# 2-11 - Create assignment 00224 Homework 11 Apr 11 at 8:10...

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Unformatted text preview: Create assignment, 00224, Homework 11, Apr 11 at 8:10 am 1 This print-out should have 17 questions, check that it is complete. Multiple-choice questions may continue on the next column or page: find all choices before making your selection. The due time is Central time. AP B 1998 MC 51 37:02, trigonometry, multiple choice, < 1 min. 001 Plane sound waves of wavelength 0 . 12 m are incident on two narrow slits in a box with nonreflecting walls, as shown. At a distance of 5 m from the center of slits, a first-order maximum occurs at point P , which is 3 m from the central maximum. Sound λ = 0.12 m 5.0 m 3.0 m The distance d between the slits is most nearly 1. d = 0 . 07 m. 2. d = 0 . 09 m. 3. d = 0 . 16 m. 4. d = 0 . 20 m. correct 5. d = 0 . 24 m. Explanation: The rules for determining interference max- imum or minimum are the same for sound waves and light waves. For the first maximum, ∆ = λ . Since sin θ ≈ ∆ d ≈ y L , therefore ∆ d ≈ y L = ⇒ d ≈ ∆ L y = λL y = (0 . 12 m)(5 m) 3 m = 0 . 2 m . Dark Fringe Distance 37:02, trigonometry, numeric, > 1 min. 002 A screen is illuminated by 500 nm light as shown in the figure below. y 7 5 m . 5mm S 1 S 2 θ viewing screen How far apart “ y 7 ” are the central bright region and the seventh dark fringe? Correct answer: 3 . 25007 cm. Explanation: Let : L = 5 m y = 3 . 25007 cm d = 0 . 5 mm 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 6 S 2 QS 1 ≈ 90 ◦ Q Basic Concepts: d sin θ = mλ, where µ m + 1 2 ¶ = 0 , ± 1 , ± 2 , ± 3 , ··· . y = L tan θ. Solution: For the third dark fringe, m = 6, so θ = arcsin µ m + 1 2 ¶ λ d Create assignment, 00224, Homework 11, Apr 11 at 8:10 am 2 = arcsin •µ 6 + 1 2 ¶ (5 × 10- 7 m) (0 . 0005 m) ‚ = 0 . 00650005 rad . The distance is y 7 = L tan θ = (5 m)tan(0 . 00650005 rad) = 0 . 0325007 m = 3 . 25007 cm . Double Slits 13 37:02, trigonometry, numeric, > 1 min. 003 The slits in a Young’s interference apparatus are illuminated with monochromatic light of wavelength λ . The third dark band is y from the axis. The two slits are d apart and the screen is L away from the slits. We want to calculate the wavelength of the light used. L d y What is y in terms of λ ,d and L ? 1. λ = 3 λL d 2. λ = 5 2 λL d correct 3. λ = 7 2 λL d 4. λ = 2 λL d 5. λ = 5 2 λd L 6. λ = 7 2 λd L 7. λ = 3 λd L 8. λ = 2 λd L 9. λ = λL d 10. λ = λd L Explanation: Basic Concepts: The condition for dark fringes (destructive interference) is d sin θ = µ m + 1 2 ¶ λ, where, m = 0 , ± 1 , ± 2 , ··· . Solution: The position of dark bands in the double slits experiments is given by d sin θ = µ m + 1 2 ¶ λ, where m = 0 , ± 1 , ± 2 , ··· . Since the angle θ is very small, we can take the following approximation θ = sin θ = tan θ = y dark L ....
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2-11 - Create assignment 00224 Homework 11 Apr 11 at 8:10...

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