English Homework for Chapter 11.In ancient times the rectilinear propagation of light was used to measure the height of objects by comparing the length of their shadows with the length of the shadow of an object of known length. A staff 2m long when held erect casts a shadow 3.4m long, while a building’s shadow is 170m long. How tall is the building?Solution. According to the law of rectilinear propagation, we get, x=100 (m)So the building is 100m tall.2.Light from a water medium with n=1.33 is incident upon a water -glass interface at an angle of 45o. The glass index is 1.50. What angle does the light make with the normal in the glass? Solution. According to the law of refraction, We get,So the light make 38.8o with the normal in the glass.3. A goldfish swims 10cm from the side of a spherical bowl of water of radius 20cm. Where does the fish appear to be? Does it appear larger or smaller? Solution. According to the equation.and n ’=1 , n=1.33, r=-20we can getSo the fish appears larger.4.32170=x ''sin sin I n I n =626968.05.145sin 33.1sin =⨯='οI ο8.38='I r nn l n l n -'=-''11416.110133.15836.8)(5836.81165.02033.01033.11>-=⨯⨯-=''=-='∴-=--+-=-'+='l n l n cm l r n n l n l βΘn′=1.50n=1.33water45oI′A4.An object is located 2cm to the left of convex end of a glass rod which has a radius of curvature of 1cm. The index of refraction of the glass is n=1.5. Find the image distance. Solution. Refer to the figure. According to the equationand n=1, n ’=1.5, l 1=-2cm, r 1=1cm , we getEnglish Homework for Chapter 21.An object 1cm high is 30cm in front of a thin lens with a focal length of 10cm. Where is the image? Verify your answer by graphical construction of the image. Solution. According to the Gauss’s equation,r n n l n l n -'=-''cm l l d l l l 2021115.15.121211='∴-∞='-=∞='∴=-+-='R 2=-20cm R 1=20cmA-10cmr 1=1cmAA′-l 1=2cml 2′′and l=-30cm f ’=10 cm.we getOthers are omitted.2.A lens is known to have a focal length of 30cm in air. An object is placed 50cm to the left of the lens. Locate the image and characterize it. Solution. According to Gauss’s equation,and f′=30cm l =-50cmwe getThe image is a real, larger one.3.The object is transparent cube, 4mm across, placed 60cm in front of 20cm focal length. Calculate the transverse and axial magnification and describe what the image looks like? Solution. From Gauss’s equation, we find for the rear surface of the cube (the face closer to the lens)that,For the front surface (the face farther away from the lens),The transverse magnification for the rear surface isBut the axial magnification isSince,the cube doesn’t look like a cube.f l l '=-'11)(15)30(10)30(10cm l f l f l =-+-⨯=+''='f l l '=-'11)(75)50(30)50(30cm l f l f l =-+-⨯=+''='5.15075-=-='=l l β)(3020)60()20()60(111cm f l f l l +=+-⨯-='+'=')(9.29204.6020)4.60(2cm l +=+-⨯-='⨯-=-+=5.06030t M ⨯+=----=∆'∆=25.0)4.60(609.2930l l M a at M M ≠-l =50cmf ′=30cm4.A biconvex lens is made out of glass of n=1.52. If one surface has twice the radius of curvature of the other, and if the focal length is 5cm, what are the two radii? Solution. Supposing r 1= -2r 2 (ρ2=-2ρ1),according to the lens equationwe get,∴r 1=7.8(cm) r 2=- 3.9(cm)返回English Homework for Chapter 41. A stop 8mm in diameter is placed halfway between an extended object and a large -diameter lens of 9cm focal length. The lens projects an image of the object onto a screen 14cm away. What is the diameter of the exit pupil?))