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Unformatted text preview: IMAGES AND OPTICAL INSTRUMENTS 31 EXERCISES Section 31.1 Images with Mirrors 17. INTERPRET This problem is about image formation in a plane mirror. DEVELOP 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 DEVELOP 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 = EVALUATE 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 by35 from the vertical to provide the customer with a floor-level view of her shoes. .2 ° ASSESS 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. INTERPRET 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. DEVELOP 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 31.2 Chapter 31 where f is the focal length of the mirror. EVALUATE (a) One can solve Equation 31.2 for , s ′ and substitute into Equation 31.1, to yield /( ) 1 5 4 h s s f s f f f M h s s s f f f ′ ′ − = = − = − = − = − = − − − (b) A negative magnification applies to a real, inverted image. ASSESS 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. INTERPRET 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....
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This note was uploaded on 10/21/2010 for the course PHYSICS 2131441 taught by Professor Pheong during the Fall '10 term at University of California, Berkeley.
- Fall '10