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Unformatted text preview: Interference and Diffraction Problem 1 (medium): A monochromatic plane wave with wavelength λ is incident on two narrow slits separated by d = 100 λ . A screen is located a distance L = 8000 λ in front of the slits. The center of the screen is y = 0, directly in front of the two slits. We count the central interference maximum as the zeroth maximum. 1: How far from the center of the screen is the fourth interference minimum? 2: How far from the center of the screen is the 65th interference maximum? 3: How many interference maxima occur between the positions y 1 = 3000 λ and y 2 = 10000 λ , measured from the center of the screen? 4: What total number of interference maxima appear on the screen? 5: What is the width of the region around the zeroth maximum for which the intensity is at least 35% of its maximum value? 6: What is the width of the region around the 80th maximum for which the intensity is at least 35% of its maximum value? &¡¡ & & ¡ ¢¡¡¡ ¡ & ¡ ¢ &¡¢££¤ ¢ Solution: We first consider part 1 . The fourth minimum occurs at an angle θ , where d sin θ = 3 + 1 2 ¶ λ = 7 2 λ sin θ = 7 λ 2 d = 7 λ 200 λ = 7 200 1 Note that the first minimum corresponds to 0+1 / 2, the second minimum to 1+1 / 2, etc.. The distance to the fourth minimum is therefore y 4 , where y 4 = L tan θ = L sin θ cos θ = L sin θ p 1 sin 2 θ = L 7 / 200 p 1 (7 / 200) 2 = (8000 λ ) 7 / 200 p 1 (7 / 200) 2 = 280 . 2 λ We see that sin θ << 1. Therefore, we could also have made the approximation y 4 = L tan θ ≈ L sin θ = 7 L 200 = (7)(8000) 200 λ = 280 λ In this case, the approximation is accurate. Note that there are two “fourth minima”, located at y = ± y 4 . This is due to the symmetry of the intensity distribution about y = 0. We now consider part 2 . The 65th maximum occurs at an angle θ , where d sin θ = 65 λ sin θ = 65 λ d = 65 λ 100 λ = 13 20 The distance to the 65th maximum is therefore y 65 , where 2 y 65 = L tan θ = L sin θ cos θ = L sin θ p 1 sin 2 θ = L 13 / 20 p 1 (13 / 20) 2 = (8000 λ ) 13 / 20 p 1 (13 / 20) 2 = 6843 λ Note that there are two “65th maxima”, located at y = ± y 65 . Let’s now try using the small angle approximation, as we did for part 1. We obtain y 65 = L tan θ ≈ L sin θ, not true = 13 L 20 = (13)(8000) 20 λ = 5200 λ, (wrong!) In this case, we see that the small angle approximation gives the wrong answer . This is because sin θ = 13 / 20, which is not small compared to 1....
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 Spring '01
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 Sin, 7l, 4 M, 2 1 m, 9.57 m

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