{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

chap32 - INTERFERENCE AND DIFFRACTION 32 EXERCISES Section...

Info icon This preview shows pages 1–3. Sign up to view the full content.

View Full Document Right Arrow Icon
INTERFERENCE AND DIFFRACTION 32 E XERCISES 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. 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 double-slit 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 Δ = + = = = 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 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)/ y L d L λ Δ = = (633 nm) 94.8 cm. = (b) For two different wavelengths, the ratio of the spacings is / / y y ; λ λ Δ Δ = therefore (5 mm)(480/633) 3.79 mm. y Δ = = 13. I NTERPRET This problem is about double-slit 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 Δ = (see Equation 31.1a). E VALUATE Substituting the values given, we obtain (0.37 mm)(0.065 )( /180 ) 420 mm. d λ θ π = Δ = ° ° = 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 gives the first dark fringe.) 0 m = Section 32.3 Multiple-Slit Interference and Diffraction Gratings 15. I NTERPRET The setup is a multi-slit interference experiment. We want to know the number of minima (destructive interferences) between two adjacent maxima. D EVELOP 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 / , m d θ λ = and minima for angles (see Equation 32.4) sin θ = / m Nd λ (excluding equal to zero or multiples of N ). m E VALUATE Between two adjacent maxima, say and ( 1) , m mN m N ′ = + there are 1 N minima. The number of integers between is and ( 1) mN m N + ( 1) 1 m N mN N 1 + = 32.1
Image of page 1

Info icon This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon