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Unformatted text preview: INTERFERENCE AND DIFFRACTION E XERCISES Section 32.2 DoubleSlit Interference 10. The experimental arrangement and geometrical approximations valid for Equation 32.2a are satisfied for the situation and data given, so bright / (7.1 cm/2.2 m)(15 m/1) 484 nm. y d mL λ μ = = = (In particular, and d λ << 2 1 3.23 10 1.85 θ = × = ° is small.) 11. I NTERPRET This problem is about doubleslit interference. We are interested in the spacing between adjacent bright fringes. D EVELOP We assume that the geometrical arrangement of the source, slits, and screen is that for which Equations 32.2a and 32.2b apply. The location of bright fringes is given by bright L y m d λ = where m is the order number. E VALUATE The spacing of bright fringes is (550 nm)(75 cm) ( 1) 1.65 cm 0.025 mm L L L y m m d d d λ λ λ ∆ = + = = = A SSESS Since , d λ << the spacing between bright fringes is much smaller than L , as it should. 12. The particular geometry of this type of doubleslit experiment is described in the paragraphs preceding Equations 32.2a and 32.2b. (a) The spacing of bright fringes on the screen is / , so (0.12 mm)(5 mm)/ y L d L λ ∆ = = (633 nm) 94.8 cm. = (b) For two different wavelengths, the ratio of the spacings is / / ; y y λ λ ′ ′ ∆ ∆ = therefore (5 mm)(480/633) 3.79 mm. y ′ ∆ = = 13. I NTERPRET This problem is about doubleslit interference. We are interested in the wavelength of the light source. D EVELOP For small angles, the interference fringes are evenly spaced, with / d θ λ ∆ = (see Equation 31.1a). E VALUATE Substituting the values given, we obtain (0.37 mm)(0.065 )( /180 ) 420 mm. d λ θ π = ∆ = ° ° = A SSESS The wavelength λ is much smaller than the slit spacing d , as expected. 14. The interference minima fall at angles given by Equation 32.1b; therefore 1 2 (4 ) /sin 4.5(546 nm)/ d λ θ = + = sin 0.113 1.25 mm. ° = (Note that m = gives the first dark fringe.) Section 32.3 MultipleSlit Interference and Diffraction Gratings 15. I NTERPRET The setup is a multislit interference experiment. We want to know the number of minima (destructive interferences) between two adjacent maxima. D EVELOP In an Nslit system with slit separation d (illuminated by normally incident plane waves), the main maxima occur for angles (see Equation 32.1a) sin / , m d θ λ = and minima for angles (see Equation 32.4) sin θ = / m Nd λ ′ (excluding m ′ equal to zero or multiples of N ). E VALUATE Between two adjacent maxima, say and ( 1) , m mN m N ′ = + there are 1 N minima. The number of integers between and ( 1) mN m N + is ( 1) 1 1 m N mN N + = 32.1 32 32.2 Chapter 32 because the limits are not included. Therefore, For 5, N = the number of minima is 4....
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 Spring '09
 Prof.Kim
 Physics, Diffraction, Wavelength, Sin, NM

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