD d R 2. 02mm 1. 333mm WD 0. 687mm Radius o Curvature R = 1.333_ mm Working Distance = 0.687_ mm