Physics Solution Manual for 1100 and 2101

# Therefore the diameter d of the central bright spot

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Unformatted text preview: the two. The phase change is equivalent to 2 λ film , where λfilm is the wavelength in the film. In the film the index of refraction is n = 1.33, while in the glass it is n = 1.52. This situation 1 is like that discussed above and, once again, the phase change is equivalent to 2 λ film , where λfilm is the wavelength in the film. Both the light reflected from the air-film interface and from the film-glass interface 1 experience phase changes, each of which is equivalent to 2 λ film . In other words, there is no 1442 INTERFERENCE AND THE WAVE NATURE OF LIGHT net phase change between the waves reflected from the two interfaces, and only the extra travel distance of the light within the film leads to the destructive interference. SOLUTION As mentioned in the REASONING, only the extra travel distance of the light within the film leads to the destructive interference. The extra distance is 2t, where t is the film thickness. The condition for destructive interference in this case is ch 1 3 5 1 2t = 2 λ film , 2 λ film , 2 λ film , ... = m + 2 λ film 123 1444 24444 4 3 Extra travel distance in film where m = 0 , 1, 2 , 3, ... Condition for destructive interference Solving for the wavelength gives λ film = 2t 1 m+ 2 m = 0, 1, 2, 3, ... According to Equation 27.3, we have λfilm = λvacuum/n. Using this substitution in our result for λfilm, we obtain 2 nt λ vacuum = m = 0, 1, 2, 3, ... 1 m+ 2 For the first four values of m and the given values for n and t, we find bg b g 2 1. 33 465 nm 2 nt = = 2470 nm 1 1 m+ 2 0+ 2 m=0 λ vacuum = m=1 λ vacuum = 825 nm m=2 λ vacuum = 495 nm m=3 λ vacuum = 353 nm The range of visible wavelengths (in vacuum) extends from 380 to 750 nm. Therefore, the only visible wavelength in which the film appears dark due to destructive interference is 495 nm . 17. REASONING In air the index of refraction is nearly n = 1, while in the film it is greater than one. 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 1 boundary between the two. The phase change is equivalent to 2 λ film , where λfilm is the wavelength in the film. This phase change occurs at the top surface of the film, where the light first strikes it. However, no phase change occurs when light that has penetrated the film reflects back upward from the bottom surface. This is because a phase change does not occur when light travels through a material with a larger refractive index toward a material Chapter 27 Problems 1443 with a smaller refractive index and reflects from the boundary between the two. Thus, to 1 evaluate destructive interference correctly, we must consider a net phase change of 2 λ film due to reflection as well as the extra distance traveled by the light within the film. SOLUTION The drawing shows the soap film and the two rays of light that represent the interfering light Incident light waves. At nearly perpendicular incidence, ray 2 travels a distance of 2t further than ray 1, where t is the thickness of the film. In addition, the net phase change for the two 1 rays is 2 λ film , as discussed in the reasoning section. We t must combine this amount with the extra travel distance to determine the condition for destructive interference. For destructive interference, the combined total must be an odd-integer number of half-wavelengths in the film: 2t { Extra distance traveled by ray 2 Subtracting the term right side, we obtain 1 2 + 1 λ 2 1 film 2 3 1 2 nair = 1.00 nsoap = n nair = 1.00 1 3 5 = 2 λ film , 2 λ film , 2 λ film , ... 1444 24444 4 3 Half wavelength net phase change due to reflection Condition for destructive interference λ film from the left side of this equation and from each term on the 2 t = 0 , film , 2 λ film , ... The minimum nonzero thickness is t = λfilm/2. But the wavelength in the film is related to the vacuum-wavelength according to Equation 27.3: λfilm = λvacuum/n. Thus, the minimum nonzero thickness is t = λvacuum/(2n). Applying this result to both regions of the film allows us to obtain the desired ratio: t magenta t yellow = bgλ = / bn g λ 2 λ vacuum, green / 2 n vacuum, green λ vacuum, blue vacuum, blue = 555 nm = 1.18 469 nm 18. REASONING When the light in the glass strikes the wedge of air, the wave reflected from the bottom surface of the glass does not experience a phase shift. This is because the light is traveling from a larger refractive index (nglass = 1.5) toward a smaller refractive index (nair = 1.0). When the light reflects from the top surface of the plastic, the wave undergoes a phase shift that is equivalent to one-half of a wavelength in the air film, since the light is traveling from a smaller refractive index (nair = 1.0) toward a larger refractive index (nplastic = 1.2). Thus, the net phase change due to reflection from the two surfaces is equivalent to one-half of a wavelength. At point A the air wedge has no thickness, so the light re...
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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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