HW35Sol - Platt, David Homework 35 Due: Dec 5 2005, 4:00 am...

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Unformatted text preview: Platt, David Homework 35 Due: Dec 5 2005, 4:00 am Inst: Ken Shih 1 This print-out should have 7 questions. Multiple-choice questions may continue on the next column or page find all choices before answering. The due time is Central time. 001 (part 1 of 2) 10 points A material with an index of refraction of 1 . 15 is used to coat glass. The index of refraction of glass is 1 . 5. What is the minimum thickness of the coat- ing that will minimize the reflection of light with a wavelength of 4730 A? Correct answer: 0 . 102826 m. Explanation: Basic Concepts: Phase Changes for Reflecting Waves. Solution: Since the coating has a refraction that is less than that for glass (but greater than that for air), we know that the reflected light from both the glass and the coating will undergo a 180 phase shift. This means that the to- tal trip inside the coating must be exactly one half of the wavelength of light inside the coating . (Hence the waves reflected from the glass and from the coating will interfere de- structively.) The wavelength of light inside the coating is n where n is the index of re- fraction of the coating. This implies that the thickness of the coating must be one fourth the wavelength of light inside this medium. Hence (calling the thickness of the material t , the wavelength of light in air , and the index of refraction of the medium n ) t = 4 n = 4 . 73 10- 7 m (4)(1 . 15) t = 1 . 02826 10- 7 m = 0 . 102826 m ....
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This note was uploaded on 04/10/2010 for the course PHY 317 taught by Professor Gositz during the Spring '08 term at University of Texas at Austin.

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HW35Sol - Platt, David Homework 35 Due: Dec 5 2005, 4:00 am...

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