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31_InstSolManual_PC

# 31_InstSolManual_PC - IMAGES AND OPTICAL INSTRUMENTS 31...

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IMAGES AND OPTICAL INSTRUMENTS E XERCISES Section 31.1 Images with Mirrors 17. I NTERPRET This problem is about image formation in a plane mirror. D EVELOP A small mirror ( M ) on the floor intercepts rays coming from a customer’s shoes ( O ), which are traveling nearly parallel to the floor. The angle to the customer’s eye ( E ) from the mirror is twice the angle of reflection, so tan 2 h d α = E VALUATE Solving for the angle, we find 1 1 1 1 140 cm tan tan 35.2 2 2 50 cm h d α - - = = = ° for the given distances. Therefore, the plane of the mirror should be tilted by 35.2 ° from the vertical to provide the customer with a floor-level view of her shoes. A SSESS The angle tilted decreases if d is increased, and vice versa. This is what we expect from our experience at shoe stores. 18. (a) Use 36 cm s = and 15 cm f = in the mirror equation (Equation 31.2) to get /( ) (15 cm)(36 cm)/ s sf s f ′ = - = (36 cm 15 cm) 25.7 cm, - = in front of the mirror. (b) The magnification (Equation 31.1) is / M s s = - = 5 7 25.7/36 /( ) 71.4% f s f - = - - = - = - (the image is reduced in size and inverted). (c) Since s is positive, the image is real. Ray tracing, as in Table 31.1, confirms these conclusions. 19. I NTERPRET We have an image formation problem involving a concave mirror. We want to know the orientation of the image as well as its height compared to the object. D EVELOP The magnification M , the ratio of the image height h to object height h , is given by Equation 31.1: h s M h s = = - where s and s are object and image distances to the mirror. The two quantities s and s are related by the mirror equation (Equation 31.2): 1 1 1 s s f + = 31.1 31

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31.2 Chapter 31 where f is the focal length of the mirror. E VALUATE (a) One can solve Equation 31.2 for , s and substitute into Equation 31.1, to yield /( ) 1 5 4 h s sf s f f f M h s s s f f f - = = - = - = - = - = - - - (b) A negative magnification applies to a real, inverted image. A SSESS The situation corresponds to the first case shown in Table 31.1. From the ray diagram, we see that the image is real, inverted, and reduced in size. Since 0, s the image is in front of the mirror. 20. (a) The mirror equation (Equation 31.2) gives /( ) (18 cm)( 40 cm)/( 58 cm) 12.4 cm s fs s f = - = - - = (positive distances are in front of the mirror, negative distances behind). (b) Equation 31.1 gives / M s s = - = 40 cm/12.4 cm 3.22. + = 21. I NTERPRET This problem is about image formation in a concave mirror. We want to know the distance the object should be placed in order to produce a full-size image. D EVELOP The magnification M , the ratio of the image height h to object height h , is given by Equation 31.1: h s M h s = = - where s and s are object and image distances to the mirror. A full-size image means that | | 1. M = E VALUATE (a) For a full-sized image, we require /( ) 1 h s sf s f f M h s s s f - = = - = - = - = - - or 2 . s f R = = That is, put the object at the center of curvature of the mirror (or at twice its focal length).
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