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Homwork Old Combined

# Homwork Old Combined - Kapoor(mk9499 oldhomework 30...

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Kapoor (mk9499) – oldhomework 30 – Turner – (60230) 1 This print-out should have 15 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 (part 1 of 2) 10.0 points FM radio bands range from 88 MHZ through 108 MHz. The speed of light is 3 × 10 8 m / s. What is the wavelength for the FM radio band at 91 MHz? Correct answer: 3 . 2967 m. Explanation: Let : f 1 = 91 MHz and c = 3 × 10 8 m / s . The wavespeed is c = λ 1 = c f 1 = 3 × 10 8 m / s 91 MHz · 1 MHz 10 6 Hz = 3 . 2967 m . 002 (part 2 of 2) 10.0 points What is the wavelength for the FM radio band at 107 MHz? Correct answer: 2 . 80374 m. Explanation: Let : f 2 = 107 MHz . λ 2 = c f 2 = 3 × 10 8 m / s 107 MHz · 1 MHz 10 6 Hz = 2 . 80374 m . 003 (part 1 of 4) 10.0 points Consider 3 polarizers #1, #2, and #3 ordered sequentially. The incident light is unpolar- ized with intensity I 0 . The intensities after the light passes through the subsequent polar- izers are labeled as I 1 , I 2 , and I 3 respectively. See the sketch. I 0 I 1 I 2 I 3 #1 #2 #3 Polarizers #1 and #3 are “crossed” such that their transmission axes are perpendic- ular to each other. Polarizer #2 is placed between the polarizers #1 and #3 with its transmission axis at 60 with respect to the transmission axis of the polarizer #1 (see the sketch). #1 #2 #3 60 After passing through the first polarizer, the intensity of the light I 1 is 1. I 1 = I 0 8 2. I 1 = 3 I 0 32 3. I 1 = 3 I 0 4 4. I 1 = I 0 32 5. I 1 = I 0 12 6. I 1 = I 0 16 7. I 1 = I 0 4 8. I 1 = 3 I 0 16 9. I 1 = 3 I 0 8 10. I 1 = I 0 2 correct Explanation:

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Kapoor (mk9499) – oldhomework 30 – Turner – (60230) 2 Basic Concepts: Malus’ law states I = I 0 cos 2 θ , where θ is the angle between the polarization of the light (before the light hits the polarizer) and the transmission axis of the polarizer. Solution: When unpolarized light falls on a polarizer, no matter what the orientation of the polarizer is, one half of the light is trans- mitted, and the polarization of the transmit- ted light is along the direction of the trans- mission axis of the polarizer (vertical in the present case). Thus the answer is I 1 = 1 2 I 0 . 004 (part 2 of 4) 10.0 points After passing through polarizer #2 the inten- sity I 2 (in terms of I 1 ) is 1. I 2 = I 1 8 2. I 2 = I 1 16 3. I 2 = 3 I 1 4 4. I 2 = 3 I 1 16 5. I 2 = I 1 2 6. I 2 = I 1 32 7. I 2 = 3 I 1 8 8. I 2 = I 1 4 correct 9. I 2 = 3 I 1 32 10. I 2 = I 1 12 Explanation: When polarized light passes through a po- larizer, the transmitted intensity is I 2 = I 1 cos 2 θ , where θ is the angle between the polarization of the light (of I 1 ) and the orien- tation of the polarizer #2 and I 1 is the result we got from Part 1. When the light passes through the polarizer #1 it is polarized verti- cally. Thus the angle between its polarization and the orientation of polarizer #2 is θ = 60 . Thus the transmitted intensity is I 2 = I 1 cos 2 θ = I 1 cos 2 (60 ) = 1 4 I 1 = 1 8 I 0 .
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