PHY 1160C Homework Chapter 26: Optical Instruments Ch 26: 2, 3, 5, 9, 13, 15, 20, 25, 27

Transcription

1 PHY 60C Homework Chapter 26: Optical Instruments Ch 26: 2, 3, 5, 9, 3, 5, 20, 25, A pin-hole camera is used to take a photograph of a student who is.8 m tall. The student stands 2.7 m in front of the camera. Film is placed 0 cm behind the pin-hole. Determine the height of the image produced..8 m d =? 2.7 m 0.0 m The triangle formed by the object (the student) and the rays from the student s head and feet going to the pin-hole and the triangle formed by the image and the rays coming from the pinhole toward the head and feet of the image are similar triangles so the ratios of corresponding sides of these triangles are equal. That is, d i 0.0 m =.8 m 2.7 m d i =.8 m 2.7 m 0.0 m di = m = 6.7 cm 26.3 The distance from the cornea to the retina of some particular eye is 2. cm. Find the effective focal length of this eye for an object located (a) at infinity, (b) for an object located 0 m away, and (c) for an object located 25 cm away. Ch 26, p

2 We can directly apply the Image Equation, f = d o + d i The image distance, di, is the distance from the cornea to the retina, 2. cm. For (a), the object distance, do, is infinite, f = d o + d i f = + 2. cm f = cm f = 2. cm For (b), the object distance is do = 0 m f = 0 m + 2. cm f =,000 cm + 2. cm f = ( ) cm f = cm f = cm 2. cm For (c), the object distance is do = 25 cm f = 25 cm + 2. cm f = ( ) cm f = 0.56 cm f =.94 cm Ch 26, p 2

3 26.5 An object sits 35 cm to the left of a converging lens with focal length of 20 cm. 75 cm to the right is a second converging lens with focal length of 5 cm. Locate and characterize the final image. A ray diagram is always a good way to start. There is usually more real understanding available in the ray diagram than in merely solving the problem numerically. In addition, a good diagram also provides a place to keep all the dimensions of the problem. F F f = 20 cm f = 20 cm f 2 = 5 cm f 2 = 5 cm d i = 46.7 cm d o = 35 cm 75 cm d o2 = 28.3 cm d i2 = 3.9 cm For the first lens, we begin with the Image Equation, f = d o + d i or = + f d o d i And we know f = 20 cm and do = 35 cm, 20 cm = 35 cm + d i = d i 20 cm 35 cm di = 46.7 cm This image (which is real, inverted, and enlarged) becomes the object for the second lens as we again apply the Image Equation, f = d o + d i F2 F2 Ch 26, p 3

4 or = + f 2 d o2 d i2 Since the lenses are 75 cm apart, the image formed 46.7 cm to the right of lens # is located 28.3 cm to the left of lens #2 so the object distance, then, is do2 = 28.3 cm 5 cm = 28.3 cm + d i2 = d i2 5 cm 28.3 cm di2 = 3.9 cm That is, the final image is located 3.9 cm to the right of the second lens. Since di2 > 0, we know this image is real. It is inverted, compared to the object for lens #2. But that was an inverted image of the original object. Therefore, this final image is upright. The total magnification is the product of the magnifications of the two lenses, Mtot = M M2 M = d i d o M = d i = 46.7 cm d o 35 cm =.33 M 2 = d i2 = 3.9 cm d o cm =.3 Mtot = (.33)(.3) Mtot =.5 Ch 26, p 4

5 26.9 A patient's eye can focus only on objects beyond 00 cm. What word characterizes this type of vision problem? What are the focal length and power of the contact lens needed to correct this problem? This patient is farsighted or hyperopic. This patient has trouble reading a book at a comfortable distance such as 25 cm so we must find a lens that will take an object at 25 cm (do = 25 cm) and create an image (a virtual image) at 00 cm (di = 00 cm). virtual image at d = 00 cm i real object at d = 25 cm o Near point: nearest point at which vision is clear 25 cm is used as a comfortable reading distance A diagram may help to clearly establish that we need do = 25 cm di = 00 cm It is important to remember or realize that the image formed is a virtual image. It is on the left side of the lens in the diagram above. This means the image distance is negative. Once we know do and di then it is easy to find the focal length of the lens from the Image Equation, f = d o + d i f = 25 cm + 00 cm f = 33.3 cm To describe the lens in terms of power (in units of diopters) we need to take the inverse of its focal length in meters, f = 0.3 m P = /f = 3 diopters Ch 26, p 5

