Homework_23

# Homework_23 - Nguyen Thanh – Homework 23 – Due 7:00 pm...

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Unformatted text preview: Nguyen, Thanh – Homework 23 – Due: Apr 29 2008, 7:00 pm – Inst: Weathers 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 1) 10 points Interference fringes are produced at point P on a screen as a result of direct rays from a 476 nm wavelength source and reflected rays off the Lloyd’s mirror as in the figure. Fringes . 54 mm apart are formed on a screen 4 m from the source S . 4 m h S viewing screen Mirror P O Find the vertical distance h of the source above the reflecting surface. Correct answer: 1 . 76296 mm. Explanation: Basic Concepts: Intensity maxima occur when the two waves have a phase difference φ = 0 , 2 π, 4 π, ··· = φ path + φ reflection , where φ reflection = π. Let : L = 4 m h = d 2 Δ y = 0 . 54 mm . Solution: If the difference in path length is δ , the phase difference due to the unequal path lengths is φ path = k δ = 2 π λ δ . (1) r 2 r 1 y L d S 1 S 2 θ = ta n- 1 ‡ y L · viewing screen δ ≈ d sin θ ≈ r 2- r 1 P O S 1 is the source and S 2 is its image. 6 S 2 QS 1 ≈ 90 ◦ d = 2 h Q R This diagram shows the similarity between the Lloyd’s mirror apparatus and Young’s double slit experiment. The difference be- tween the two experiments is that there is a 180 ◦ phase change at the reflecting mirror. sin θ = δ 2 h = y L . (2) In the small angle approximation θ ≈ δ 2 h ≈ y L h = δ 2 y L, (3) where the height of the light source above the mirror is h = d 2 ....
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## This note was uploaded on 10/24/2011 for the course PHYS 2220 taught by Professor Littler during the Spring '00 term at North Texas.

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Homework_23 - Nguyen Thanh – Homework 23 – Due 7:00 pm...

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