HW6 - PHYS 4D Solution to HW 6 February 10, 2011 Problem...

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PHYS 4D Solution to HW 6 February 10, 2011 Problem Giancoli 34-4 (II) Monochromatic light falls on two very narrow slits 0 . 048 mm apart. Successive fringes on a screen 6 . 00 m away are 8 . 5 cm apart near the center of the pattern. Determine the wavelength and frequency of the light. Solution: For constructive interference, the path difference is a multiple of the wavelength, as given by Eq. 34-2a. The location on the screen is given by x = l tan θ , as seen in Fig. 34-7(c). For small angles, we have sin θ tan θ x/l . Adjacent fringes will have ∆ m = 1. d x l = d sin θ = x = λml d . x 1 = λml d , x 2 = λ ( m + 1) l d x = x 2 x 1 = λl d λ = d x l = (0 . 048 mm ) × (8 . 5 cm ) 6 . 00 m = 6 . 8 × 10 7 m f = c λ = 3 . 0 × 10 8 m/s 6 . 8 × 10 7 m = 4 . 4 × 10 14 Hz Problem Giancoli 34-8 (II) Light of wavelength 680 nm falls on two slits and produces an interference pattern in which the third-order bright fringe is 38 mm from the central fringe on a screen 2 . 6 m away. What is the separation of the two slits? Solution: d x l = d = λml x = (680 nm )3(2 . 6 m ) 38 mm = 1 . 4 × 10 4 m. Problem Giancoli 34-12 (II) In a double-slit experiment it is found that blue light of wavelength 480 nm gives a second-order maximum at a certain location on the screen. What wavelength of visible light would have a minimum at the same location? Solution: We equate the expression from Eq. 34-2a for the second order blue light to Eq. 34-2b, since the slit separation and angle must be the same for the two conditions to be met at the same location. d
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HW6 - PHYS 4D Solution to HW 6 February 10, 2011 Problem...

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