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# CH37 - CHAPTER 37 INTERFERENCE AND DIFFRACTION ActivPhysics...

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CHAPTER 37—INTERFERENCE AND DIFFRACTION ActivPhysics can help with these problems: Activities in Section 16, Physical Optics Section 37 - 2:—Double-Slit Interference Problem 1. A double-slit system is used to measure the wavelength of light. The system has slit spacing d = 15 m μ and slit-to- screen distance L m = = 2 2 1 . m. If the maximum in the interference pattern occurs 71 . cm from screen center, what is the wavelength? Solution The experimental arrangement and geometrical approximations valid for Equation 37-2a are satisfied for the situation and data given, so λ μ = = = y d mL bright cm m m nm. = = = ( . . )( ) 71 2 2 15 1 484 (In particular, λ θ ¿ d and 1 2 3 23 10 185 = × = ° - . . is small.) Problem 2. A double-slit experiment with d L = = 0 025 75 . mm and cm uses 550-nm light. What is the spacing between adjacent bright fringes? Solution Assume that the geometrical arrangement of the source, slits, and screen is that for which Equations 37-2a and b apply. The spacing of bright fringes is y L d = = = λ = = ( )( ) ( . ) . 550 75 0 025 165 nm cm mm cm. Problem 3. A double-slit experiment has slit spacing 012 . mm. (a) What should be the slit-to-screen distance L if the bright fringes are to be 5 0 . mm apart when the slits are illuminated with 633-nm laser light? (b) What will be the fringe spacing with 480-nm light? Solution The particular geometry of this type of double-slit experiment is described in the paragraphs preceding Equations 37-2a and b. (a) The spacing of bright fringes on the screen is y L d L = = = λ = = , ( . )( ) ( ) . so mm mm nm cm. 012 5 633 94 8 (b) For two different wavelengths, the ratio of the spacings is = y y = = λ λ ; therefore ∆ ′ = = y ( )( ) . 5 480 633 3 79 mm mm. = Problem 4. With two slits separated by 0 37 . mm, the interference pattern has bright fringes with angular spacing 0.065 ° . What is the wavelength of the light illuminating the slits? Solution For small angles, the interference fringes are evenly spaced, with θ λ = = d (see Equation 37-1a). Thus, λ π = ° ° = ( . )( . )( ) 0 37 0 065 180 420 mm nm. = Problem 5. The green line of gaseous mercury at 546 nm falls on a double-slit apparatus. If the fifth dark fringe is at 0.113 ° from the centerline, what is the slit separation? Solution The interference minima fall at angles given by Equation 37-1b; therefore d = + = ° = ( ) sin . ( ) sin . 4 4 5 546 0113 1 2 λ θ = = nm 125 . mm. (Note that m = 0 gives the first dark fringe.)

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CHAPTER 37 Problem 6. What is the angular position θ of the second-order bright fringe in a double-slit system with 1.5- μ m slit spacing if the light has wavelength (a) 400 nm or (b) 700 nm? Solution (a) From Equation 37-1a, θ λ μ λ = = × = ° ° = - - sin ( ) sin ( . ) . , 1 1 2 400 15 32 2 700 m d = = nm m and (b) 69.0 for nm. Problem 7. Light shines on a pair of slits whose spacing is three times the wavelength. Find the locations of the first- and second- order bright fringes on a screen 50 cm from the slits. Hint: Do Equations 37-2 apply?
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