Mirrors and Lenses285(b) The virtual image case is shown in the second diagram. Note that in this case, ( )12.9 cmqp=−+, so the thin lens equation gives 11112.9 cm2.44 cmpp−=+(or )2212.9 cm31.5 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 cmp=. V±²³´µl¶·µ¸¹±º»º¼½¾¾º¿·À¼½9º¿·ÁÂÃ¹¿³²³23.34We 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 0f<>>2qMp=+, giving 2= −. Then, from the thin lens equation,
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