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1034_Physics ProblemsTechnical Physics

# 1034_Physics ProblemsTechnical Physics - Chapter 36...

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Chapter 36 375 P36.70 (a) For the light the mirror intercepts, P = = I A I R a 0 0 2 π 350 1 000 2 W W m 2 = e j π R a and R a = 0 334 . m or larger . (b) In 1 1 1 2 p q f R + = = we have p → ∞ so q R = 2 M h h q p = = − so ′ = − F H G I K J = − F H G I K J ° ° F H G I K J L N M O Q P = − F H G I K J h q h p R R 2 0 533 2 9 30 . . π rad 180 m rad a f where h p is the angle the Sun subtends. The intensity at the image is then I h I R h I R R a a = = = × P π π π 2 0 2 2 0 2 2 3 2 4 4 4 2 9 30 10 b g e j . rad 120 10 16 1 000 9 30 10 3 2 2 3 2 × = × W m W m rad 2 2 e j e j R R a . so R R a = 0 025 5 . or larger . P36.71 In the original situation, p q 1 1 1 50 + = . m . In the final situation, p p 2 1 0 900 = + . m and q q p 2 1 1 0 900 0 600 = = . . m m . Our lens equation is 1 1 1 1 1 1 1 2 2 p q f p q + = = + . Substituting, we have 1 1 1 50 1 0 900 1 0 600 1 1 1 1 p p p p + = + + . . . m . Adding the fractions, 1 50 1 50 0 600 0 900 0 900 0 600
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