Homework 3 KEY

# Homework 3 KEY - Golshani(ag5873 Homework 03 Yao(57045 This...

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Golshani (ag5873) – Homework 03 – Yao – (57045) 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 A screen is illuminated by monochromatic light whose wave length is λ , as shown. The distance from the slits to the screen is L . y L d S 1 S 2 θ viewing screen At the sixth dark fringe on the screen (po- sition y on the screen), δ (the corresponding path length difference) and φ (the phase angle difference) is given by 1. δ = 5 2 λ and φ = 5 π 2. δ = 9 2 λ and φ = 9 π 3. δ = 4 λ and φ = 8 π 4. δ = 6 λ and φ = 12 π 5. δ = 11 2 λ and φ = 11 π correct 6. δ = 7 λ and φ = 14 π 7. δ = 2 λ and φ = 4 π 8. δ = 13 2 λ and φ = 13 π 9. δ = 3 2 λ and φ = 3 π 10. δ = 7 2 λ and φ = 7 π Explanation: For dark fringes δ = d sin θ = parenleftbigg m + 1 2 parenrightbigg λ , where m = 0 , ± 1 , ± 2 , ± 3 , so for the sixth dark fringe ( m = 5), δ = parenleftbigg m + 1 2 parenrightbigg λ = 11 2 λ and δ λ = φ 2 π = 11 2 φ = 11 π . 002 10.0 points A double slit arrangement produces an inter- ference fringe for yellow sodium light. To produce narrower-spaced fringes, should red light or blue light be used? 1. Red 2. Blue correct 3. Either Explanation: Blue light has a shorter wavelength than both yellow and red light and will produce narrower-spaced fringes. 003 10.0 points A screen is illuminated by monochromatic light as shown in the figure below. The distance from the slits to the screen is 3 m . 1 . 5 cm 3 m 0 . 56 mm S 1 S 2 θ viewing screen What is the wave length if the distance from the central bright region to the sixth dark fringe is 1 . 5 cm . Correct answer: 509 . 085 nm. Explanation:

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Golshani (ag5873) – Homework 03 – Yao – (57045) 2 Basic Concepts: For bright fringes, we have d sin θ = m λ , and for dark fringes, we have d sin θ = parenleftbigg m + 1 2 parenrightbigg λ , where m = 0 , ± 1 , ± 2 , ± 3 , · · · . From geometry, we have y = L tan θ . Let : y = 1 . 5 cm = 0 . 015 m , L = 3 m , and d = 0 . 56 mm = 0 . 00056 m . r 2 r 1 y L d S 1 S 2 θ = tan 1 parenleftBig y L parenrightBig viewing screen δ d sin θ r 2 - r 1 P O negationslash S 2 Q S 1 90 Q r 2 r 1 d S 1 S 2 θ = tan 1 parenleftBig y L parenrightBig θ δ d sin θ r 2 - r 1 negationslash S 2 Q S 1 90 Q Solution: The angle θ
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