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1030_Physics ProblemsTechnical Physics

# 1030_Physics ProblemsTechnical Physics - Chapter...

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Chapter 36 371 *P36.61 (a) 1 1 1 1 5 1 7 5 15 1 1 1 1 q f p q = = = cm cm cm . M q p M M M M M q p p q p q f q q q p p q p q 1 1 1 1 2 2 2 2 2 2 2 2 2 2 2 2 2 2 1 2 2 15 2 1 2 1 2 2 1 1 1 1 2 1 1 10 15 30 7 5 15 30 15 67 5 = − = − = − = = − = − = − = + = + = = = + + + = + + + = cm 7.5 cm cm cm, cm cm cm cm cm cm 1 a f . . (b) 1 1 1 1 5 1 1 1 + = = p q f cm Solve for q 1 in terms of p 1 : ′ = ′ − q p p 1 1 1 5 5 (1) ′ = − = − ′ − M q p p 1 1 1 1 5 5 , using (1). ′ = ′ = = − ′ − = − M M M M M M p q p 1 2 2 1 1 2 2 3 5 5 b g ∴ ′ = ′ − q p p 2 2 1 3 5 5 b g (2) Substitute (2) into the lens equation 1 1 1 1 10 2 2 2 + = = p q f cm and obtain p 2 in terms of p 1 : ′ = ′ − ′ − p p p 2 1 1 10 3 10 3 5 b g b g . (3) Substituting (3) in (2), obtain q 2 in terms of p 1 : ′ = ′ − q p 2 1 2 3 10 b g . (4) Now, ′ + ′ + ′ + ′ = p q p q 1 1 2 2 a constant. Using (1), (3) and (4), and the value obtained in (a): ′ + ′ − + ′ − ′ − + ′ − = p p p p p p 1 1 1 1 1 5 5 10 3 10 3 5 2 3 10 67 5 b g b g b g . . This reduces to the quadratic equation 21 322 5 1 212 5 0 1 2 1 ′ + = p p . . , which has solutions ′ = p 1 8 784 . cm and 6.573 cm. Case 1: ′ = p 1 8 784 . cm
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