# OHW 40 - terry(ect328 oldhomework 40 Turner(59130 This...

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terry (ect328) – oldhomework 40 – Turner – (59130) 1 This print-out should have 10 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 λ , 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 θ . y L d S 1 S 2 θ viewing screen θ δ At the fourth minimum on the screen what is φ (the phase angle difference of the two rays from slit S 1 and slit S 2 ) and δ (the corre- sponding path difference)? 1. φ = 7 π and δ = 7 2 λ correct 2. φ = 4 π and δ = 2 λ 3. φ = 7 π and δ = 7 λ 4. φ = 6 π and δ = 3 λ 5. φ = 6 π and δ = 6 λ 6. φ = 8 π and δ = 4 λ 7. φ = 5 π and δ = 5 λ 8. φ = 4 π and δ = 4 λ 9. φ = 8 π and δ = 8 λ 10. φ = 5 π and δ = 5 2 λ Explanation: In general, the phase angle difference for minima is given by φ = (2 n + 1) π, with n = 0 , 1 , 2 · · · . The fourth minimum corresponds to n = 3, so φ = 7 π and δ = parenleftbigg λ 2 π parenrightbigg φ = λ 2 π (7 π ) = 7 2 λ . 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 = λ 2. y = 2 λ L d 3. y = d L λ 4. y = λ L 2 d 5. y = 2 d L λ 6. y = 2 λ d L 7. y = d L 2 λ 8. y = λ d 2 L 9. y = λ d L 10. y = λ L d correct Explanation: The first maximum occurs when the path difference is λ , so θ = λ d = y L y = λ L d . 003 (part 3 of 3) 10.0 points Find the minimum positive θ value such that

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terry (ect328) – oldhomework 40 – Turner – (59130) 2 I I 0 = 1 4 , where I 0 and I are the intensities of light at 0 and at θ , respectively. Use the small angle approximation.
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