Physics Solution Manual for 1100 and 2101

The proper lifetime is equal to the contracted

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: ating undergoes the temperature change ∆T = +100.0 C°, its expansion increases the slit separation from d1 to d2. From Equation 27.7, then we have sin θ1 = 7λ d1 and sin θ 2 = 7λ d2 (1) where θ1 and θ2 are the angles of the seventh-order principal maximum before and after the temperature changes. Because the grating expands, d2 is greater than d1, and we see from Equations (1) that θ2 is less than θ1. This means that the seventh-order bright fringe moves closer to the central maximum as the grating heats up. Therefore, we expect the algebraic sign of the angle change ∆θ = θ2 − θ1 to be negative. To determine the new slit separation d2, we will make use of ∆L = α L0 ∆T (Equation 12.2), where α = 1.30×10−4 (C°)−1 is the coefficient of thermal expansion of the diffraction grating, L0 = d1 is the initial slit separation, ∆T is the temperature change, and ∆L = d2 − d1 is the amount by which the slit separation increases. Making these substitutions into Equation 12.2 yields d 2 − d1 = α d1∆T (2) SOLUTION Solving Equation (2) for d2, we obtain d 2 = d1 + α d1∆T = d1 (1 + α∆T ) (3) Substituting Equation (3) into the second of Equations (1) yields sin θ2 = 7λ 7λ = d2 d1 (1 + α∆T ) Taking the inverse sine of Equation (4) and the first of Equations (1), then, we have (4) 1466 INTERFERENCE AND THE WAVE NATURE OF LIGHT 7λ −1 7 λ ∆θ = θ 2 − θ1 = sin −1 − sin d d1 (1 + α ∆T ) 1 = sin ( ) ) ( 7 656.0 × 10 −9 m 7 656.0 × 10 −9 m − sin −1 1.250 × 10 −5 m 1.30 × 10 −4 o −5 1.250 × 10 m 1 + 100.0 C Co −1 ( ( ) ) = −0.29o 53. REASONING AND SOLUTION The last maximum formed by the grating corresponds to θ = 90.0°, so that m= where we have used d = d λ sin 90.0 ° = d λ = 1.78 × 10 −6 m = 3.78 471 × 10 −9 m 1 = 1. 78 × 10 −4 cm = 1.78 × 10 −6 m . 5620 lines / cm Thus, the last maximum formed by the grating is m = 3. This maximum lies at L F 1 IO F λI θ = sin −1 G J = sin −1 MG JP 52 .5° mJ 3 MG JP= G dK M3 P H NH .78 K Q The distance from the center of the screen to the m = 3 maximum is y = (0.750 m) tan 52.5° = 0.977 m The screen must have a width of 2y = 1.95 m 54. REASONING The wavelength in the film is λ film = λ vacuum / nfilm (Equation 27.3), so that the Incident light refractive index of the film is nfilm = λ vacuum λ film 1 2 nair = 1.00 (1) t nfilm ngas = 1.40 The wavelength in a vacuum is given, and we can determine the wavelength in the film by considering the constructive interference that occurs. The drawing shows the thin film and Chapter 27 Problems 1467 two rays of light shining on it. At nearly perpendicular incidence, ray 2 travels a distance of 2t farther than ray 1, where t is the thickness of the film. In addition, ray 2 experiences a phase shift of 1 λ 2 film upon reflection at the bottom film surface, while ray 1 experiences the same phase shift at the upper film surface. This is because, in both cases, the light is traveling through a region where the refractive index is lower toward a region where it is higher. Therefore, there is no net phase change for the two reflected rays, and only the extra travel distance determines the type of interference that occurs. For constructive interference the extra travel distance must be an integer number of wavelengths in the film: 2t + 123 Extra distance traveled by ray 2 This result is equivalent to 0 13 2 Zero net phase change due to reflection = λ film , 2λ film , 3λ film ,... 1444 24444 4 3 Condition for constructive interference 2t = mλ film m = 1, 2, 3, ... The wavelength in the film, then, is λ film = 2t m m = 1, 2, 3, ... (2) SOLUTION Substituting Equation (2) into Equation (1) gives nfilm = λ vacuum λ film = λ vacuum 2t m Since the given value for t is the minimum thickness for which constructive interference can occur, we know that m = 1. Thus, we find that nfilm = mλ vacuum 2t = (1) ( 625 nm ) = 1.29 2 ( 242 nm ) 55. SSM REASONING The angles θ that determine the locations of the dark and bright fringes in a Young’s double-slit experiment are related to the integers m that identify the fringes, the wavelength λ of the light, and the separation d between the slits. Since values are given for m, λ, and d, the angles can be calculated. 1468 INTERFERENCE AND THE WAVE NATURE OF LIGHT SOLUTION The expressions that specify θ in terms of m, λ, and d are as follows: λ Bright fringes sin θ = m Dark fringes sin θ = m + ( m = 0, 1, 2, 3, ... 1 2 )λ d (27.1) m = 0, 1, 2, 3, ... d (27.2) Applying these expressions gives the answers that we seek. ( a. sin θ = m + b. sin θ = m c. sin θ = m + d. ( or θ = sin −1 0 + 1 2 θ = sin −1 (1) or θ = sin −1 1 + 1 2 or θ = sin −1 ( 2 ) λ d ( sin θ = m )d λ or 1 2 1 2 )d λ λ d ) 520 × 10 −9 m = 11° 1.4 × 10 −6 m 520 × 10−9 m = 22° 1.4 × 10−6 m () 520 × 10−9 m = 34° 1.4 × 10−6 m 520 × 10 −9 m = 48° 1.4 × 10 −6 m 56. REASONING In a Young’s double-slit experiment a dark fringe is located at an angle θ that is determined according to ( sin θ = m + 1 2 )λ d m = 0, 1, 2, 3, ... (27.2) where λ is the wavelength of the light and d is the separation of the slits. Here, neither λ nor d is known. However, we know the angle for the dark fringe for which m = 0. Using this angle and m = 0 in Equation 27.2 will allow us to determine the ratio λ/d, w...
View Full Document

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.

Ask a homework question - tutors are online