chap32 - INTERFERENCE AND DIFFRACTION 32 EXERCISES Section...

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INTERFERENCE AND DIFFRACTION 32 EXERCISES Section 32.2 Double-Slit 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. yd m L λ μ == = (In particular, and d << 2 1 3.23 10 1.85 θ =×=° is small.) 11. INTERPRET This problem is about double-slit interference. We are interested in the spacing between adjacent bright fringes. DEVELOP 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 ym d = where m is the order number. EVALUATE The spacing of bright fringes is (550 nm)(75 cm) (1 ) 1 . 6 5 c m 0.025 mm LL L m dd d λλ Δ= + = = = ASSESS Since , d the spacing between bright fringes is much smaller than L , as it should. 12. The particular geometry of this type of double-slit 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)/ yL d L Δ (633 nm) 94.8 cm. = (b) For two different wavelengths, the ratio of the spacings is // yy ; ′′ Δ Δ= therefore (5 mm)(480/633) 3.79 mm. y Δ= = 13. INTERPRET This problem is about double-slit interference. We are interested in the wavelength of the light source. DEVELOP For small angles, the interference fringes are evenly spaced, with / d Δ = (see Equation 31.1a). EVALUATE Substituting the values given, we obtain (0.37 mm)(0.065 )( /180 ) 420 mm. d θπ =Δ= ° °= ASSESS The wavelength 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 λθ =+ = sin 0.113 1.25 mm. (Note that gives the first dark fringe.) 0 m = Section 32.3 Multiple-Slit Interference and Diffraction Gratings 15. INTERPRET The setup is a multi-slit interference experiment. We want to know the number of minima (destructive interferences) between two adjacent maxima. DEVELOP In an N -slit system with slit separation d (illuminated by normally incident plane waves), the main maxima occur for angles (see Equation 32.1a) sin / , md = and minima for angles (see Equation 32.4)sin = / mN d (excluding equal to zero or multiples of N ). m EVALUATE Between two adjacent maxima, say and ( 1) , mm N m N ′ = + there are 1 N minima. The number of integers between is and ( 1) mN m N + ) 1 m N N 1 + −− = 32.1
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32.2 Chapter 32 because the limits are not included. Therefore, For 5, N = the number of minima is 4. ASSESS The interference pattern resembles that shown in Figure 32.8. 16. The first minimum (next to the central peak) occurs at sin /3 , a d θ λ = and the first (primary) maximum at sin / ; b d = 11 therefore sin (3sin ) sin (3sin 5 ) 15.2 . ba θθ −− == ° = ° ± 17. INTERPRET In this problem we want to locate certain maxima and minima in a multi-slit interference experiment.
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This note was uploaded on 10/21/2010 for the course PHYSICS 2131441 taught by Professor Pheong during the Fall '10 term at University of California, Berkeley.

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chap32 - INTERFERENCE AND DIFFRACTION 32 EXERCISES Section...

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