19_Ch 23 College Physics ProblemCH23 Mirrors and Lenses

19_Ch 23 College Physics ProblemCH23 Mirrors and Lenses -...

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Mirrors and Lenses 285 (b) The virtual image case is shown in the second diagram. Note that in this case, ( ) 12.9 cm qp = −+ , so the thin lens equation gives 11 1 12.9 cm 2.44 cm pp −= + ( or ) 22 12.9 cm 31.5 cm +− 0 = 2.10 cm or The quadratic formula then gives 15.0 cm = =− Since the object is real, the negative solution must be rejected leaving 2.10 cm p = . V ±²³´µl ¶·µ¸¹ ± º»º¼½¾¾º¿· À¼½9º¿· ÁÂù¿³ ² ³ 23.34 We must first realize that we are looking at an upright, magnified, virtual image. Thus, we have a real object located between a converging lens and its front-side focal point, so . The magnification is 0, 0, and 0 f <> > 2 q M p =+ , giving 2 = − . Then, from the thin lens equation,
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