6 26.0 A certain eye can focus only on objects closer than 50.0 cm. What word characterizes this type of vision problem? What sort of contact lens (described both in focal length and power) will correct this problem? This eye is myopic or nearsighted. We need to find a lens that will take an object at infinity (do = ) and produce a virtual image at 50.0 cm (di = 50.0 cm = 0.50 m). real object at d = o virtual image at d = 50 cm i The object distance is taken as infinity so the patient can see distant scenes or road signs. Far point: 50 cm from the eye is the farthest this eye can focus clearly. Now we know the image and object distances do = di = 50.0 cm It is important to remember or realize that the image formed is a virtual image. It is on the left side of the lens in the diagram above. This means the image distance is negative. Once we know do and di then it is easy to find the focal length of the lens from the Image Equation, f = d o + d i f = cm f = 50.0 cm f = 0.50 m P = /f = 2 diopters Note: This problems started out with a far point of 500 cm or 5.0 m. As we will see, that does not require very much correction to be able to see something at infinity. This does not Ch 26, p 6

7 seem like a reasonable or interesting problem. Therefore, I changed the problem. However, let s go ahead and solve this for do = di = 500 cm f = d o + d i f = cm f = 500 cm = 5 m P = 0.20 diopters 26. A particular eye can focus only on objects more distant than 50 cm. What word characterizes this type of vision problem? What are the focal length and power of the corrective lens needed to correct this problem? The lens is to be worn 2.0 cm in front of the eye? virtual image at d = 48 cm i real object at d = 23 cm o Near point: nearest point at which vision is clear. Being 50 cm from the eye means it is 48 cm from the lens. When the object is 25 cm from the eye it is 23 cm from the lens. This is used as a comfortable reading distance Using a lens that is 2 cm from the eye is but a slight variation from problem 26.9 that used a contact lens. For this situation we have do = 23 cm di = 48 cm It is important to remember or realize that the image formed is a virtual image. It is on the left side of the lens in the diagram above. This means the image distance is negative. Once we Ch 26, p 7

8 know do and di then it is easy to find the focal length of the lens from the Image Equation, f = d o + d i f = 23 cm + 48 cm f = 27.2 cm = m P = /f = 3.7 diopters A diamond is viewed with a jeweler's loupe that has a focal length of 5.0 cm. Where must the diamond be placed to provide a virtual image at infinity? What will be the angular magnification of this simple magnifier? The diamond should be placed at the focal point of the lens (the jeweler s loupe), that is, 5.0 cm from the lens. The magnification is given by M = 25 cm f M = M = 5 25 cm 5 cm Ch 26, p 8

9 26.25 A microscope has its objective and eyepiece 8 cm apart. If fobj = 0.40 cm and feye = 5.0 cm, where must a specimem be located to produce a final virtual image at infinity? What is the total magnification of this microscope? objective eyepiece 8.0 cm f ob f ob d o d i f eye To have the final image at di =, we need the real image formed by the objective to be located at the focal point of the eyepiece. With the objective and eyepiece lenses 8.0 cm apart this means, di + feye = 8.0 cm di cm = 8.0 cm di = 3.0 cm Now we can again apply the Image Equation, f = d o + d i d o = f _ d i = d o 0.40 cm _ 3.0 cm do = 0.4 cm Ch 26, p 9

10 26.27 An astronomical telescope is used to view an object at infinity. The objective lens has a focal length of 5.0 cm. Where must the 0.5 cm eyepiece be placed to form an image at infinity? What is the total angular magnification? f obj = d i f eye F eye F ob h ' Feye virtual image at infinity The real image formed by the objective lens will be located at the focal point for that lens, di = fobj = 5.0 cm from that lens. To produce a virtual image at infinity, the eyepiece must be located so that this image which acts as an object for the eyepiece is located at the focal point for the eyepiece lens, do = feye = 0.5 cm. Therefore, the eyepiece must be 5.5 cm from the objective lens. The total magnification is given by M = f obj f eye = M = 30 5 cm 0.5 cm = 30 Ch 26, p 0