# 09HW9 - Niu(qn269 – Homework#9 – antoniewicz –(58855...

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Unformatted text preview: Niu (qn269) – Homework #9 – antoniewicz – (58855) 1 This print-out should have 16 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 as shown in the figure below. The distance from the slits to the screen is 7 . 3 m . 1 . 9cm 7 . 3 m . 97mm S 1 S 2 θ viewing screen What is the wave length if the distance from the central bright region to the fifth dark fringe is 1 . 9 cm . Correct answer: 561 . 033 nm. Explanation: 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 . 9 cm = 0 . 019 m , L = 7 . 3 m , and d = 0 . 97 mm = 0 . 00097 m . r 2 r 1 y L d S 1 S 2 θ = ta n − 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 = ta n − 1 parenleftBig y L parenrightBig θ δ ≈ d s i n θ ≈ r 2- r 1 negationslash S 2 Q S 1 ≈ 9 ◦ Q Solution: The angle θ from the slits’ mid- point to the y position on the screen is θ = arctan bracketleftBig y L bracketrightBig = arctan bracketleftbigg (0 . 019 m) (7 . 3 m) bracketrightbigg = 0 . 00260273 rad . The wavelength of the light for the fifth dark fringe, m = 4, is λ = d sin θ parenleftbigg m + 1 2 parenrightbigg = (0 . 00097 m) sin(0 . 00260273 rad) (4 . 5) = 5 . 61033 × 10 − 7 m = 561 . 033 nm . 002 10.0 points A pair of narrow parallel slits separated by a distance of 0.208 mm are illuminated by the green component from a mercury vapor lamp ( λ = 546 . 6 nm). What is the angle from the central maxi- mum to the second dark fringe on either side of the central maximum? Correct answer: 0 . 225851 ◦ . Explanation: Niu (qn269) – Homework #9 – antoniewicz – (58855) 2 Basic Concept: d (sin θ ) = parenleftbigg m + 1 2 parenrightbigg λ, m = 0 , ± 1 , ± 2 , ··· Given : d = 0 . 208 mm · 1 m 1000 mm = 2 . 08 × 10 − 4 m λ = 546 . 6 nm · 1 m 10 9 nm = 5 . 466 × 10 − 7 m m = 1 , second dark fringe Solution: sin θ = parenleftbigg m + 1 2 parenrightbigg λ d θ = sin − 1 parenleftbigg m + 1 2 parenrightbigg λ d = sin − 1 bracketleftbigg 1 . 5(5 . 466 × 10 − 7 m) . 000208 m bracketrightbigg = . 225851 ◦ . 003 (part 1 of 2) 10.0 points Light of wavelength 747 nm illuminates a sin- gle slit....
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09HW9 - Niu(qn269 – Homework#9 – antoniewicz –(58855...

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