36 - Kapoor (mk9499) oldhomework 36 Turner (60230) 1 This...

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Unformatted text preview: Kapoor (mk9499) oldhomework 36 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 A standard setup of the double slit experiment as shown in the sketch. The distance between the two slits is d , the wave length of the incident wave is , the distance between the slits and the screen is L . A point on the screen is specified by its y-coordinate, or the corresponding angle . y L d S 1 S 2 viewing screen At the fourth minimum on the screen, (i) , the phase angle difference of the two rays from slit S 1 and slit S 2 , and (ii) , the corresponding path difference are given by 1. = 6 and = 6 . 2. = 7 and = 7 2 . correct 3. = 8 and = 8 . 4. = 4 and = 2 . 5. = 5 and = 5 . 6. = 6 and = 3 . 7. = 5 and = 5 2 . 8. = 4 and = 4 . 9. = 7 and = 7 . 10. = 8 and = 4 . Explanation: In general, the phase angle difference for minima is given by = (2 n + 1) , with n = 0 , 1 , 2 . The fourth minimum corresponds to n = 3, so = 7 . The corresponding path difference is given by = parenleftbigg 2 parenrightbigg = 7 2 . 002 (part 2 of 3) 10.0 points What is the vertical distance y for the first maximum (which is adjacent to the central maximum)? Use the small angle approxima- tion. 1. y = d L . 2. y = 2 d L . 3. y = 2 L d . 4. y = 2 d L . 5. y = d L 2 . 6. y = d L . 7. y = L 2 d . 8. y = L d . correct 9. y = d 2 L . 10. y = . Explanation: The first maximum occurs when the path difference is , so = d = y L or y = L d . 003 (part 3 of 3) 10.0 points Kapoor (mk9499) oldhomework 36 Turner (60230) 2 Find the minimum positive value such that I I = 1 4 , where I and I are the intensities of light at 0 and at , respectively. Use the small angle approximation. 1. = d 3 . 2. = 5 d 24 . 3. = 8 d . 4. = 3 d . correct 5. = 6 d ....
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36 - Kapoor (mk9499) oldhomework 36 Turner (60230) 1 This...

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