21. REASONING The drawing shows a top view of the slit and screen, as well as the position of the central bright fringe and the third dark fringe. The distance y can be obtained from the tangent function as y = L tan θ. Since L is given, we need to find the angle θ before y can be determined. According to Equation 27.4, the angle θ is related to the wavelength λ of the light and the width W of the slit by sin /mWθλ= , where m = 3 since we are interested in the angle for the third dark fringe. SOLUTION We will first compute the angle between the central bright fringe and the third dark fringe using Equation 27.4 (with m = 3): () 9 11 6 3 668 10 m sin sin 17.3 6.73 10 m m W λ θ − −− − g170g186 ×g167g183 == =°g171g187 g168g184 g169g185 ×g172g188 The vertical distance is () tan 1.85 m tan17.3 0.576 myL θ== °= ________________________________________________________________________ ______ θ L y Third dark fringe Central bright fringe Slit 22. REASONING The width W of the slit and the wavelength λ of the light are related to the angle θ defining the location of a dark fringe in the single-slit diffraction pattern, so we can determine the width from values for the wavelength and the angle. The wavelength is given. We can obtain the angle from the width given for the central bright fringe on the screen and the distance between the screen and the slit. To do this, we will use trigonometry and the fact that the width of the central bright fringe is defined by the first dark fringe on either side of the central bright fringe. SOLUTION The angle that defines the location of a dark fringe in the diffraction pattern can be determined according to sin 1, 2, 3,...mm W λ θ == (27.4) Recognizing that we need the case for m = 1 (the first dark fringe on either side of the central bright fringe determines the width of the central bright fringe) and solving for W give sin W λ θ = (1) Referring to Figure 27.24 in the text, we see that the width of the central bright fringe is 2y, where trigonometry shows that () 11 2 0.050 m 2 2 tan or tan tan 2.4 2 2 0.60 m y yL L θθ −− g170g186 g167g183 === =° g171g187g168g184 g169g185 g172g188 Substituting this value for θ into Equation (1), we find that 9 5 510 10 m 1.2 10 m sin sin 2.4 W λ θ − − × == =× ° 23. SSM WWW REASONING According to Equation 27.4, the angles at which the dark fringes occur are given by sin /θ λ= mW, where W is the slit width. In Figure 27.24, we see from the trigonometry of the situation that tan /θ = yL. Therefore, the latter expression can be used to determine the angle θ , and Equation 27.4 can be used to find the wavelength λ . SOLUTION The angle θ is given by θ = F H G I K J = ×F H G I K J =° −− tan tan . . 11 35 10 0 050 y L m 4.0 m –3 The wavelength of the light is λθ==× °=× −− W sin . sin . .5 6 10 0 050 4 9 10 490 47 mm=nmch ________________________________________________________________________ ______ 24. REASONING The angle θ that specifies the location of the m th dark fringe is given by sin /mWθλ= (Equation 27.4), where λ is the wavelength of the light and W is the width of the slit. When θ has its maximum value of 90.0°, the number of dark fringes that can be produced is a maximum. We will use this information to obtain a value for this number. SOLUTION Solving Equation 27.4 for m, and setting θ = 90.0°, we have () 6 9 5.47 10 m sin 90.0sin 90.0 8.40 651 10 m W m λ − − ×°° == = × Therefore, the number of dark fringes is 8 . ________________________________________________________________________ ______ sw Microsoft Word - prob_021.doc