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# chap31 - IMAGES AND OPTICAL INSTRUMENTS 31 EXERCISES...

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IMAGES AND OPTICAL INSTRUMENTS 31 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 from the vertical to provide the customer with a floor-level view of her shoes. .2 ° 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 and in the mirror equation (Equation 31.2) to get s s 36 cm s = 15 cm f = /( ) (15 cm)(36 cm)/ f 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

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