IMAGES AND OPTICAL INSTRUMENTS 31EXERCISESSection 31.1 Images with Mirrors 17. INTERPRETThis problem is about image formation in a plane mirror. DEVELOPA 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 2hdα=EVALUATESolving for the angle, we find 1111140 cmtantan35.22250 cmhdα−−⎛⎞⎛⎞===⎜⎟⎜⎟⎝⎠⎝⎠°for the given distances. Therefore, the plane of the mirror should be tilted by35from the vertical to provide the customer with a floor-level view of her shoes. .2°ASSESSThe angle tilted decreases if dis increased, and vice versa. This is what we expect from our experience at shoe stores. 18.(a)Useandin the mirror equation (Equation 31.2) to getss36 cms=15 cmf=/()(15 cm)(36 cm)/′fsf=−=(36 cm15 cm)25.7 cm,−=in front of the mirror. (b)The magnification (Equation 31.1) is/Ms s′= −=5725.7/36/()71.4%fsf−= −−= −= −(the image is reduced in size and inverted). (c)Sinces′is positive, the image is real. Ray tracing, as in Table 31.1, confirms these conclusions. 19. INTERPRETWe 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. DEVELOPThe magnification M, the ratio of the image heighth′to object height h, is given by Equation 31.1: hsMhs′′== −where sands′are object and image distances to the mirror. The two quantities sands′are related by the mirror equation (Equation 31.2): 111ssf+=′31.1
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