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Unformatted text preview: Flores, Michael – Exam 5 – Due: May 4 2004, 11:00 pm – Inst: Sonia Paban 1 This printout should have 14 questions. Multiplechoice questions may continue on the next column or page – find all choices before answering. The due time is Central time. 001 (part 1 of 2) 10 points Consider the setup of a Lloyd mirror shown in the figure. The interference pattern on the vertical screen is due to the superposition of a direct ray and a reflected ray where at reflection there is a change in the phase angle of π . Assume: The wavelength of the light rays is λ . y L Mirror h S 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 = 3 λ and φ = 6 π . 2. δ = 5 2 λ and φ = 5 π . 3. δ = 3 2 λ and φ = 3 π . 4. δ = 2 λ and φ = 4 π . 5. δ = 4 λ and φ = 8 π . 6. δ = 6 λ and φ = 12 π . correct 7. δ = 9 2 λ and φ = 9 π . 8. δ = 7 λ and φ = 14 π . 9. δ = 7 2 λ and φ = 7 π . 10. δ = 11 2 λ and φ = 11 π . Explanation: Basic Concepts: The Lloyd’s mirror set up is the same as a double slit diffraction set up (see figure in next part), with the exception that the mirror will cause a 180 ◦ phase change ( i.e. , π phase change) in the reflected ray. This will produce the reverse conditions for bright and dark fringes, and the seperation of the ’slits’ is the vertical distance between the source and its image in the mirror: d = 2 h For dark fringes, we have δ = d sin θ = m λ, (1) and for bright fringes, we have δ = d sin θ = µ m + 1 2 ¶ λ, (2) where m = 0 , ± 1 , ± 2 , ± 3 , ··· . From geometry, we have y = L tan θ . Solution: Using the above equation for the sixth dark fringe; i.e. , for m = 6, we have δ = mλ = 6 λ. Since δ λ = φ 2 π , we have φ = 2 π δ λ = 2 π m λ λ = 2 π (6) = 12 π . 002 (part 2 of 2) 10 points Let L be the distance from the source to the screen and h be the distance from the source to the mirror. Using the small angle approximation ( θ = sin θ = tan θ ) , if the distance from the central dark region to the sixth dark fringe is y , what is the wave length of the light, given that d = 2 h is the vertical distance between the source from its image in the mirror. Flores, Michael – Exam 5 – Due: May 4 2004, 11:00 pm – Inst: Sonia Paban 2 1. λ = 1 6 L y d 2. λ = 1 7 d y L 3. λ = 2 11 L y d 4. λ = 1 6 d L y 5. None of these. 6. λ = 2 11 d L y 7. λ = 1 7 L y d 8. λ = 1 7 d L y 9. λ = 2 11 d y L 10. λ = 1 6 d y L correct Explanation: r 2 r 1 y L d S 1 S 2 θ = ta n 1 ‡ y L · viewing screen δ ≈ d sin θ ≈ r 2 r 1 P O S 1 is the source and S 2 is its image....
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 Spring '04
 Panab
 Light, Wavelength, Doubleslit experiment, Thomas Young, Sonia Paban, Michael – Exam

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