# oldhw 31 - nguyen(jmn727 oldhomework 31 Turner(59070 This...

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nguyen (jmn727) – oldhomework 31 – Turner – (59070) 1 This print-out should have 12 questions. Multiple-choice questions may continue on the next column or page – fnd all choices beFore answering. 001 (part 1 oF 3) 10.0 points Two polarizers are aligned with the frst 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 frst 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 frst polarizer, the intensity oF the light is: 1. I 0 2 cos 2 θ 1 2. I 0 2 correct 3. I 0 cos 2 θ 1 4. I 0 5. none oF these 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 di±erence 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 2 θ 1 ) 3. I 0 2 cos 2 θ 1 cos 2 ( θ 2 θ 1 ) 4. I 0 cos 2 θ 1 cos 2 θ 2 5. I 0 2 cos 2 ( θ 2 θ 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 α = 30 to the transmis- sion axis oF the frst polarizer (see the second sketch). IF initially the light is unpolarized with intensity I 0 , aFter passing through all three polarizers, what is the fnal intensity? 1. I 0 4 2. I 0 2 3. I 0 16 4. 3 8 I 0 5. 3 32 I 0 correct

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nguyen (jmn727) – oldhomework 31 – Turner – (59070) 2 6. I 0 8 Explanation: See the sketch below. After the Frst 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 ±ight is to place a perfectly re±ecting aluminized sheet into Earth’s orbit and use the light from the Sun to push this solar sail. Suppose a sail of area 2 . 906 × 10 5 m 2 and mass 8221 kg is placed in orbit facing the Sun.
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oldhw 31 - nguyen(jmn727 oldhomework 31 Turner(59070 This...

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