(where lambda is the wavelength in the film), is satisfied. However, when one of
ID: 2301978 • Letter: #
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(where lambda is the wavelength in the film), is satisfied. However, when one of the two waves has a half-cycle reflection phase shift, the same equation is the condition for destructive interference. Similarly, if neither wave, or both, has a half-cycle phase shift, the condition for destructive interference in the reflected waves is 2t = (m + l/2)A (m = 0,1, 2,). (26.8) However, if one wave has a half-cycle phase shift, the same equation is the condition for constructive interference. Destructive interference Two very flat planes of glass touch at one end, but are held apart at the other end by a tiny filament of nylon placed between them. When the planes of glass are viewed by reflection from a sodium lamp, 23 evenly spaced dark bands are counted. The width of the filament is 23 wavelengths of sodium light. is 11 wavelengths of sodium light. cannot be determined without a knowledge of the length of the glass plates. solution Equation 26.7 gives the thickness of the film of air between the two glass plates at the locations of the dark bands. The first dark band occurs where the thickness is zero-that is, where the glass planes touch. The second occurs where the air thickness is lambda /2, the third where it is 2 lambda /2, and the fourth where it is 3 lambda /2. Continuing this pattern, we see that, at the 23rd dark line, the thickness is 22 lambda /2, or 11 wavelengths of light. If the filament is just beyond the 23rd dark line, it will actually be just a little thicker than 11 wavelengths of light. example 26.3 Interference fringes Suppose the two glass plates in Figure 26.10 are two microscope slides 10 cm long. At one end, they are in contact; at the other end, they are separated by a piece of paper 0.020 mm thick. What is the spacing of the interference fringes seen by reflection? Is the fringe at the line of contact bright or dark? Assume monochromatic light with lambda 0 = 500 nm.Explanation / Answer
Apply 2nt = m(wavelength)
2(1)(2 X 10-5) = m(500 X 10-9)
m = 80
That means there are 80 bright fringes in the 10 cm distance of the plates
10/80 = .125 cm
Thus the spacing between similar fringes is .125 cm or 1.25 mm (Between a bright and its neighboring dark, it is half that, or .0625 cm (.625 mm))
Also, the point of contact is a dark fringe.
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