15exam4 - Husain, Zeena Exam 4 Due: Apr 23 2004, noon Inst:...

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Unformatted text preview: Husain, Zeena Exam 4 Due: Apr 23 2004, noon Inst: Sonia Paban 1 This print-out should have 7 questions. Multiple-choice questions may continue on the next column or page find all choices before making your selection. The due time is Central time. 001 (part 1 of 2) 10 points Given a light bulb S emitting light isotropi- cally, ( i.e. uniformly in all directions), with a power of 45 W. The paper is 3 . 07 m away and has an area of A = 0 . 02 m 2 , with the coefficient of reflection = 1 3 ; i.e. , 1 3 of the light intensity is reflected, and 2 3 of the light intensity is absorbed. r S Q A Determine the intensity at point Q . Correct answer: 0 . 37995 W / m 2 . Explanation: Basic Concepts: Radiation intensity due to a point source is given by I = P 4 r 2 . Radiation pressure on a perfect absorbing sur- face is P ab = I c , while the radiation pressure on a perfect re- flective surface is P ref = 2 I c . Solution: Since the light bulb emits light isotropically, ( i.e. , uniformly in all directions), it can be treated as a point source. I = P 4 r 2 = (45 W) 4 (3 . 07 m) 2 = 0 . 37995 W / m 2 . 002 (part 2 of 2) 10 points Determine the radiation pressure P at Q , where I is the light intensity at Q . See the previous question for the setup of this prob- lem. 1. P = 1 2 I c 2. P = 3 I c 3. P = 2 3 I c 4. P = 0 5. P = 2 I c 6. P = 4 3 I c correct 7. P = 1 3 I c 8. P = 7 3 I c 9. P = 5 3 I c 10. P = I c Explanation: Since 1 3 of light is reflected, the radiation pressure due to this portion of light is given by P ref = 2 I ref c = 2 1 3 I c = 2 3 I c ....
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15exam4 - Husain, Zeena Exam 4 Due: Apr 23 2004, noon Inst:...

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