Physics Solution Manual for 1100 and 2101

# 22 d where is the wavelength of the light being

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Unformatted text preview: wave in the film 1λ 12 24 4 film 3 = λfilm , 2λfilm , 3λfilm , K 14444 244444 4 3 Condition for constructive interference Half-wavelength net phase change due to reflection We will use this relation to find the two smallest non-zero film thicknesses for which constructive interference occurs in the reflected light. SOLUTION The smallest film thickness occurs when the condition for constructive interference is λfilm. Then, the relation above becomes 2t + 1 λfilm = λfilm 2 Since λfilm = λvacuum nfilm or t = 1 λfilm 4 (Equation 27.3), we have that λ t = 1 λfilm = 1 vacuum 4 4 nfilm 1 691 nm 2 = 1.30 ×10 nm = 4 1.33 The next smallest film thickness occurs when the condition for constructive interference is 2λfilm. Then, we have 2t + 1 λfilm = 2λfilm 2 or λ 691 nm 2 t = 3 λfilm = 3 vacuum = 3 = 3.90 ×10 nm 4 4 4 nfilm 1.33 14. REASONING The vacuum wavelength λvacuum of the light is related to its wavelength λvacuum (Equation 27.3), where nfilm is the index of nfilm refraction of the soap film. Consider when monochromatic light is incident on the soap film λfilm in the soap film by λfilm = 1440 INTERFERENCE AND THE WAVE NATURE OF LIGHT from above. Part of this light, ray 1, reflects from the upper surface, traveling from a smaller to a larger refractive index (nair = 1.00 to nfilm = 1.33), incurring a phase change of one-half of a wavelength. Some of the incident light, ray 2, passes into the soap film and reflects from its lower surface, traveling from a larger to a smaller refractive index (nfilm = 1.33 to nair = 1.00). This reflection incurs no phase change. Ray 2 travels back through the film, enters the air, and combines with ray 1. The second ray travels twice through the thickness t of the film, a total distance of 2t farther than the first ray. Therefore, the total phase change between the two rays is equal to the sum of 2t and 1 λfilm . The two rays undergo destructive 2 interference, so that the total phase change between them is 2t + odd number of half-wavelengths: 1λ and 2 film is equal to an 3 5 2t + 1 λfilm = 1 λfilm , 2 λfilm , 2 λfilm ,K 2 2 (1) We will use Equation 27.3 and Equation (1), together with the condition that the second smallest nonzero thickness for which destructive interference occurs is 296 nm, and determine the vacuum wavelength λvacuum of the light reflecting from the film. SOLUTION Subtracting 1 λfilm from both sides of Equation (1), we find that the condition 2 for destructive interference becomes 2t = 0, 1 λfilm , 3 λfilm , K 2 2 (2) All of the values on the right side of Equation (2) give destructive interference. The first value gives a thickness of t = 0 m. The second value is the smallest nonzero thickness for destructive interference, so the next value, 3 λfilm corresponds to the second smallest 2 nonzero thickness: 2t = 3 λfilm (3) 2 Substituting λfilm = λvacuum nfilm (Equation 27.3) into Equation (3) and solving for λvacuum yields 2t = 3 λfilm = 2 3λvacuum 2nfilm or λvacuum = 4tnfilm 4 ( 296 nm ) (1.33) = = 525 nm 3 3 15. 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 in the film, since the light travels from a smaller refractive index (nair = 1.00) toward a larger refractive index (nfilm = 1.43). When the light reflects from the bottom surface of the film Chapter 27 Problems 1441 the wave undergoes another phase shift that is equivalent to one-half of a wavelength in the film, since this light also travels from a smaller refractive index (nfilm = 1.43) toward a larger refractive index (nglass = 1.52). Thus, the net phase change due to reflection from the two surfaces is equivalent to one wavelength in the film. This 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 destructive interference is + 12t3 2 Extra distance traveled by wave in the film 1λ +1λ = 1 λfilm , 3 λfilm , 5 λfilm , K 2 film 2 film 2 2 244444 2 144 2444 144444 4 3 3 One wavelength net phase change due to reflection (1) Condition for destructive interference We will use this relation to find the longest possible wavelength of light that will yield destructive interference. SOLUTION Note that the left-hand side of Equation (1) is greater than λfilm. Thus, the right-hand side of this equation must also be greater than λfilm. The smallest value that is greater than λfilm is the term 2t + 1 λfilm + 1 λfilm = 2 2 3λ . 2 film 3λ 2 film Therefore, we have that or λfilm = 4t = 4 (1.07 × 10 −7 m ) = 4.28 × 10 −7 m Since λvacuum = nfilmλfilm (Equation 27.3), the wavelength in vacuum is λvacuum = nfilm λfilm = (1.43) ( 4.28 ×10−7 m ) = 6.12 ×10−7 m Other terms on the right-hand side of Equation (1) that are greater than smaller values of λvacuum . 3λ lead 2 film to 16. REASONING In air the index of refraction is nearly n = 1, while in the film it is n = 1.33. A phase change occurs whenever light travels through a material with a smaller refractive index toward a material with a larger refractive index and reflects from the boundary 1 between...
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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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