1、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 build
2、ings shadow is 170m long. How tall is the building?Solution. According to the law of rectilinear propagation, we get, 4.32170x 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 angl
3、e does the light make with the normal in the glass?Solution. According to the law of refraction, We get,siniII6298.05.143sin8.ISo 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? Do
4、es it appear larger or smaller?Solution. According to the equation.rnland n=1 , n=1.33, r=-20we can get 146.103.586.)(3. 5.2.0.11 lncmlrnn=1.50n=1.33water45oIASo the fish appears larger.4.An object is located 2cm to the left of convex end of a glass rod which has a radius of curvature of 1cm. The in
5、dex of refraction of the glass is n=1.5. Find the image distance.Solution. Refer to the figure. According to the equation R2=-20cmR1=20cmA-10cmrnl and n=1, n=1.5, l1=-2cm, r1=1cm , we get cmldl2015.1English Homework for Chapter 21.An object 1cm high is 30cm in front of a thin lens with a focal lengt
6、h of 10cm. Where is the image? Verify your answer by graphical construction of the image.Solution. According to the Gausss equation,fl1and l=-30cm f=10 cm.we get r1=1cmA A-l1=2cm l2-l=30cmf=10cmy=1cm)(15)30(cmlfl Others are omitted.2.A lens is known to have a focal length of 30cm in air. An object i
7、s placed 50cm to the left of the lens. Locate the image and characterize it.Solution. According to Gausss equation,fl1and f=30cm l=-50cmwe get )(75)0(3cmlfl 5.107lThe image is a real, larger one.3.The object is transparent cube, 4mm across, placed 60cm in front of 20cm focal length. Calculate the tr
8、ansverse and axial magnification and describe what the image looks like?Solution. From Gausss equation, we find -l=50cmf=30cmfor the rear surface of the cube (the face closer to the lens) that, )(302)6(01 cmfl For the front surface (the face farther away from the lens),)(9.204.6)(2cmlThe transverse
9、magnification for the rear surface is 5.063tMBut the axial magnification is 25.0)4.6(093lMaSince atM,the cube doesnt look like a cube.4.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 radi
10、i?Solution. Supposing r1= -2r2 ( 2=-2 1),according to the lens equation )(12n we get, )(152.(28.01564.02r 1=7.8(cm) r2=-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 im
11、age of the object r1 -r2onto a screen 14cm away. What is the diameter of the exit pupil?Solution. Refer to the figure. First, from the known focal length and the image distance, we find the object distance.fl1and l=14 f=9 l=-25.2(cm)The stop is one-half that distance is front of the lens, so ls=12.6
12、(cm) l-lImageLensStopObjectl s=31.5(cm)2.531sstopexlD )(8.02cmex2. 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
13、 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? F1(F2)-l2L1Stopf=12.5cm8cmSolution. Refer to the figure. For the system to be a focal, the focal points of the two lenses must coincide. Since f1=12.5cm, and the two lenses are 8cm apart, so f2=-4.5cm. The entrance pupil is the image of stop formed by the first lens. According to Gausss equation, 11fl and l1=4cm, f1=12.5cm. We get)(8.5.4211 cmlfl )(0.24.1DstopentracThe exit pupils location is
