19_Chapter 23 HomeworkCH23 Mirrors and Lenses

# 19_Chapter 23 HomeworkCH23 Mirrors and Lenses - 285 Mirrors...

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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 q p =− + , so the thin lens equation gives 1 1 1 12.9 cm 2.44 cm p p = + ( or ) 2 2 12.9 cm 31.5 cm p p + 0 = 2.10 cm or p p The quadratic formula then gives 15.0 cm = = − Since the object is real, the negative solution must be rejected leaving 2.10 cm p = . 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 q p f < > > 2 q M p =− =+ , giving 2 q p = − . Then, from the thin lens equation, 1 1 1 1 2 2 p p p f = + = or ( ) 2 2 2.84 cm
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