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Unformatted text preview: CHAPTER 37INTERFERENCE 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 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 = = = y d mL bright cm m m nm. = = = ( . . )( ) 71 2 2 15 1 484 (In particular, 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 = = = = = ( )( ) ( . ) . 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 = = = = = , ( . )( ) ( ) . so mm mm nm cm. 012 5 633 94 8 (b) For two different wavelengths, the ratio of the spacings is = y y = = ; 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 = = d (see Equation 37-1a). Thus, = = ( . )( . )( ) 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 = = nm 125 . mm. (Note that m = 0 gives the first dark fringe.) CHAPTER 37 Problem 6. What is the angular position of the second-order bright fringe in a double-slit system with 1.5- m slit spacing if the light has wavelength (a) 400 nm or (b) 700 nm?...
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