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Unformatted text preview: HW10 Solutions 26. REASONING The magnification m is given by m = – d i / d o (Equation 25.4), where d i and d o are the image and object distances, respectively. The object distance is known, and we can obtain the image distance from the mirror equation: o i 1 1 1 d d f + = (25.3) SOLUTION Solving the mirror equation (Equation 25.3) for the image distance d i gives 1 1 1 1 1 1 d d f d f d d f fd d fd d f o i i o o o i o o or = or + = =-- =- Substituting this result into the magnification equation (Equation 25.4) gives m d d fd d f d f f d = - = -- =- i o o o o o / c h Using this result with the given values for the focal length and object distances, we find Smaller object distance Greater object distance m f f d m f f d =- =--- = =- =--- = o o cm cm cm cm cm cm 27 0 27 0 9 0 0 750 27 0 27 0 18 0 0 600 . . . . . . . . b gb g b gb g 29. SSM REASONING We need to know the focal length of the mirror and can obtain it from the mirror equation, Equation 25.3, as applied to the first object: 1 1 1 d d f f o1 i1 1 14.0 cm 1 –7.00 cm or –14.0 cm + = + = = According to the magnification equation, Equation 25.4, the image height h i is related to the object height h o as follows: h mh d d h i o i o o = = – / c h . 2 THE REFLECTION OF LIGHT: MIRRORS SOLUTION Applying this result to each object, we find that h h i2 i1 = , or – – d d h...
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This note was uploaded on 10/04/2009 for the course PHYS 750:204 taught by Professor Rollino during the Spring '09 term at Rutgers.
- Spring '09