WHomework set 16_solns_2011

WHomework set 16_solns_2011 - Homework set # 16 Problems:...

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Homework set # 16 Problems: 23.13, 11, 5, 28 Solutions: 23.5 Since the mirror is convex, 0 R . Thus, 0.550 m R   . With a real object, 0 p , so 10.0 m p  . The mirror equation then gives the image distance as 121 2 1 0.550 m 10.0 m qRp  , or 0.268 m q   Thus, the image is virtual and located 0.268 m behind the mirror The magnification is 0.268 m 0.026 8 10.0 m q M p  Therefore, the image is upright   since 0 M and diminished in size   since 1 M 23.11 The image is upright, so 0 M , and we have 2.0 q M p , or   2.0 2.0 25 cm 50 cm qp The radius of curvature is then found to be 211 1 1 2 1 25 cm 50 cm 50 cm Rpq  , or 0.50 m 2 1.0 m +1 R    
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23.13 The focal length of the mirror may be found from the given object and image distances as 111 f pq  , or    152 cm 18.0 cm 16.1 cm 152 cm 18.0 cm f   For an upright image twice the size of the object, the magnification is 2.00 q M p  giving qp Then, using the mirror equation again, p qf becomes 1 21 1 p p pf
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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 University of Georgia Athens.

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WHomework set 16_solns_2011 - Homework set # 16 Problems:...

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