Lect06 - Lecture 6, p. 1 Lecture 6: Waves Review,...

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Unformatted text preview: Lecture 6, p. 1 Lecture 6: Waves Review, Crystallography, and Examples Lecture 6, p. 2 Single-Slit Diffraction Example W = 1 cm L = 2 m a θ Suppose that when we pass red light ( λ = 600 nm ) through a slit of unknown width a , the width of the spot (the distance between the first zeros on each side of the bright peak) is W = 1 cm on a screen that is L = 2 m behind the slit. How wide is the slit? Lecture 6, p. 3 Solution W = 1 cm L = 2 m a θ The angle to the first zero is: θ = ± λ /a W = 2L tan θ 2245 2L θ = 2L λ /a a = 2L λ /W = (4m)(6 × 10-7 m) /(10-2 m) = 2.4 × 10-4 m = 0.24 mm Suppose that when we pass red light ( λ = 600 nm ) through a slit of unknown width a , the width of the spot (the distance between the first zeros on each side of the bright peak) is W = 1 cm on a screen that is L = 2 m behind the slit. How wide is the slit? Lecture 6, p. 4 Multiple Slit Interference (from L4) The positions of the principal maxima occur at φ = 0, ± 2 π , ± 4 π , ... where φ is the phase between adjacent slits. θ = 0, ±λ /d , ± 2 λ /d, ... The intensity at the peak of a principal maximum goes as N 2 . 3 slits: A tot = 3A 1 ⇒ I tot = 9I 1 . N slits: I N = N 2 I 1 . Between two principal maxima there are N-1 zeros and N-2 secondary maxima ⇒ The peak width ∝ 1/N . The total power in a principal maximum is proportional to N 2 (1/N) = N. 2π-2π I 4 16I 1 N=4 2π-2π I 5 25I 1 N=5 2π-2π I 3 9I 1 N=3-λ/ d 0 λ/ d φ θ φ θ-λ/ d 0 λ/ d-λ/ d 0 λ/ d φ θ Lecture 6, p. 5 Multiple-slit Example Three narrow slits with equal spacing d are at a distance L = 1.4 m away from a screen. The slits are illuminated at normal incidence with light of wavelength λ = 570 nm . The first principal maximum on the screen is at y = 2.0 mm . 1. What is the slit spacing , d ? 2. If the wavelength, λ , is increased, what happens to the width of the principal maxima? L y d θ Lecture 6, p. 6 Solution Three narrow slits with equal spacing d are at a distance L = 1.4 m away from a screen. The slits are illuminated at normal incidence with light of wavelength λ = 570 nm . The first principal maximum on the screen is at y = 2.0 mm . 1. What is the slit spacing , d ? The first maximum occurs when the path difference between adjacent slits is λ . This happens at sin θ = λ / d . We are told that tan θ = y / L = 1.43X10-3 , so the small angle approximation is OK. Therefore, d ≈ λ / θ = 0.40 mm. 2. If the wavelength, λ , is increased, what happens to the width of the principal maxima? The relation between θ and φ is φ /2 π = δ / λ = d sin θ / λ . Therefore, for every feature that is described by φ (peaks, minima, etc .) sin θ is proportional to λ . The width increases....
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This note was uploaded on 04/04/2011 for the course PHYSICS 214 taught by Professor Mestre during the Spring '11 term at University of Illinois at Urbana–Champaign.

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Lect06 - Lecture 6, p. 1 Lecture 6: Waves Review,...

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