hw14sol - Flores Michael Homework 14 Due May 3 2004 4:00 am...

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Flores, Michael – Homework 14 – Due: May 3 2004, 4:00 am – Inst: Sonia Paban 1 This print-out should have 13 questions. Multiple-choice questions may continue on the next column or page – find all choices before making your selection. The due time is Central time. 001 (part 1 of 1) 10 points Consider the setup of a single slit experiment shown in the figure. The first minimum for 490 nm light is at y 1 . The first minimum for 585 nm light is at y 2 . y L a S 1 S 2 θ viewing screen What is the ratio y 2 y 1 ? Correct answer: 1 . 19388 . Explanation: The first minimum is at y = λL a For the first light, y 1 = λ 1 L a For the second light, y 2 = λ 2 L a Therefore y 2 y 1 = λ 2 λ 1 = (585 nm) (490 nm) = 1 . 19388 . 002 (part 1 of 1) 10 points Hint: Use a small angle approximation; e.g. , sin θ = tan θ = θ . Light of wavelength 668 nm falls on a dou- ble slit as shown in the schematic figure below. Note: The dimensions y d and y s are not to scale (for obvious reasons)! 0 . 06 mm 0 . 096 mm y s y d 3 . 2 m S 1 S 2 θ What is the ratio y s y d of the half-width of the central maxima of single-slit diffraction pattern to that of the half-width of double- slit interference pattern? Correct answer: 3 . 2 . Explanation: Basic Concepts: For a single-slit diffrac- tion pattern, the full-width of the central maxima is y single = 2 λ L a . For a double-slit interference pattern, the full- width of the central maxima is y double = λ L d . Solution: For the single-slit diffraction pat- tern, the full-width of the central fringe is y single = 2 λ L a = 2 (668 nm)(3 . 2 m) 0 . 06 mm = 2 (6 . 68 × 10 - 7 m)(3 . 2 m) 6 × 10 - 5 m = 0 . 0712533 m . For the double-slit interference pattern, the width of the central maxima is y double = λ L d = (668 nm) (3 . 2 m) (0 . 096 mm) = (6 . 68 × 10 - 7 m) (3 . 2 m) (9 . 6 × 10 - 5 m) = 0 . 0222667 m .
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Flores, Michael – Homework 14 – Due: May 3 2004, 4:00 am – Inst: Sonia Paban 2 The ratio of single-slit central maxima to double-slit central maxima is y s y d = 2 y single 2 y double = (0 . 0712533 m) (0 . 0222667 m) = 3 . 2 , where the full-width is twice the half-width. Alternative Solution: y s y d = 2 λ L a λ L d = 2 d a = 2 (0 . 096 mm) (0 . 06 mm) = 3 . 2 . 003 (part 1 of 1) 10 points The resolution of a lens can be estimated by treating the lens as a circular aperture. The resolution is the smallest distance between two point sources that produce distinct im- ages. This is similar to the resolution of a single slit, related to the distance from the middle of the central bright band to the first- order dark band; however, the aperture is circular instead of a rectangular slit which introduces a scale factor. Suppose the Hubble Space Telescope, 2 . 4 m in diameter, is in orbit 108 km above Earth and is turned to look at Earth. If you ignore the effect of the atmosphere, what is the resolution of this telescope for light of wavelength 441 nm? Correct answer: 2 . 42109 cm. Explanation: L x w For single slit diffraction x L = λ w , For a circular aperture the resolution differs by a factor of 1.22. Therefore x = 1 . 22 L λ w = 1 . 22 (108000 m) (4 . 41 × 10 - 7 m) (2 . 4 m) × (100 cm / m) = 2 . 42109 cm .
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