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Unformatted text preview: terry (ect328) oldhomework 31 Turner (59130) 1 This printout should have 12 questions. Multiplechoice questions may continue on the next column or page find all choices before answering. 001 (part 1 of 3) 10.0 points Two polarizers are aligned with the first ori ented at an angle 1 clockwise from the verti cal, and the second at an angle 2 clockwise from the vertical, with 0 < 1 < 2 (see the first sketch). A beam of unpolarized light with intensity I is incident normally on the two polarizers from the left. #1 #2 #3 #1 #2 2 1 After passing through the first polarizer, the intensity of the light is: 1. I 2 cos 2 1 2. none of these 3. I 4. I 2 correct 5. I cos 2 1 Explanation: Basic concepts: Malus law states: I = I cos 2 Solution: When unpolarized light falls on a polarizer, no matter what the orientation of the polarizer is, one half of the light is transmitted. Thus the answer is I 2 . the result we got from When the light passes through the second polarizer it is polarized at an angle 1 . Thus the difference between it and the second polarizer is = 2 1 . Thus the transmitted intensity is: I = I cos 2 = I 2 cos 2 ( 2 1 ) . 002 (part 2 of 3) 10.0 points After passing through both polarizers the in tensity is: 1. I cos 2 ( 2 1 ) 2. I cos 2 1 cos 2 ( 2 + 1 ) 3. I 2 cos 2 1 cos 2 ( 2 1 ) 4. I cos 2 1 cos 2 2 5. I 2 cos 2 ( 2 1 ) correct Explanation: 003 (part 3 of 3) 10.0 points Suppose that the polarizers are crossed so that no light can be transmitted through the second polarizer. Now a third polarizer is in serted between the crossed polarizers with its transmission axis at = 30 to the transmis sion axis of the first polarizer (see the second sketch). If initially the light is unpolarized with intensity I , after passing through all three polarizers, what is the final intensity? 1. I 4 2. I 8 3. I 16 4. 3 8 I 5. 3 32 I correct terry (ect328) oldhomework 31 Turner (59130) 2 6. I 2 Explanation: See the sketch below. After the first polar izer: I 1 = I 2 After the third (which was inserted): I 3 = I 1 cos 2 30 = 3 4 I 1 After the former second: I 2 = I 3 cos 2 60 = 1 4 I 3 = 1 4 3 4 1 2 I = 3 32 I #1 #2 #3 1 2 3 004 10.0 points A possible means of space flight is to place a perfectly reflecting aluminized sheet into Earths orbit and use the light from the Sun to push this solar sail. Suppose a sail of areato push this solar sail....
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This note was uploaded on 02/17/2010 for the course PHY 59130 taught by Professor Turner during the Fall '09 term at University of Texas at Austin.
 Fall '09
 Turner

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