(1(21ρρϕ--=n ))(152.1(5121ρρ+-=1282.01=∴ρ2564.02-=ρr 1-r 2l ’-lImageLensStopObjectSolution. Refer to the figure. First, from the known focal length and the image distance,we find the object distance.and l ’=14 f ’=9 l =-25.2(cm)The stop is one -half that distance is front of the lens, so l s =12.6(cm) ∴l s ’=31.5(cm)∴2. Two lenses, a lens of 12.5cm focal length and a minus lens of unknown power, are mounted coaxially and 8 cm apart. The system is a focal, that is light entering the system parallel at one side emerges parallel at the other. If a stop 15mm in diameter is placed halfway between the lenses: 1) Where is the entrance pupil? 2) Where is the exit pupil? 3) What are their diameters?Solution. Refer to the figure. For the system to be a focal, the focal points of the two lenses mustcoincide. Since f 1’=12.5cm, and the two lenses are 8cm apart, so f 2’=-4.5cm. The entrance pupil is the image of stop formed by the first lens.f l l '=-'111Θ22.255.31-='==ss stop ex l l D D βΘ)(28.05.2cm D ex =⨯=F 1’(F 2)-l 2’L 1’Stopf =12.5cm8cmAccording to Gauss’s equation,and l 1’=4cm, f 1’=12.5cm. We getThe exit pupil’s location is返回English Homework for Chapter 71. A person wants to look at the image of his or her own eyes, without accommodation, using a concave mirror of 60cm radius of curvature. How far must the mirror be from the eye if the person has 1) Normal vision?2) 4diopter myopia, without correction? 3) 4diopter hyperopia, without correction? Solution.1) When the person has normal vision, according to the following scheme 1, we getso,2) According to the following scheme 2,111111f l l '=-'())(88.55.845.1211111cm l f l f l =⨯='-'''=)(05.22488.5151mm D D stopentrance =⨯==β)(95.715412.2)(12.25.818)4()5.4()4()5.4(222222mm D D cm f l l f l stop exit =⨯=•=-=-=-+--⨯-='+'='β∞='l cm rl 302==l'=∞lScheme 1and, orSo the mirror must be 75cm or 10cm from the eye. 3) According to the following scheme 3,and, or (Since the object is real, so we can give up this answer)So the mirror must be 50cm from the eye.141-=m l rcml l r 25-=='r l l 211=+'Θ)(25cm l l +'=cm r 60=265852253048585025308522±=⨯⨯-±==⨯+-l l l Θ⎩⎨⎧==∴)(50')(7511cm l cm l ⎩⎨⎧-==)(15')(1022cm l cm l r l l 211=+'Θ)(25'cm l l +=cm r 60=265352253043535025303522±=⨯⨯+±==⨯--l l l Θ⎩⎨⎧==∴)(75')(5011cm l cm l ⎩⎨⎧=-=)(10')(1522cm l cm l ll'Scheme 225l'lScheme 3252. Discussion: What differences between the following situations:1) a microscope is used for projection;2) the microscope is used for visual observation.返回工程光学（上）期末考试试卷一．问答题：（共12分，每题3分）1．摄影物镜的三个重要参数是什么？它们分别决定系统的什么性质？2．为了保证测量精度，测量仪器一般采用什么光路？为什么？3．显微物镜、望远物镜、照相物镜各应校正什么像差？为什么？4．评价像质的方法主要有哪几种？各有什么优缺点？二．图解法求像或判断成像方向：（共18分，每题3分）1．求像A'B'2．求像A'B'3．求物AB经理想光学系统后所成的像，并注明系统像方的基点位置和焦距4．判断光学系统的成像方向5．求入瞳及对无穷远成像时50%渐晕的视场6．判断棱镜的成像方向三．填空：（共10分，每题2分）1．照明系统与成像系统之间的衔接关系为：①________________________________________________②________________________________________________2．转像系统分____________________和___________________两大类，其作用是：_________________________________________ 3．一学生带500度近视镜，则该近视镜的焦距为_________________,该学生裸眼所能看清的最远距离为_________________。