Physics Solution Manual for 1100 and 2101

500c 2 120 light years a2 a1 1 1 0500 light yearsyear

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Unformatted text preview: hich can then be used to find the angle that locates the dark fringe for m = 1. SOLUTION Applying Equation 27.2 to the dark fringes for m = 0 and m = 1, we have ( )λ d sin θ 0 = 0 + 1 2 and ( )λ d sin θ1 = 1 + 1 2 Dividing the expression for sin θ1 by the expression for sin θ 0 gives Chapter 27 Problems sin θ1 sin θ 0 1469 (1 + ) λ = 3.0 d = (0 + ) λ d 1 2 1 2 The angle that locates the dark fringe for m = 1 can now be found: sin θ1 = 3.0 sin θ 0 θ1 = sin −1 ( 3.0 sin θ 0 ) = sin −1 ( 3.0 sin15° ) = 51° or 57. SSM REASONING The slit separation d is given by Equation 27.1 with m = 1; namely d = λ / sin θ . As shown in Example 1 in the text, the angle θ is given by θ = tan −1 ( y / L ) . SOLUTION The angle θ is θ = tan −1 0 F.037 m I = 0.47 ° G4 .5 m J HK Therefore, the slit separation d is d= λ sin θ = 490 × 10 –9 m = sin 0.47 ° 6.0 × 10 –5 m 58. REASONING The drawing shows a top Third dark view of the slit and screen, as well as the Slit fringe position of the central bright fringe and the y third dark fringe. The distance y can be θ obtained from the tangent function as y = L Central bright fringe tan θ. Since L is given, we need to find the L 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 θ = mλ / W , 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 ( m) mλ −1 3 668 × 10 = 17.3° = sin W 6.73 × 10−6 m θ = sin −1 The vertical distance is y = L tan θ = (1.85 m ) tan17.3° = 0.576 m 1470 INTERFERENCE AND THE WAVE NATURE OF LIGHT 59. REASONING For the hunter’s eye to resolve the squirrels as separate objects, the Rayleigh criterion must be satisfied. The Rayleigh criterion relates the angle θmin (the angle subtended at the eye by the objects) to the wavelength λ of the light and the diameter D of the pupil. Thus, we can calculate the diameter if values are available for the wavelength and the angle. The wavelength is given. To obtain the angle, we will use the concept of the radian and express the angle approximately as an arc length (approximately the separation distance s between the squirrels) divided by a radius (the distance L between the hunter and the squirrels), as discussed in Section 8.1. SOLUTION a. The Rayleigh criterion specifies the angle θmin in radians as θ min ≈ 1.22 Solving for D gives D≈ λ (27.6) D 1.22λ (1) θ min The drawing at the right shows the hunter’s eye and the squirrels. According to the discussion in Section 8.1, the angle θmin can be expressed approximately in radians as θmin ≈ s L Squirrel 1 Hunter’s eye L θmin (8.1) s Squirrel 2 This expression for the angle is approximate because we are assuming that the distance s between the squirrels is nearly the same as the arc length on a circle of radius L centered on the hunter’s eye. Substituting Equation 8.1 into Equation (1), we find that D≈ D≈ ( 1.22λ )( θ min ≈ 1.22λ 1.22λ L = s s L 1.22 498 × 10−9 m 1.6 × 103 m 0.10 m ) = 9.7 ×10−3 m or 9.7 mm b. A diameter of 9.7 mm is beyond the normal range for the human eye. Moreover, the human eye would increase the diameter of its pupil to a maximum only under dark conditions. Hunting in the dark is difficult. The hunter’s claim is not reasonable. Chapter 27 Problems 1471 60. REASONING The drawing shows th m bright fringe Double a top view of the double slit and the slit screen, as well as the position of the central bright fringe (m = 0) and the θ th Central bright m bright fringe. The angle θ to the th fringe (m = 0) m bright fringe is given by Equation 27.1 as sin θ = mλ / d , where λ is the wavelength of the light and d is the separation between the slits. The largest that the angle can be is θ = 90.0°. This condition will tell us how many bright fringes can be formed on either side of the central bright fringe. SOLUTION Substituting θ = 90.0° into Equation 27.1 and solving for m gives m= d sin 90.0° λ = ( 3.76 ×10−6 m ) sin 90.0° = 6.02 625 × 10−9 m Therefore, the greatest number of bright fringes is 6 . 61. SSM REASONING The angle θ that locates the first-order maximum produced by a grating with 3300 lines/cm is given by Equation 27.7, sin θ = mλ / d , with the order of the fringes given by m = 0 , 1, 2, 3, . . . Any two of the diffraction patterns will overlap when their angular positions are the same. SOLUTION Since the grating has 3300 lines/cm, we have d= 1 = 3.0 × 10 –4 cm = 3.0 × 10 –6 m 3300 lines / cm a. In first order, m = 1; therefore, for violet light, θ = sin –1 –9 –1 –6 Similarly for red light, θ = sin 4 F λ I = sin L1) F10 × 10 m ( Gd J MG × 10 H K NH 3.0 –1 6 F λ I = sin L1) F60 × 10 m ( Gd J MG × 10 H K NH 3.0 –9 –1 –6 m m IO 7.9 ° J= P K Q m m IO 13° J= P K Q 1472 INTERFERENCE AND THE WAVE NATURE OF LIGHT b. Repeating the calculation for the second order maximum ( m = 2 ), we find that ( m = 2) for violet for red θ = 16 ° θ = 26 ° c. Repea...
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This note was uploaded on 04/30/2013 for the course PHYS 1100 and 2 taught by Professor Chastain during the Spring '13 term at LSU.

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