Chapter 27: Optical InstrumentsAnswers to Even-Numbered Conceptual Questions2. No. The lens will still show a complete image, though you may have to move your head more from side toside to see it all.4. The reason things look blurry underwater is that there is much less refraction of light when it passes fromwater to your cornea than when it passes from air to your cornea. Therefore, your eyes simply aren’tconverging light enough when they are in water. Since farsightedness is caused when your eyes don’tconverge light as much as they should (see Figure 27-11), this can be considered as an extreme case offarsightedness.6. Yes, it matters. A simple magnifier is nothing more than a convex lens. As we can see from Figure 26-35, aconvex lens forms an enlarged (magnified) image only when the object is closer to the lens than its focallength.8. No. Chromatic aberration occurs in lenses because light of different frequency refracts by different amounts.In the case of a mirror, however, all light—regardless of its frequency—obeys the same simple law ofreflection; namely, that the angle of reflection is equal to the angle of incidence. Since light of all colors isbent in the same way by a mirror, there is no chromatic aberration.Solutions to Problems and Conceptual Exercises1. Picture the Problem: An octopus moves its rigid lens back and forth inside its eye in order to change the distance from the lens to the retina and bring an object into focus.Strategy: Use the principles concerning the operation of converging lenses to answer the conceptual question.Solution: 1. (a) As an object moves closer to the front of an octopus eye, the image it forms moves farther behind the eye. The situation is similar to that in active example 27-1 and Figure 26-35 (a). To keep the image on the retina, therefore, it is necessary to move the lens itself farther from the retina.2. (b) The best explanation is II. When the object moves closer to the eye the image produced by the lens will be fartherbehind the lens; therefore, the lens must move farther from the retina. Statement I is incorrect. Insight: Likewise, the lens-retina distance is a minimum when the object is infinitely far away. In that case the image appears right at the focal plane of the lens.2. Picture the Problem: The figure shows a 1.9-meter-tall person (represented by the arrow) standing do = 3.2 meters from the eye lens. The image is formed on the retina di = 2.5 cm from the lens.Strategy: Use equations 26-17 and 26-18 to calculate the image height.Solution: 1. (a) Combine equations 26-17 and 26-18 to calculate the image height: hi=mho=−didoho=0.025 m3.2 m1.9 m( )=1.5 cmCopyright © 2010 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist. Noportion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.27 – 1Chapter 27: Optical Instruments James S. Walker, Physics, 4th Edition2. (b) Change the object distance to 4.2 m: hi=0.025 m4.2 m1.9 m( )=1.1 cmInsight: As an object moves away from you, the size of the image it forms on your retina decreases.Copyright © 2010 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist. Noportion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.27 – 2Chapter 27: Optical Instruments James S. Walker, Physics, 4th Edition3. Picture the Problem: The figure shows an object a distance do from the eye. The image is formed on the retina 2.5 cm behind the lens.Strategy: Use equations 26-17 and 26-18 to calculate the image height for the two different objects that are specified in the problem statement.Solution: 1. Combine equations26-17 and 26-18 to find the image height of the tree: hi=mho=−didoho=2.5 cm210 ft43 ft( )=0.51 cm2. Calculate the image height of the flower: hi=2.5 cm2.0 ft1.0 ft( )=1.3 cm3. The flower forms the larger image.Insight: The size of the image on the retina is determined by the angular size of an object as viewed by the eye. The flower has a larger angular size (even though it is physically smaller than the tree) because it is nearby.4. Picture the Problem: The figure shows an object at the eye’s near point. The lens has a focal length 2.20 cm and the image is produced on the retinadi = 2.60 cm behind the lens.Strategy: Solve equation 26-16 for the object distance, which in this case is also the near-point distance, N.Solution: Set the object distance in equation 26-16 equal to the near point: do=N =1f−1di⎛⎝⎜⎞⎠⎟−1=12.20 cm−12.60 cm⎛⎝⎜⎞⎠⎟−1=14 cmInsight: The near point is a function of the size of the eye and the minimum focal length of the lens. If the eye were thinner, say di = 2.40 cm, this lens would produce a near point of 26 cm. 5. Picture the Problem: The eye produces an image di = 2.60 cm behind the lens of objects that are located at 285 cm and 28.5 cm in front of the eye.Strategy: Solve equation 26-16 for the focal length. Solution: 1. (a) Find f for do = 285 cm: f =1di+1do⎛⎝⎜⎞⎠⎟−1=12.60 cm+1285 cm⎛⎝⎜⎞⎠⎟−1= 2.58 cm2. (b) Find f for do = 28.5 cm: f =12.60 cm+128.5 cm⎛⎝⎜⎞⎠⎟−1= 2.38 cmInsight: As the object moves toward the eye, the focal length decreases so that the image distance remains constant.Copyright © 2010 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist. Noportion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.27 – 3Chapter 27: Optical Instruments James S. Walker, Physics, 4th Edition6. Picture the Problem: The f-number of a camera lens is the ratio of the focal length to the aperture diameter.Strategy: Divide the focal length by the f-number to calculate the aperture diameter for each lens. Then rank the lenses by aperture diameter.Solution: 1. Calculate the diameter of lens A: D =ff -number=150 mm1.2=130 mm2. Calculate the diameter of lens B: D =ff -number=150 mm5.6=27 mm3. Calculate the diameter of lens C: D =ff -number=35 mm1.2=29 mm4. Calculate the diameter of lens D: D =ff -number=35 mm5.6=6.3 mm5. Rank the lenses from largest to smallest diameter: A, C, B, DInsight: Note that lenses B and C have nearly the same diameters.7.
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