32 - Kapoor(mk9499 homework32 Turner(60230 This print-out...

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Kapoor (mk9499) – homework32 – Turner – (60230) 1 This print-out should have 12 questions. Multiple-choice 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 θ 1 clockwise from the verti- cal, and the second at an angle θ 2 clockwise from the vertical, with 0 < θ 1 < θ 2 (see the first sketch). A beam of unpolarized light with intensity I 0 is incident normally on the two polarizers from the left. α #1 #2 #3 #1 #2 θ 2 θ 1 After passing through the first polarizer, the intensity of the light is: 1. none of these 2. I 0 2 correct 3. I 0 cos 2 θ 1 4. I 0 2 cos 2 θ 1 5. I 0 Explanation: Basic concepts: Malus’ law states: I = I 0 cos 2 θ 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 0 2 . the result we got from When the light passes through the second polarizer it is polarized at an angle θ 1 . Thus the difference between it and the second polarizer is θ = θ 2 θ 1 . Thus the transmitted intensity is: I = I 0 cos 2 θ = I 0 2 cos 2 ( θ 2 θ 1 ) . 002 (part 2 of 3) 10.0 points After passing through both polarizers the in- tensity is: 1. I 0 cos 2 θ 1 cos 2 ( θ 2 + θ 1 ) 2. I 0 cos 2 θ 1 cos 2 θ 2 3. I 0 2 cos 2 θ 1 cos 2 ( θ 2 θ 1 ) 4. I 0 2 cos 2 ( θ 2 θ 1 ) correct 5. I 0 cos 2 ( θ 2 θ 1 ) 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 α = 30 to the transmis- sion axis of the first polarizer (see the second sketch). If initially the light is unpolarized with intensity I 0 , after passing through all three polarizers, what is the final intensity? 1. I 0 4 2. 3 8 I 0 3. I 0 8 4. 3 32 I 0 correct 5. I 0 2 6. I 0 16
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Kapoor (mk9499) – homework32 – Turner – (60230) 2 Explanation: See the sketch below. After the first polar- izer: I 1 = I 0 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 0 = 3 32 I 0 #1 #2 #3 Ι Ι Ι 0 1 Ι 2 3 004 10.0 points A possible means of space flight is to place a perfectly reflecting aluminized sheet into Earth’s orbit and use the light from the Sun to push this solar sail. Suppose a sail of area 74300 m 2 and mass 5672 kg is placed in orbit facing the Sun.
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