HW6 solution

HW6 solution - Problem 1 Problem 10.6 β = ±π sin θ = ±(a Lθ ≈ ± L b b ≈θ(b Lθ ≈ ± f 2 b Problem 10.7 For far field it must

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Unformatted text preview: Problem 1 Problem 10.6 β = ±π , sin θ = ± (a) Lθ ≈ ± L λ b λ b ≈θ ; (b) Lθ ≈ ± f 2 λ b Problem 10.7 For far field, it must satisfy R > sin θ 1 = λ b2 b2 , = λ λ (10 −4 m) 2 = 0.02 , so it is far field. 4.619 × 10 −7 m λ ⇒ θ 1 = sin −1 ( ) = 0.26 o so angular width is 2θ1 = 0.52 o b b Problem 2 Problem10.9 λ = (20cm) × sin 36.87 o = 12cm Problem 3 Problem 10.10 ka kb α = sin θ , β = sin θ , a = mb, α = mβ , α = m2π 2 2 α m2π N= number of fringes = = = 2m π π Problem 4 Since q1 = 1.22 Rλ (3.76 × 10 8 m)(6.328 × 10 −7 m) = 1.22 × = 1.45 × 10 5 m −3 2a 2 × 10 m Problem 5 a sin θ m = mλ ; sin θ1 = y y Rλ (2.0m)(6.943 × 10 −7 m) ⇒a =λ⇒ y= = = 0.46m R R a 3.0 × 10 −6 m Problem 6 ym y mRλ ⇒ a = λ ⇒ ym = R R a 6 −6 10000lines / cm = 10 lines / m ⇒ a = 10 m 1 × 5.895923 × 10 −7 × 1 y1 (589.5923nm) = = 0.5895923m 10 − 6 1 × 5.889953 × 10 −7 × 1 y1' (588.9953nm) = = 0.5889953m 10 − 6 Separation = y1 − y1' = 5.97 × 10 −4 m Since a sin θ m = mλ ; sin θ m = Problem 7 ...
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This note was uploaded on 10/16/2011 for the course ELEC 308,315,10 taught by Professor Prof.shenghuisong during the Spring '11 term at CUHK.

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HW6 solution - Problem 1 Problem 10.6 β = ±π sin θ = ±(a Lθ ≈ ± L b b ≈θ(b Lθ ≈ ± f 2 b Problem 10.7 For far field it must

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