Physics Solution Manual for 1100 and 2101

Since the proper time interval always has the same

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Unformatted text preview: red from the middle of the central bright spot, so that the angle θspot subtended by the entire bright spot is twice as large: θspot = 2θ (2) λ The first circular dark fringe is located by the angle θ = sin −1 1.22 (Equation 27.5), D where D is the diameter of the spotlight and λ is the wavelength of the light. SOLUTION Substituting Equation (2) into Equation (1) yields d = rθspot = 2rθ (3) λ Substituting θ = sin −1 1.22 (Equation 27.5) into Equation (3), we obtain D λ 694.3 × 10−9 m 3 d = 2rθ = 2r sin −1 1.22 = 2 ( 3.77 × 108 m ) sin −1 1.22 = 3.2 × 10 m D 0.20 m 41. SSM WWW REASONING Assuming that the angle θmin is small, the distance y between the blood cells is given by (8.1) y = f θ min where f is the distance between the microscope objective and the cells (which is given as the focal length of the objective). However, the minimum angular separation θmin of the cells is given by the Rayleigh criterion as θmin = 1.22 λ/D (Equation 27.6), where λ is the wavelength of the light and D is the diameter of the objective. These two relations can be used to find an expression for y in terms of λ. Blood cells y f θmin Microscope objective D Chapter 27 Problems 1459 SOLUTION a. Substituting Equation 27.6 into Equation 8.1 yields 1.22λ y = f θ min = f D Since it is given that f = D, we see that y = 1.22 λ . b. Because y is proportional to λ, the wavelength must be shorter closer together. to resolve cells that are 42. REASONING AND SOLUTION a. Equation 27.6 θ min = 1.22 λ / D gives the minimum angle θmin that two point objects can subtend at an aperture of diameter D and still be resolved. The angle must be measured in radians. For a separation s between the two circles and a distance L between the concentric arrangement and the camera, Equation 8.1 gives the angle in radians as θ min = s / L . Therefore, we find that c h 1.22 λ s = D L or L= sD 1.22 λ Since s = 0.040 m – 0.010 m = 0.030 m, we calculate that bg c bg c h h 0.030 m 12 .5 × 10 –3 m sD L= = = 550 m 1.22 λ 1.22 555 × 10 –9 m b. The calculation here is similar to that in part a, except that the separation s is between one side of a diameter of the small circle and the other side, or s = 0.020 m: sD L= = 1.22 λ 0 b.020 m g12.5 × 10 mh= c 1 b.22 g555 × 10 mh c –3 –9 370 m 43. SSM REASONING AND SOLUTION According to Equation 27.7, the angles that correspond to the first-order (m = 1) maximum for the two wavelengths in question are: a. for λ = 660 nm = 660 × 10 –9 m , 1460 INTERFERENCE AND THE WAVE NATURE OF LIGHT 660 F λ I = sin L1) F × 10 ( Gd J M G × 10 H K NH 1.1 θ = sin –1 m –9 –1 –6 m m IO 37 ° J= P K Q m m IO 22 ° J= P K Q b. for λ = 410 nm = 410 × 10 –9 m , 410 F λ I = sin L1) F × 10 ( Gd J MG × 10 H K NH 1.1 θ = sin –1 m –9 –1 –6 44. REASONING The number of lines per centimeter that a grating has is the reciprocal of the spacing between the slits of the grating. We can determine the slit spacing by considering the angle θ that defines the position of a principal bright fringe. This angle is related to the order m of the fringe, the wavelength λ of the light, and the spacing d between the slits. Thus, we can use the values given for θ, m, and λ to determine d. SOLUTION The number of lines per centimeter that the grating has is N and is the reciprocal of the spacing d between the slits: N= 1 d (1) where d must be expressed in centimeters. The relationship that determines the angle defining the position of a principal bright fringe is sin θ = m λ d m = 0, 1, 2, 3, ... (27.7) Solving this equation for d and applying the result for a second-order fringe (m = 2) gives ( ) 2 495 ×10−9 m mλ d= = = 6.10 × 10−6 m sin θ sin 9.34° or 6.10 × 10−4 cm Substituting this result into Equation (1), we find that N= 1 1 = = 1640 lines/cm d 6.10 ×10−4 cm Chapter 27 Problems 1461 45. SSM REASONING AND SOLUTION The geometry of the situation is shown below. First dark fringe y θ d Midpoint of central bright fringe L From the geometry, we have tan θ = y 0.60 mm = = 0.20 L 3.0 mm θ = 11.3° or Then, solving Equation 27.7 with m = 1 for the separation d between the slits, we have c h (1) 780 × 10 –9 m mλ d= = = sin θ sin 11.3° 4.0 × 10 –6 m 46. REASONING The drawing shows the angle θ that locates a principal maximum on the screen, along with the separation L between the grating and the screen and the distance y from the midpoint of the screen. It follows from the drawing that y = L tan θ , which becomes y = L sin θ , since we are dealing with small angles ( tan θ ≈ sin θ ) . For a diffraction grating, the angle θ that locates a principal maximum can be found using sin θ = mλ/d (Equation 27.7), where λ is the wavelength, d is the separation between the grating slits, and the order m is m = 0, 1, 2, 3, …. y θ L Grating We will use the above relations to obtain an expression for the separation between adjacent principal maxima. 1462 INTERFERENCE AND THE WAVE NATURE OF LIGHT SOLUTION The dist...
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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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