# PHY4 - Howard(clh2528 – homework#4 – shubeita –(57885...

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Unformatted text preview: Howard (clh2528) – homework #4 – shubeita – (57885) 1 This print-out should have 14 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 10.0 points Is diffraction more pronounced through a small or a large opening? 1. small correct 2. It depends on the frequency of the light. 3. large Explanation: Diffraction is more pronounced through a small opening. 002 10.0 points Light from a He-Ne laser of wavelength 635 . 1 nm is incident on a single slit. What is the minimum width for which no diffraction minima are observed? Correct answer: 635 . 1 nm. Explanation: Given : λ = 635 . 1 nm . The first minimum is at a sin θ = λ . This equation has no solution if λ a > 1. Hence the minimum width of the single slit for which no diffraction minima are observed is a min = λ = 635 . 1 nm . 003 (part 1 of 2) 10.0 points Consider the setup of a single slit experiment. The wavelength of the incident light is λ . The slit width and the distance between the slit and the screen is specified in the figure. y 1 L a S 1 S 2 θ viewing screen Find the position y = y 1 of the first inten- sity minimum. Use a small angle approxima- tion; e.g. , sin θ = tan θ . 1. y 1 = 2 a L λ 2. y 1 = 2 λ L a 3. y 1 = λ a L 4. y 1 = a L 2 λ 5. y 1 = λ a 2 L 6. y 1 = a L λ 7. y 1 = λ L a correct 8. y 1 = 2 λ a L 9. y 1 = λ L 2 a Explanation: δ ≡ a sin θ ≈ a bracketleftBig y L bracketrightBig For single slit diffraction, destructive in- terference occurs when, a 2 sin θ = λ 2 , or sim- ply when, δ ≡ a sin θ = λ . Thus, between the two end rays which correspond to the first minimum, the phase angle difference is β 1 = 2 π and the path length difference is δ 1 = λ . The small angle approximation gives us y 1 L = tan θ 1 ≈ θ 1 ≈ sin θ 1 = δ 1 a , or y 1 = δ 1 a L = λ L a . 004 (part 2 of 2) 10.0 points Howard (clh2528) – homework #4 – shubeita – (57885) 2 Denote the intensity on the screen at y 2 by I 2 and the intensity on the screen at y = 0 by I . Let the fist dark fringe fall at y 1 , and let R = y 2 y 1 . y 2 L a S 1 S 2 θ viewing screen Find the intensity ratio I 2 I . 1. I 2 I = bracketleftbigg sin( π y 2 /δ 2 ) π y 2 /δ 2 bracketrightbigg 2 2. I 2 I = bracketleftbigg sin( π R L/y 2 ) π R L/y 2 bracketrightbigg 2 3. I 2 I = bracketleftbigg sin( π R y 2 /δ 2 ) π R y 2 /δ 2 bracketrightbigg 2 4. I 2 I = bracketleftbigg sin( π R y 2 /L ) π R y 2 /L bracketrightbigg 2 5. I 2 I = bracketleftbigg sin( π R ) π R bracketrightbigg 2 correct 6. I 2 I = bracketleftbigg sin( π R δ 2 /y 2 ) π R δ 2 /y 2 bracketrightbigg 2 7....
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PHY4 - Howard(clh2528 – homework#4 – shubeita –(57885...

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