1030_Physics ProblemsTechnical Physics

1030_Physics ProblemsTechnical Physics - Chapter 36 *P36.61...

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Chapter 36 371 *P36.61 (a) 111 1 5 1 75 15 1 qfp q =−= ∴= cm cm cm . M q p MMM M M q p pq f qq qp pqpq 1 1 1 12 2 2 2 2 22 222 122 15 2 1 2 2 1 2 11 10 15 30 7 5 15 30 15 67 5 =− =∴ = = = += = = +++= + + + = cm 7.5 cm cm cm, cm cm cm cm cm cm 1 af .. (b) 1 5 + == f cm Solve for q 1 in terms of p 1 : ′ = ′− q p p 1 1 1 5 5 (1) ′=− M q pp 1 1 5 5 , using (1). ′ = ′′ ′ = −= M M M p q p 2 1 1 2 2 3 5 5 bg ∴ ′ = p 1 3 5 5 (2) Substitute (2) into the lens equation 1 10 + f cm and obtain p 2 in terms of p 1 : ′= ′ − p p p 2 1 1 10 3 10 35 . (3) Substituting (3) in (2), obtain q 2 in terms of p 1 : ′ = ′ − 21 23 10 . (4) Now, ′+ ′+ ′+ ′= 1122 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 2 3 10 67 5 This reduces to the quadratic equation 21 322 5 1 212 5 0 1 2 1 = , which has solutions p 1 8784 . cm and 6.573 cm. Case 1: p 1 . c m ∴′− = = 128 . cm cm cm. From (4): q 2 32 7 cm = 32 7 15 17 7 cm cm cm. Case 2: p 1 6573 c m 0927 . cm cm From (4): q
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This note was uploaded on 12/14/2011 for the course PHY 203 taught by Professor Staff during the Fall '11 term at Indiana State University .

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