# ch23-sol - Chapter 23 Solution 23.5 Since the mirror is...

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Chapter 23 Solution 23.5 Since the mirror is convex, 0 R < . Thus, 0.550 m R = − . With a real object, 0 p > , so . The mirror equation then gives the image distance as 10.0 m p =+ 121 2 1 0.550 m1 0 . 0 m qRp =−= , or 0.268 m q = − Thus, the image is virtual and located 0.268 m behind the mirror The magnification is 0.026 8 q M p =− =− = Therefore, the image is upright ( ) since 0 M > and diminished in size () 1 M < 23.7 The radius of curvature of a concave mirror is positive, so 20.0 cm R = + . The mirror equation then gives 10.0 cm 1 1 p p p −= , or ( ) p q p = (a) If 40. , 0 cm p = 13.3 cm q and 0.333 40.0 cm q M p = − The image is 13.3 cm in fron inverted t of the mirror, real, and (b) When 20. , p = q = + and 1.00 q M p The image is 20.0 cm in fron t of the (c) If 10 , .0 cm p = ( ) q =→ and no image is form ed. Parallel rays leave the

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23.8 (a) Since the object is in front of the mirror, 0 p > . With the image behind the mirror, . The mirror equation gives the radius of curvature as 0 q < 211 1 1 1 1.00 cm 10.0 cm 10.0 Rpq = + =−= 0 - 1 cm or 22 . 22 cm 9 R ⎛⎞ == + ⎜⎟ ⎝⎠ 23.11 The magnified , virtual images formed by a concave mirror are upright, so 0 M > .
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ch23-sol - Chapter 23 Solution 23.5 Since the mirror is...

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