Physics Solution Manual for 1100 and 2101

# Therefore the angle may be found from y tan 1 equation

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Unformatted text preview: er blue fringes. Therefore, the screen must be moved toward the slits so that the orange fringes will appear on the screen. The distance between the screen and the slits is L, and the amount by which the screen must be moved toward the slits is L blue – Lorange . We know that L blue = 0.500 m , and must, therefore, determine L orange . We begin with Equation 27.1, as it applies to first-order fringes, that is, sin θ = λ/d. Furthermore, as discussed in Example 1 in the text, tan θ = y/L, where y is the distance from the center of the screen to a fringe. Since the angle θ locating the fringes is small, sin θ ≈ tan θ , and we have that λ d ≈ y L or L= yd λ Writing this result for L for both colors and dividing the two equations gives Chapter 27 Problems L blue yd / λ blue = Lorange yd / λ orange = 1437 λ orange λ blue where we have recognized that y is one-half the screen width (the same for both colors), and that the slit separation d is the same for both colors. Using the result above, we find that Fλ I L = L F– λ I 1 GJ Gλ J λ HK H K L 471 nm O 0.115 m = b.500 m g – = 0 1 M 611 nm P N Q L blue – Lorange = L blue – blue blue blue blue orange orange 11. SSM WWW REASONING The light that travels through the plastic has a different path length than the light that passes through the unobstructed slit. Since the center of the screen now appears dark, rather than bright, destructive interference, rather than constructive interference occurs there. This means that the difference between the number 1 of wavelengths in the plastic sheet and that in a comparable thickness of air is 2 . SOLUTION The wavelength of the light in the plastic sheet is given by Equation 27.3 as λplastic = λvacuum = n 586 × 10−9 m = 366 × 10−9 m 1.60 The number of wavelengths contained in a plastic sheet of thickness t is N plastic = t λ plastic = t 366 × 10 −9 m The number of wavelengths contained in an equal thickness of air is N air = t λ air = t 586 × 10 −9 m where we have used the fact that λair ≈ λvacuum. Destructive interference occurs when the 1 difference, Nplastic − Nair, in the number of wavelengths is 2 : 1 N plastic − N air = 2 t t 1 − = −9 −9 366 × 10 m 586 × 10 m 2 1438 INTERFERENCE AND THE WAVE NATURE OF LIGHT –9 Solving this equation for t yields t = 487 × 10 m = 487 nm . 12. REASONING To solve this problem, we must express the condition for destructive interference in terms of the film thickness t and the wavelength λ film of the light as it passes through the magnesium fluoride coating. We must also take into account any phase changes that occur upon reflection. SOLUTION Since the coating is intended to be nonreflective, its thickness must be coating chosen so that destructive interference (film) occurs between waves 1 and 2 in the 2 air camera drawing. For destructive interference, the lens combined phase difference between the two 1 waves must be an odd integer number of half wavelengths. The phase change for wave 1 is equivalent to one-half of a wavelength, since this light travels from a smaller refractive index ( n air = 1.00 ) toward a larger refractive index ( n film = 1.38 ). Similarly, there is a phase change when wave 2 reflects from the right surface of the film, since this light also travels from a smaller refractive index ( n film = 1.38 ) toward a larger one ( n lens = 1.52 ). Therefore, a phase change of one-half wavelength occurs at both boundaries, so the net phase change between waves 1 and 2 due to reflection is zero. Since wave 2 travels back and forth through the film and, and since the light is assumed to be at nearly normal incidence, the extra distance traveled by wave 2 compared to wave 1 is twice the film thickness, or 2 t . Thus, in this case, the minimum condition for destructive interference is 1 2 t = 2 λ film The wavelength of light in the coating is λ film = λ vacuum n = 565 nm = 409 nm 1.38 (27.3) Solving the above expression for t, we find that the minimum thickness that the coating can have is 1 1 t = 4 λ film = 4 (409 nm) = 102 nm Chapter 27 Problems 1439 13. SSM REASONING When the light strikes the film from above, the wave reflected from the top surface of the film undergoes a phase shift that is equivalent to one-half of a wavelength, since the light travels from a smaller refractive index (nair = 1.00) toward a larger refractive index (nfilm = 1.33). On the other hand, there is no phase shift when the light reflects from the bottom surface of the film, since the light travels from a larger refractive index (nfilm = 1.33) toward a smaller refractive index (nair = 1.00). Thus, the net phase change due to reflection from the two surfaces is equivalent to one-half of a wavelength in the film. This half-wavelength must be combined with the extra distance 2t traveled by the wave reflected from the bottom surface, where t is the film thickness. Thus, the condition for constructive interference is + 12t3 2 Extra distance traveled by...
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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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