1034_Physics ProblemsTechnical Physics

1034_Physics ProblemsTechnical Physics - Chapter 36 P36.70...

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Chapter 36 375 P36.70 (a) For the light the mirror intercepts, P == IA I R a 00 2 π 350 1 000 2 W W m 2 = ej R a and R a = 0334 . m o r l a r g e r . (b) In 111 2 pq 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 hq h p RR 2 0533 2 930 .. rad 180 m rad af where h p is the angle the Sun subtends. The intensity at the image is then I h IR h R a a = = = × 2 0 2 2 0 2 2 3 2 4 4 4 2 9 30 10 bg r a d 120 10 16 1 000 930 10 3 2 23 2 ×= × Wm rad 2 2 R R a . so R R a = 00255 o r l a r g e r . P36.71 In the original situation, pq 11 150 += m . In the final situation, pp 21 0900 =+ m and qq p 1 0600 = m m . Our lens equation is 11111 22 f . Substituting, we have 1 1 1 1 p p + = + + . m . Adding the fractions, 0900 0600 . . m m −+ = −++ +− b g . Simplified, this becomes p p 1 1 1 50 0 900 0 600
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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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