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HW5solution - Version PREVIEW HW 5 Savrasov(39824 This...

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Version PREVIEW – HW 5 – Savrasov – (39824) 1 This print-out should have 11 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. Bright Band Separation 001 (part 1 of 2) 10.0 points Two slits are illuminated by a 533 nm light. The angle between the zeroth-order bright band at the center of the screen and the fourth-order bright band is 17 . 4 . If the screen is 167 cm from the double-slit, how far apart is this bright band from the central peak? Correct answer: 52 . 3346 cm. Explanation: Basic Concept: y = L tan θ. Solution: y = L tan θ = (167 cm) tan(17 . 4 ) = 52 . 3346 cm . 002 (part 2 of 2) 10.0 points What is the distance between the two slits? Correct answer: 0 . 00712946 mm. Explanation: Basic Concept: d sin θ = Solution: The fourth-order bright band oc- curs when m = 4; therefore, d = m λ sin θ = 4 (533 nm) sin(17 . 4 ) = 0 . 00712946 mm . Double Slit Wave Length 01 003 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 5 m . 1 . 9 cm 5 m 0 . 9 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 . 9 cm . Correct answer: 621 . 814 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 = 5 m , and d = 0 . 9 mm = 0 . 0009 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
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Version PREVIEW – HW 5 – Savrasov – (39824) 2 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 from the slits’ mid- point to the y position on the screen is θ = arctan bracketleftBig y L bracketrightBig = arctan bracketleftbigg (0 . 019 m) (5 m) bracketrightbigg = 0 . 00379998 rad . The wavelength of the light for the sixth dark fringe, m = 5, is λ = d sin θ parenleftbigg m + 1 2 parenrightbigg = (0 . 0009 m) sin(0 . 00379998 rad) (5 . 5) = 6 . 21814 × 10 7 m = 621 . 814 nm .
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