HW 4 solutions (Ch 27)

HW 4 solutions (Ch 27) - Physics 6C HW#4 solution May 1,...

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Unformatted text preview: Physics 6C HW#4 solution May 1, 2008 27.14 Light reflecting from the first surface suffers phase re- versal. Light reflecting from the second surface does not, but passes twice through the thickness t of the film. So, for constructive interference, we require D = λ n 2 + 2 t = mλ n For the first max, m = 1 and the wavelength in the film is λ n = λ n . Then, 2 t = λ n 2 = λ 2 n λ = 4 nt = 612 nm 27.19 For destructive interference in the air, the path dif- ference D satisfies a following condition. D = λ 2 + 2 t = ( m + 1 2 ) λ 2 t = mλ For 30 dark fringes, including the one where the plates meet, m =29 t = 29 × (600 nm ) 2 = 8 . 70 × 10- 6 m Therefore, the radius of the wire is, r = t 2 = 4.35 μ m 27.21 y m L = tan θ ≈ sin θ = mλ a Here, Δ y = y 3- y 1 = 3 . 00 × 10- 3 nm Δ y = 3 λL a- 1 λL a = 2 λL a a = 2 λL Δ y = 2 . 30 × 10- 4 m 27.25 The rectangular patch on the wall is wider than it is tall. The aperture will be taller than it is wide. For 1 horizontal spreading, we have.horizontal spreading, we have....
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This note was uploaded on 07/02/2008 for the course PHYS 6C taught by Professor Staff during the Spring '01 term at UCLA.

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HW 4 solutions (Ch 27) - Physics 6C HW#4 solution May 1,...

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