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Unformatted text preview: Physics 315: Oscillations and Waves Homework 9: Solutions ψ opaque screen d/2 slits projection screen R ρ lightray optical axis θ 1. Let d be the spacing of the slits, λ the wavelength of the light illuminating them, and R the distance of the (cylindrical) projection screen behind them. Consider a light ray which propagates from one of the slits to a point on the projection screen whose angular position (w.r.t. the optical axis) is θ . See the above diagram. Let ρ be the pathlength of the ray. It follows from application of the cosine formula of trigonometry to the triangle in the diagram that ρ = bracketleftbig R 2 − 2 R ( d/ 2) cos ψ + ( d/ 2) 2 bracketrightbig 1 / 2 = R bracketleftbigg 1 − d R sin θ + d 2 4 R 2 bracketrightbigg 1 / 2 , (1) since ψ = 90 ◦ − θ . Assuming that d/R ≪ 1, and making use of the binomial expansion, we obtain ρ ≃ R bracketleftbigg 1 − d 2 R sin θ + O parenleftbigg d 2 R 2 parenrightbiggbracketrightbigg = R − d 2 sin θ + O parenleftbigg d 2 R parenrightbigg . (2) The change in phase of the light as it propagates from the slit to the screen is thus ∆ φ = k ρ = k R − k d 2 sin θ + O parenleftbigg k d 2 R parenrightbigg , (3) where k = 2 π/λ is the wavenumber. The Fraunhoffer, or far field, limit corresponds to the situation in which the phase shift generated by the final term in the above expansion is negligible, so that ∆ φ ≃ k R − k d 2 sin θ. (4) This is the case provided k d 2 R ≪ 2 π, (5) or R ≫ d 2 λ . (6) Suppose that d = 2 . 5 × 10 − 4 m and λ = 5 × 10 − 7 m. It follows that R ≫ (2 . 5 × 10 − 4 ) 2 5 × 10 − 7 = 0 . 125 m = 12 . 5 cm . (7) ψ d θ ∆ 2. Let d be the spacing of the slits, λ the wavelength of the light illuminating them, and R the distance of the (cylindrical) projection screen behind them.the distance of the (cylindrical) projection screen behind them....
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This note was uploaded on 11/02/2010 for the course PHY 59372 taught by Professor Fitzpatrick during the Spring '10 term at University of Texas.
 Spring '10
 Fitzpatrick
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