WHomework set 17_solns_2011

# WHomework set 17_solns_2011 - Homework set 17 Problems...

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Homework set # 17 Problems: 23.32, 42, 7, 8, 29 Solutions: 23.7 The radius of curvature of a concave mirror is positive, so 20.0 cm R   . The mirror equation then gives  121 1 1 1 0 . 0 c m 10.0 cm 10.0 cm p qRp p p   , or   10.0 cm 10.0 cm p q p (a) If 40.0 cm p , 13.3 cm q  and 13.3 cm 0.333 40.0 cm q M p    The image is 13.3 cm in front of the mirror, real, and inverted (b) When 20.0 cm p , 20.0 cm q   and 20.0 cm 1.00 20.0 cm q M p The image is 20.0 cm in front of the mirror, real, and inverted (c) If 10.0 cm p ,    10.0 cm 10.0 cm 10.0 cm 10.0 cm q  and no image is formed. Parallel rays leave the mirror 23.8 (a) Since the object is in front of the mirror, 0 p . With the image behind the mirror, 0 q . The mirror equation gives the radius of curvature as 211 1 1 1 0 - 1 1.00 cm 10.0 cm 10.0 cm Rpq  or 10.0 cm 22 . 2 2 c m 9 R     (b) The magnification is   10.0 cm 10.0 1.00 cm q M p  

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23.29 From the thin lens equation, 11 1 p qf , the image distance is found to be
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## This note was uploaded on 05/12/2011 for the course PHYS 1112 taught by Professor Seaton during the Spring '08 term at UGA.

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WHomework set 17_solns_2011 - Homework set 17 Problems...

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