Physics Solution Manual for 1100 and 2101

0 c its expansion increases the slit separation from

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Unformatted text preview: m = m − 1 vacuum (2) 2 2 nfilm ( ) ( ) ′ Setting Equations (1) and (2) equal to each other and solving for λvacuum yields λ′ 640.0 nm = m − 1 vacuum 2 nfilm nfilm ( ) or ′ λvacuum = 640.0 nm m− 1 2 ′ ′ ′ SOLUTION For m = 1, λvacuum =1280 nm; for m = 2, λvacuum = 427 nm; for m = 3, λvacuum ′ = 256 nm. Values of m greater than 3 lead to values of λvacuum that are smaller than 256 nm. Thus, the only wavelength in the visible spectrum (380 to 750 nm) that will give constructive interference is 427 nm . 21. REASONING AND SOLUTION The condition for destructive interference is 2t = mλ. The thickness t of the "air wedge" at the 100th dark fringe is related to the radius R of the fringe and the radius of curvature r of the curved glass in the following way. If s is the length of the circular arc along the curved glass from the center to the fringe, then the angle subtended by this arc is θ = s/r, as shown in Drawing A. Since r is large, the arc is almost straight and is the hypotenuse of a right triangle in the air wedge of sides t and R, and angle (1/2)θ opposite t, as shown in Drawing B. In drawing B, the angle opposite t is labeled α. It can be seen that α = (1/2)θ , because α + β = 90° and (1/2)θ + β = 90°. θ θ/2 θ/2 r r r r R s t β α R R Drawing A s Drawing B t Chapter 27 Problems 1447 In the right triangle in Drawing B, we have (1/2)θ ≈ tan (1/2)θ = t /R so R = 2tr/s Now if (1/2)θ is very small, then s ≈ R, so R2 = 2tr = (100)λr = (100)(654 × 10–9 m)(10.0 m) so R = 0.0256 m 22. REASONING The bright rings occur wherever the thickness of the drop causes constructive interference between a ray of light reflected from the top surface of the drop and a ray reflected from the bottom surface of the drop. The edge of the drop, where the thickness is zero and phase change can only occur due to reflection, is bright. Constructive interference at zero thickness shows that the reflections of the two rays produce no net phase change. At the other bright rings (and the bright spot at the center), then, where the thickness t is not zero, the phase change must be entirely due to the path difference 2t between the two rays. For constructive interference to occur, this phase change must be equal to an integer number of wavelengths λoil of light in the oil, so we have that 2t = 0, λoil , 2λoil , 3λoil ,K (1) The first of the 56 bright rings appears at the edge of the drop, where the thickness is zero. This corresponds to the first value (0 wavelengths) on the right side of Equation (1). For blue light, there are 55 other bright rings, plus the bright spot at the center, where the thickness tcenter of the drop is greatest. Therefore, we can say that the path difference 2tcenter at the center of the drop is equal to 55 + 1 = 56 wavelengths of blue light in the oil: 2tcenter = 56λoil, blue (2) When red light shines on the drop, there is a bright spot at the center, so the path difference 2tcenter is equal to an integer number m of wavelengths of red light in the oil: 2tcenter = mλoil, red (3) By comparison with Equation (2), we see that there will be m bright rings in red light. The wavelength of light in the oil is given by λoil = λvacuum noil (Equation 27.3), where λvaccuum is the vacuum wavelength of the light reflecting from the drop, and noil is the 1448 INTERFERENCE AND THE WAVE NATURE OF LIGHT refractive index of the oil. Applying Equation 27.3 to the two colors of light, and noting that the index of refraction is the same for both the blue and red wavelengths, we obtain λoil, blue = λvacuum, blue λoil, red = and noil λvacuum, red noil (4) SOLUTION Setting the right sides of Equations (2) and (3) equal, since the thickness tcenter at the center of the drop is the same for both wavelengths of light, we see that 2tcenter = mλoil, red = 56λoil, blue (5) Substituting Equations (4) into Equation (5), we obtain λvacuum, red m noil λ = 56 vacuum, blue noil m= or 56λvacuum, blue λvacuum, red = 56 ( 455 nm ) = 40 637 nm Therefore, there are 40 bright rings when the light is red. 23. REASONING For both sound waves and light waves, the angle θ that locates the first minimum or dark fringe in the diffraction pattern can be calculated from the wavelength λ of the waves and the width W of the doorway through which the waves pass. The width of the doorway for the sound is given, and we can determine the wavelength since it is related to the given speed and frequency of the sound. Once the angle is obtained for the sound waves, we will be able to use the relationship between θ, λ, and W a second time to obtain the width of the doorway for the light, since the wavelength of the light is given. SOLUTION a. The angle θ that locates the first minimum or dark fringe in the diffraction pattern can be obtained from Equation 27.4 (with m = 1): sin θ = λ W or λ W θ = sin −1 According to Equation 16.1, the wavelength of the sound wave is λ= v f where v and f are, respectively, t...
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