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Unformatted text preview: 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)! . 06mm . 096mm 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 λL a . For a double-slit interference pattern, the full- width of the central maxima is y double = λL d . Solution: For the single-slit diffraction pat- tern, the full-width of the central fringe is y single = 2 λL a = 2 (668 nm)(3 . 2 m) . 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 = λL 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 . 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 λL a λL 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....
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This note was uploaded on 02/18/2010 for the course PHYS 303l taught by Professor Panab during the Spring '04 term at North Texas.
- Spring '04