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

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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 μ 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 λ 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 λ , and the index of refraction of the medium n ) t = λ 4 n = 4 . 73 × 10 - 7 m (4) (1 . 15) t = 1 . 02826 × 10 - 7 m = 0 . 102826 μ m . 002 (part 2 of 2) 10 points Now assume that the coating’s index of re- fraction is 1 . 57. Assume that the rest of the system (from the previous question) remains the same.
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